Free math worksheets

- Clear a variable denominator by multiplying both sides. - Equal fractions with equal denominators have equal numerators. - A fraction equals zero when its numerator is zero.

1. For a), \(12=4x\), so \(x=3\). 2. For b), equal fractions with the same nonzero denominator have equal numerators, so \(x=7\). 3. For c), the numerator must be zero: \(x-5=0\), so \(x=5\). 4. For d), cross-multiply: \(8=-x\), so \(x=-8\). 5. Each solution keeps the original denominators nonzero.

a) \(x=3\) b) \(x=7\) c) \(x=5\) d) \(x=-8\)

- Start with the denominator rather than solving the equation. - Which input makes the rational expression undefined?

1. The denominator cannot equal zero. 2. Since \(x-3=0\) at \(x=3\), that input is excluded.

\(x=3\) must be excluded.

- Factor each denominator if needed. - Include each distinct denominator factor once.

1. The two variable denominators are \(x-2\) and \(x+2\). 2. They have no common nonconstant factor. 3. Their product \((x-2)(x+2)\) is the least common denominator.

\((x-2)(x+2)\)

- Check the original denominator before considering any algebraic work. - A candidate that is in the domain is not automatically a solution. - Verify an allowed candidate in the original equation.

1. The original denominator is zero at \(x=4\), so \(x=4\) is immediately invalid. 2. The value \(x=7\) is in the domain, so it is not ruled out by a denominator restriction. 3. Substitute \(x=7\) into the original equation: both sides equal \(\frac{10}{3}\), so it is valid.

Reject \(x=4\) immediately because it makes the denominator zero. The candidate \(x=7\) is allowed by the domain but must still be checked in the original equation; it is valid.

- Set the two function expressions equal. - Keep the domain restriction in mind before multiplying by \(x\). - Substitute the solved input into either function to find the output.

1. Set the function values equal: \(\frac{15}{x}=\frac{3}{x}+4\). 2. Subtract \(\frac{3}{x}\): \(\frac{12}{x}=4\). 3. Since \(x\ne0\), multiply by \(x\): \(12=4x\), so \(x=3\). 4. Find the corresponding output: \(f(3)=\frac{15}{3}=5\). 5. The graphs intersect at \((3, 5)\). 6. The input \(x=0\) is not in either function’s domain because division by zero is undefined.

a) \((3, 5)\) b) \(x=0\) is excluded from both domains, so neither graph has a point there.

- Exclude the value that makes the denominator zero. - Clear the denominator to obtain a linear equation. - Equal function values represent an intersection.

1. The denominator requires \(x\ne1\). 2. Multiply by \(x-1\): \(6=3(x-1)\). 3. Solve: \(6=3x-3\), so \(9=3x\) and \(x=3\). 4. Since \(3\ne1\), the solution is valid. 5. Graphically, \(x=3\) is the x-coordinate of the intersection of \(f\) and \(g\). The intersection point is \((3, 3)\).

a) Domain restriction: \(x\ne1\). Solution set: \(\{3\}\). b) The graphs intersect at \((3, 3)\).

- Multiply by a common denominator to clear fractions. - Combine fractions with the same denominator first. - Check that each solution satisfies the original domain restrictions.

1. For a), \(12=3(x+1)\), so \(x=3\). The restriction is \(x\ne-1\). 2. For b), \(5x-10=4x\), so \(x=10\). 3. For c), combine the fractions: \(\frac{8}{x}=2\). Then \(8=2x\), so \(x=4\), with \(x\ne0\). 4. For d), \(2x=10\), so \(x=5\), with \(x\ne0\).

a) \(x=3\) b) \(x=10\) c) \(x=4\) d) \(x=5\)

- Which value of \(x\) would make a denominator zero? - What least common denominator will clear both fractions? - After clearing the fractions, isolate \(x\).

1. Because \(x\) is in a denominator, the domain is \(D=\mathbb{R}\setminus\{0\}\). 2. Multiply every term by the least common denominator \(2x\): \(10-x=6\). 3. Solve: \(-x=-4\), so \(x=4\). 4. Since \(4\) is in the domain, it is a valid solution.

\(D=\mathbb{R}\setminus\{0\}\); \(x=4\)

- Identify the values that make either denominator zero. - Cross-multiply to clear the denominators. - Distribute carefully before combining like terms.

1. The denominators require \(x\ne1\) and \(x\ne-\frac12\). 2. Cross-multiply: \(2(2x+1)=5(x-1)\). 3. Expand: \(4x+2=5x-5\). 4. Solve: \(7=x\), so \(x=7\). 5. Since \(7\) is not excluded, the solution set is \(\{7\}\).

\(\{7\}\)

- Which value makes the denominator zero? - Multiply both sides by the denominator. - Substitute your result into the original equation.

1. The denominator requires \(x\ne4\), so \(D=\mathbb{R}\setminus\{4\}\). 2. Multiply by \(x-4\): \(18=6(x-4)\). 3. Expand and solve: \(18=6x-24\), so \(42=6x\) and \(x=7\). 4. Check: \(\frac{18}{7-4}=\frac{18}{3}=6\). The solution is valid.

\(D=\mathbb{R}\setminus\{4\}\); \(x=7\)

- Identify the value excluded by the denominator. - Clear the denominator to obtain a linear equation. - Check that the result is in the domain.

1. The denominator requires \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2\}\). 2. Multiply by \(x+2\): \(2x+10=3(x+2)\). 3. Expand and solve: \(2x+10=3x+6\), so \(x=4\). 4. Since \(4\) is in the domain, the solution set is \(\{4\}\).

\(D=\mathbb{R}\setminus\{-2\}\); \(S=\{4\}\)

- Identify the values that make either denominator zero. - Cross-multiply to clear both denominators. - Substitute your result into both sides of the original equation.

1. The denominators require \(x\ne1\) and \(x\ne-1\), so \(D=\mathbb{R}\setminus\{-1, 1\}\). 2. Cross-multiply: \(4(x+1)=6(x-1)\). 3. Expand and solve: \(4x+4=6x-6\), so \(10=2x\) and \(x=5\). 4. Check: \(\frac{4}{5-1}=1\) and \(\frac{6}{5+1}=1\). The solution is valid.

\(D=\mathbb{R}\setminus\{-1, 1\}\); \(x=5\)

- Rewrite each negative exponent as a reciprocal. - Which value is excluded by the denominator? - Clear the denominator by multiplying by \(x\).

1. Since \(x^{-1}=\frac{1}{x}\), both equations require \(x\ne0\). Thus, the domain for each is \(D=\mathbb{R}\setminus\{0\}\). 2. a) Rewrite as \(\frac{8}{x}=16\). Multiply by \(x\): \(8=16x\), so \(x=\frac{1}{2}\). 3. b) Rewrite as \(\frac{4}{x}+5=7\). Then \(\frac{4}{x}=2\). Multiply by \(x\): \(4=2x\), so \(x=2\). 4. Both values are in the domain.

a) \(D=\mathbb{R}\setminus\{0\}\); \(S=\left\{\frac{1}{2}\right\}\) b) \(D=\mathbb{R}\setminus\{0\}\); \(S=\{2\}\)

- Identify the values that make either function undefined. - Set the function expressions equal and clear the denominators. - Substitute the resulting \(x\)-value into either function.

1. The denominators require \(x\ne0\) and \(x\ne1\), so \(D=\mathbb{R}\setminus\{0, 1\}\). 2. Set the functions equal: \(\frac{3}{x}=\frac{2}{x-1}\). 3. Cross-multiply: \(3(x-1)=2x\). 4. Solve: \(3x-3=2x\), so \(x=3\). 5. Find the corresponding output: \(f(3)=\frac{3}{3}=1\). 6. The graphs intersect at \((3, 1)\).

\(D=\mathbb{R}\setminus\{0, 1\}\); the intersection is \((3, 1)\).

- Set the two function expressions equal. - Combine the terms with the same denominator. - Substitute the resulting input into either function.

1. Set the functions equal: \(\frac4x=\frac1x+1\). 2. Subtract \(\frac1x\): \(\frac3x=1\). 3. Multiply by \(x\): \(3=x\), so \(x=3\). 4. Since \(3\) is in the domain, it is valid. 5. The common output is \(f(3)=\frac43\).

The functions have the same output at \(x=3\), and the common output is \(\frac43\).

- Find the zeros of both denominators. - Cross-multiply to clear the fractions. - Check that the result is not excluded.

1. The denominators require \(x\ne-2\) and \(x\ne-1\), so \(D=\mathbb{R}\setminus\{-2, -1\}\). 2. Cross-multiply: \(5(x+1)=3(2x+4)\). 3. Expand: \(5x+5=6x+12\). 4. Solve: \(x=-7\). 5. Since \(-7\) is in the domain, the solution set is \(\{-7\}\).

\(D=\mathbb{R}\setminus\{-2, -1\}\); \(S=\{-7\}\)

- Set the function expressions equal to find their intersection. - At an x-intercept, the function value is zero. - Keep the domain restriction \(x\ne0\) in mind.

1. Both functions require \(x\ne0\). 2. For the intersection, set the functions equal: \(\frac6x-3=\frac2x+1\). 3. Simplify: \(\frac4x=4\), so \(x=1\). 4. The common output is \(f(1)=3\), so the intersection is \((1, 3)\). 5. For the x-intercept of \(f\), solve \(\frac6x-3=0\). Then \(\frac6x=3\), so \(x=2\). The intercept is \((2, 0)\). 6. For the x-intercept of \(g\), solve \(\frac2x+1=0\). Then \(\frac2x=-1\), so \(x=-2\). The intercept is \((-2, 0)\).

Intersection: \((1, 3)\) x-intercept of \(f\): \((2, 0)\) x-intercept of \(g\): \((-2, 0)\)

- An equation with no solution means the graphs do not intersect. - A point on the y-axis has x-coordinate 0. - Clear the denominator in each equation.

1. (A) The domain excludes \(x=-3\). Multiply by \(x+3\): \(x=x+3\), which gives the contradiction \(0=3\). Therefore, the graphs have no point in common, so (A) matches 1. 2. (B) The domain excludes \(x=2\). Multiply by \(x-2\): \(4=-2(x-2)\). This simplifies to \(4=-2x+4\), so \(x=0\). A point with \(x=0\) lies on the y-axis, so (B) matches 2.

(A) \(\rightarrow\) 1 (B) \(\rightarrow\) 2

- Identify the excluded values first. - Cross-multiply to clear both denominators. - Check that the result is not excluded.

1. The denominators require \(z\ne0\) and \(z\ne-1\). 2. Cross-multiply: \(12(z+1)=15z\). 3. Expand: \(12z+12=15z\). 4. Solve: \(12=3z\), so \(z=4\). 5. Since \(4\) is allowed, the solution set is \(\{4\}\).

\(S=\{4\}\)

- A rational expression is zero when its numerator is zero and its denominator is nonzero. - Identify the excluded input first. - Check that the zero of the numerator is allowed.

1. The denominator requires \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2\}\). 2. A rational expression equals zero when its numerator equals zero and its denominator does not. 3. Solve \(3x-9=0\): \(3x=9\), so \(x=3\). 4. Since \(3\) is in the domain, the solution set is \(\{3\}\).

\(D=\mathbb{R}\setminus\{-2\}\) \(S=\{3\}\)

- Identify both excluded inputs. - Cross-multiply and expand carefully. - The quadratic terms may cancel.

1. The denominators require \(x\ne-2\) and \(x\ne1\), so \(D=\mathbb{R}\setminus\{-2, 1\}\). 2. Cross-multiply: \((2x-3)(x-1)=(2x+1)(x+2)\). 3. Expand: \(2x^2-5x+3=2x^2+5x+2\). 4. Subtract \(2x^2\) and solve: \(-5x+3=5x+2\), so \(1=10x\) and \(x=\frac1{10}\). 5. The solution is in the domain.

\(D=\mathbb{R}\setminus\{-2, 1\}\); \(S=\left\{\frac1{10}\right\}\)

- At an intersection, the functions have equal outputs. - Clear the denominator and solve for \(x\). - Use the constant function to identify the y-coordinate.

1. The rational function requires \(x\ne2\). 2. Set the function values equal: \(5=\frac{2x+8}{x-2}\). 3. Multiply by \(x-2\): \(5(x-2)=2x+8\). 4. Expand and solve: \(5x-10=2x+8\), so \(3x=18\) and \(x=6\). 5. Since \(f(x)=5\), the corresponding output is 5. 6. The intersection is \((6, 5)\).

\((6, 5)\)

- Find the value that makes the denominator zero. - Clear the denominator to obtain a linear equation. - Check the result against the domain.

1. The denominator requires \(2x-5\ne0\), so \(x\ne\frac52\). Thus, \(D=\mathbb{R}\setminus\left\{\frac52\right\}\). 2. Multiply by \(2x-5\): \(x+2=3(2x-5)\). 3. Expand and solve: \(x+2=6x-15\), so \(17=5x\) and \(x=\frac{17}{5}\). 4. Since \(\frac{17}{5}\) is in the domain, it is the solution.

\(D=\mathbb{R}\setminus\left\{\frac52\right\}\); \(S=\left\{\frac{17}{5}\right\}\)

- Cross-multiply to clear the fractions. - Distribute carefully across both binomials. - Check the result against the excluded values.

1. Cross-multiply: \(6(x+1)=4(x-1)\). 2. Expand: \(6x+6=4x-4\). 3. Solve: \(2x=-10\), so \(x=-5\). 4. Since \(-5\) belongs to the stated domain, it is valid.

\(S=\{-5\}\)

- A rational function equals zero when its numerator is zero and its denominator is nonzero. - Factor out the greatest power of \(x\) from the numerator. - Does the remaining quadratic match a perfect-square trinomial? - Check whether the denominator can be zero for real \(x\).

1. The denominator \(x^2+1\) is positive for every real \(x\), so it does not exclude any real zeros. 2. Factor the numerator: \(x^4-6x^3+9x^2=x^2(x^2-6x+9)=x^2(x-3)^2\). 3. Therefore, the zeros are \(x=0\) and \(x=3\). 4. Each corresponding factor is squared, so each zero has multiplicity \(2\).

\(x=0\) has multiplicity \(2\), and \(x=3\) has multiplicity \(2\).

- Find the excluded input first. - Clear the common denominator only after stating the domain. - Check the candidate against the domain.

1. The denominator requires \(x\ne1\), so \(D=\mathbb{R}\setminus\{1\}\). 2. For \(x\) in the domain, multiply by \(x-1\): \(2=x+1\). 3. This gives \(x=1\). 4. The candidate is excluded from the domain, so the equation has no solution.

\(S=\varnothing\), because the only candidate is \(x=1\), which is excluded.

- Use substitution by setting the two expressions for \(y\) equal. - Clear the denominator before solving the power equation. - Substitute each \(x\)-value into either original equation to find \(y\). - Compare even and odd exponents.

1. Set the two expressions for \(y\) equal: \(x^3=\frac{16}{x}\). Since \(x\ne0\), multiply by \(x\): \(x^4=16\). 2. The real solutions are \(x=\pm2\). 3. Substitute into \(y=x^3\). When \(x=2\), \(y=8\). When \(x=-2\), \(y=-8\). 4. The intersection points are \((2, 8)\) and \((-2, -8)\). 5. In the comparison system, substitution produces \(x^3=8\), which has one real solution because the exponent is odd. Here, \(x^4=16\) has two real solutions because the exponent is even and the right side is positive.

a) \(x=\pm2\) b) \((2, 8)\) and \((-2, -8)\) c) The equation \(x^4=16\) has two real solutions, while \(x^3=8\) has only one.

- Substitute the expression for \(y\) into the other equation. - Combine the powers of \(x\). - Recall how many real solutions \(x^3=a\) has.

1. Substitute \(y=2x\) into \(yx^2=54\): \((2x)x^2=54\). 2. Simplify: \(2x^3=54\), so \(x^3=27\) and \(x=3\). 3. Substitute into \(y=2x\): \(y=6\). The solution is \((3, 6)\). 4. The equation \(x^3=27\) has exactly one real solution because an odd power function takes each real value exactly once.

1. \((3, 6)\) 2. The odd-power equation \(x^3=27\) has exactly one real solution.

- Set the numerator equal to zero. - Check every candidate in the original denominator. - Factor the difference of squares in part 2.

1. A rational expression equals zero when its numerator is zero and its denominator is nonzero. 2. For \(\frac{x-12}{3x}\), the numerator is zero at \(x=12\), and the denominator is then nonzero. 3. For \(\frac{x^2-25}{x+5}\), factor the numerator as \((x-5)(x+5)\). Its zeros are \(x=5\) and \(x=-5\), but \(x=-5\) makes the denominator zero. Only \(x=5\) is valid. 4. For \(\frac{x(x+2)}{x-2}\), the numerator is zero at \(x=0\) and \(x=-2\). Neither value makes the denominator zero.

1) \(x=12\) 2) \(x=5\) 3) \(x=0\) or \(x=-2\)

- Express each runner's time as distance divided by speed. - Ethan's speed is \(1\,\text{mph}\) greater than Henry's. - After solving, check that the speed keeps both denominators nonzero and is physically meaningful.

1. Henry's speed is \(v\,\text{mph}\), so Ethan's speed is \((v+1)\,\text{mph}\). 2. Equal running times give \(\frac{7}{v+1}=\frac{6}{v}\), with \(v\ne0,-1\). 3. Cross-multiply: \(7v=6(v+1)\). 4. Solve: \(7v=6v+6\), so \(v=6\). 5. The value \(v=6\) is in the equation's domain and represents a positive speed, so it is valid.

\(\frac{7}{v+1}=\frac{6}{v}\), and Henry runs \(6\,\text{mph}\).

- Remember that an equation of the form \(x^2=a\) may have two real solutions. - Consider what happens to a fraction when its denominator grows. - Compare the two function values at a positive value greater than \(10\).

1. The domain of \(f\) excludes \(x=0\). Set the functions equal: \(\frac{18}{x}=2x\). 2. Multiply by \(x\): \(18=2x^2\), so \(x^2=9\). 3. Therefore, \(x=3\) or \(x=-3\). The corresponding y-values are \(6\) and \(-6\), so the intersections are \((3,6)\) and \((-3,-6)\). 4. As positive \(x\) increases, \(f(x)=\frac{18}{x}\) approaches \(0\), while \(g(x)=2x\) increases without bound. 5. For \(x>10\), \(2x>\frac{18}{x}\) because \(2x^2>18\). Therefore, the graph of \(g\) is above the graph of \(f\).

a) \((-3,-6)\) and \((3,6)\) b) The graph of \(g\) is above the graph of \(f\) for \(x>10\). As \(x\) increases, \(f(x)\) approaches \(0\), while \(g(x)\) increases without bound.

- What does it mean for a point to lie on a graph? - How can substitution be used to find an unknown parameter? - What function value defines a zero?

1. Substitute \(P(4, 5)\): \(5 = \frac{a}{4 - 2} + 3\). 2. Then \(2 = \frac{a}{2}\), so \(a = 4\). 3. To find the zero, solve \(0 = \frac{4}{x - 2} + 3\). 4. Then \(-3 = \frac{4}{x - 2}\), so \(-3(x - 2) = 4\). This gives \(-3x + 6 = 4\), \(-3x = -2\), and \(x = \frac{2}{3}\).

a) \(a = 4\) b) \(x = \frac{2}{3}\)

- Substitute the given point to find the unknown parameter. - What must the function value equal at an x-intercept? - How can you clear the denominator when solving the equation?

1. Substitute \(A(3, -2)\): \(-2 = \frac{a}{3 - 1} - 4\). 2. Then \(2 = \frac{a}{2}\), so \(a = 4\). Thus, \(g(x) = \frac{4}{x - 1} - 4\). 3. For an x-intercept, solve \(0 = \frac{4}{x - 1} - 4\). 4. Then \(4 = \frac{4}{x - 1}\), so \(x - 1 = 1\) and \(x = 2\). Since \(x = 2\) is in the domain, the x-intercept is \((2, 0)\).

a) \(a = 4\) b) Yes. The x-intercept is \((2, 0)\).

- A fraction is zero only when its numerator is zero and its denominator is not. - Compare the numerators of the two rational expressions. - An x-intercept occurs where the function value is \(0\).

1. A rational expression equals zero when its numerator is zero and its denominator is nonzero. 2. For equation I, \(x+4=0\), so \(x=-4\). The denominator is \(2(-4)-6=-14\ne0\), so the solution is valid. 3. Equation II has constant numerator \(10\), which can never equal zero. Therefore, the equation has no solution. 4. Since \(h(x)=0\) has no solution, the graph has no x-intercept.

a) \(x=-4\) b) The numerator is always \(10\), so the rational expression cannot equal \(0\). c) The graph has no x-intercepts.

- First identify every value that makes a denominator zero. - When one rational expression equals another, cross-multiplication may be useful. - To add rational expressions, use a common denominator. - A rational expression is zero when its numerator is zero and its denominator is not zero.

1. a) The denominator requires \(x\ne0\). Multiply by \(x\): \(x+6=3x\). Thus \(6=2x\), so \(x=3\). 2. b) The denominators require \(x\ne2\) and \(x\ne-1\). Cross-multiply: \(3(x+1)=6(x-2)\). Then \(3x+3=6x-12\), so \(x=5\). 3. c) The denominators require \(x\ne0\). Multiply by \(3x\): \(3+1=12x\). Thus \(x=\frac13\). 4. d) The denominator requires \(x\ne-2\). A rational expression equals zero when its numerator is zero, so \(3x-12=0\). Thus \(x=4\), which is allowed.

a) \(x=3\), with \(x\ne0\) b) \(x=5\), with \(x\ne2,-1\) c) \(x=\frac13\), with \(x\ne0\) d) \(x=4\), with \(x\ne-2\)

- Only the denominator changes. - Write an equation that sets the new fraction equal to \(\frac13\). - Clear the fractions, then solve for the unknown.

1. Let \(x\) be the number added to the denominator. Then \(\frac{5}{12+x}=\frac13\), with \(x\ne-12\). 2. Cross-multiply: \(5\cdot3=12+x\). 3. Solve: \(15=12+x\), so \(x=3\).

\(3\)

- Represent the numerator and denominator using one variable. - Write the fraction obtained after adding 3 to both parts. - Set that fraction equal to \(\frac34\) and solve.

1. Let \(n\) be the denominator. Then the numerator is \(n-5\), so the original fraction is \(\frac{n-5}{n}\), with \(n\ne0\). 2. After adding 3 to both parts, \(\frac{n-2}{n+3}=\frac34\), where \(n\ne-3\). 3. Cross-multiply: \(4(n-2)=3(n+3)\). 4. Expand and solve: \(4n-8=3n+9\), so \(n=17\). 5. The numerator is \(17-5=12\). Therefore, the original fraction is \(\frac{12}{17}\).

\(\frac{12}{17}\)

- Express the denominator in terms of the numerator. - Write an equation for the fraction after the stated changes. - A fraction equal to 1 has equal numerator and denominator, provided the denominator is nonzero.

1. Let \(z\) be the numerator. Then the denominator is \(3z+1\), so the original fraction is \(\frac{z}{3z+1}\), with \(z\ne-\frac13\). 2. The changed fraction gives \(\frac{2z}{3z-4}=1\), with \(z\ne\frac43\). 3. Multiply by \(3z-4\): \(2z=3z-4\). 4. Solve: \(z=4\). 5. The original denominator is \(3\cdot4+1=13\), so the original fraction is \(\frac{4}{13}\).

\(\frac{4}{13}\)

- Find the zeros of both denominators first. - Cross-multiply, then expand both products carefully. - Compare the quadratic terms on both sides before solving.

1. The denominators require \(x\ne-3\) and \(x\ne\frac12\), so \(D=\mathbb{R}\setminus\left\{-3, \frac12\right\}\). 2. Cross-multiply: \((2x-1)^2=(x+3)(4x+5)\). 3. Expand: \(4x^2-4x+1=4x^2+17x+15\). 4. Subtract \(4x^2\): \(-4x+1=17x+15\). 5. Solve: \(-14=21x\), so \(x=-\frac23\). 6. Since \(-\frac23\) is in the domain, the solution set is \(\left\{-\frac23\right\}\).

a) \(D=\mathbb{R}\setminus\left\{-3, \frac12\right\}\) b) \(S=\left\{-\frac23\right\}\)

- Find every value that makes either denominator zero. - A value produced after clearing denominators must still belong to the original domain. - Check the candidate in the original equation.

1. The denominators require \(x^2-16\ne0\) and \(x+2\ne0\). Therefore, \(x\ne-4,-2,4\), so \(D=\mathbb{R}\setminus\{-4, -2, 4\}\). 2. Cross-multiplication gives \((x-4)(x+2)=x^2-16\). 3. Expand and simplify: \(x^2-2x-8=x^2-16\), so \(-2x=-8\) and \(x=4\). 4. However, \(x=4\) is excluded from the domain because it makes \(x^2-16=0\). 5. The only candidate is extraneous, so the solution set is empty.

\(D=\mathbb{R}\setminus\{-4, -2, 4\}\). The candidate \(x=4\) is excluded because it makes \(x^2-16=0\), so \(S=\varnothing\).

- Pay attention to whether the negative exponent applies to a quantity in parentheses or to \(x\) alone. - Rewrite the decimals as fractions. - What sign does \(\frac{1}{x^2}\) have for nonzero real \(x\)?

1. a) Rewrite the equation as \(\frac{1}{x-5}=0.2\), with \(x\ne5\). Since \(0.2=\frac{1}{5}\), cross-multiplication gives \(x-5=5\), so \(x=10\). 2. b) Rewrite as \(\frac{1}{x}-\frac{1}{5}=\frac{1}{5}\), with \(x\ne0\). Then \(\frac{1}{x}=\frac{2}{5}\), so \(x=\frac{5}{2}\). 3. c) For every nonzero real \(x\), \(x^{-2}=\frac{1}{x^2}>0\). Therefore, it cannot equal \(-1\).

a) \(D=\mathbb{R}\setminus\{5\}\); \(S=\{10\}\) b) \(D=\mathbb{R}\setminus\{0\}\); \(S=\left\{\frac{5}{2}\right\}\) c) \(x^{-2}=\frac{1}{x^2}\) is positive for every nonzero real \(x\), so there is no real solution.

- Rewrite the expressions without negative exponents. - Use the least common denominator to clear the fractions in part b. - In part c, cross-multiply after stating both restrictions.

1. a) \(T_1(x)=\frac{1}{3x}\) and \(T_2(x)=\frac{3}{x}\). Both require \(x\ne0\), so both have domain \(\mathbb{R}\setminus\{0\}\). 2. b) Solve \(\frac{1}{3x}=\frac{3}{x}+4\). Multiply by \(3x\): \(1=9+12x\). Thus, \(-8=12x\), so \(x=-\frac{2}{3}\), which is allowed. 3. c) Solve \(\frac{1}{3x}=\frac{1}{x+4}\), with \(x\ne0,-4\). Cross-multiplication gives \(x+4=3x\), so \(x=2\), which is allowed.

a) No. Both domains are \(\mathbb{R}\setminus\{0\}\). b) \(S=\left\{-\frac{2}{3}\right\}\) c) \(x=2\)

- Set the function expressions equal. - Clear the denominators using their product. - Use the difference-of-squares pattern when expanding.

1. The equation \(f(x)=g(x)\) has domain \(D=\mathbb{R}\setminus\{0,2\}\). 2. Set the expressions equal: \(\frac{x+2}{x}=\frac{x}{x-2}\). 3. Multiply by \(x(x-2)\): \((x+2)(x-2)=x^2\). 4. Use the difference of squares: \(x^2-4=x^2\). 5. Subtract \(x^2\) from both sides to obtain \(-4=0\), a contradiction. 6. Therefore, the equation has no solution, so the graphs do not intersect.

The graphs do not intersect because the equation reduces to the contradiction \(-4=0\).

- Factor the denominator on the right. - Multiply every term by the least common denominator. - Check the resulting candidate against the original domain.

1. Since \(x^2-4=(x-2)(x+2)\), the domain excludes \(x=-2\) and \(x=2\). 2. Multiply by \((x-2)(x+2)\): \((x+2)+(x-2)=4\). 3. Simplify: \(2x=4\), so \(x=2\). 4. The candidate \(x=2\) is excluded from the original domain. 5. Therefore, the solution set is empty.

\(S=\varnothing\)

- Clear the denominator, remembering that \(x\ne0\). - Isolate \(x\) in terms of \(k\). - Compare the resulting expression for \(x\) with the domain restriction.

1. The original equation requires \(x\ne0\). 2. a) Multiply by \(x\): \(2x+k=5x\). Thus, \(k=3x\), so \(x=\frac{k}{3}\). 3. b) Set \(\frac{k}{3}=2\). Then \(k=6\). 4. c) Since the solution formula is \(x=\frac{k}{3}\) and the original equation requires \(x\ne0\), the parameter must satisfy \(k\ne0\). 5. When \(k=0\), the equation becomes \(2=5\) for every allowed \(x\), so it has no solution.

a) \(x=\frac{k}{3}\) b) \(k=6\) c) \(k\ne0\)

- Factor the quadratic denominator first. - Multiply every term by the least common denominator. - Check the candidate against the original domain.

1. Factor \(x^2-x=x(x-1)\). The domain excludes \(x=0\) and \(x=1\). 2. Multiply every term by \(x(x-1)\): \(x(x+1)-x(x-1)=2\). 3. Expand and simplify: \(x^2+x-x^2+x=2\), so \(2x=2\) and \(x=1\). 4. The candidate \(x=1\) is excluded from the original domain. 5. Therefore, the equation has no solution.

\(D=\mathbb{R}\setminus\{0, 1\}\); \(S=\varnothing\)

- Multiply each term by the least common denominator. - Recognize the difference-of-squares product. - Check the result against the stated domain.

1. Multiply every term by the least common denominator \(x(x-4)\): \(2x^2-(x+4)(x-4)=x(x-4)\). 2. Use the difference of squares and expand the right side: \(2x^2-(x^2-16)=x^2-4x\). 3. Simplify: \(x^2+16=x^2-4x\). 4. Subtract \(x^2\): \(16=-4x\), so \(x=-4\). 5. Since \(-4\) is in the stated domain, the solution set is \(\{-4\}\).

\(S=\{-4\}\)

- Represent an unreduced fraction equivalent to \(\frac34\) using a positive integer multiplier. - Add 12 to both parts and set the result equal to \(\frac45\). - Cross-multiply and solve for the multiplier.

1. Write the original fraction as \(\frac{3k}{4k}\), where \(k\) is a positive integer. 2. The second condition gives \(\frac{3k+12}{4k+12}=\frac45\). 3. Cross-multiply: \(5(3k+12)=4(4k+12)\). 4. Expand: \(15k+60=16k+48\). 5. Solve: \(k=12\). 6. The original numerator is \(3\cdot12=36\), and the denominator is \(4\cdot12=48\).

The original fraction is \(\frac{36}{48}\).

- Represent the original fraction using a common scale factor. - Find the reciprocal of \(\frac32\). - Apply the stated changes and set the new fraction equal to that reciprocal.

1. Write the original fraction as \(\frac{3k}{2k}\), where \(k\ne0\). 2. The reciprocal of \(\frac32\) is \(\frac23\). 3. Set up the equation \(\frac{3k-5}{2k+5}=\frac23\), with \(2k+5\ne0\). 4. Cross-multiply: \(3(3k-5)=2(2k+5)\). 5. Expand and solve: \(9k-15=4k+10\), so \(5k=25\) and \(k=5\). 6. The original fraction is \(\frac{3\cdot5}{2\cdot5}=\frac{15}{10}\).

\(\frac{15}{10}\)

- Represent a fraction equal to \(\frac12\) as \(\frac{k}{2k}\). - A fraction equals 1 when its numerator and denominator are equal. - Examine whether the resulting value of \(k\) is valid.

1. Represent the original fraction as \(\frac{k}{2k}\), where \(k\) is a positive integer. 2. The condition gives \(\frac{k+n}{2k+n}=1\). 3. Multiply by \(2k+n\): \(k+n=2k+n\). 4. Subtract \(n\): \(k=2k\), so \(k=0\). 5. This contradicts the requirement that \(k\) be positive and would make the original denominator zero. 6. Therefore, no such fraction exists.

No. The equation forces \(k=0\), which does not define a fraction with value \(\frac12\).

- When equal rational expressions have the same nonzero denominator, compare their numerators. - A rational function is zero when its numerator is zero and its denominator is not zero. - Identify the excluded input first.

1. Both functions require \(x\ne-2\). 2. For the intersection, solve \(\frac3{x+2}=\frac{x-1}{x+2}\). Since the denominators are equal and nonzero, \(3=x-1\), so \(x=4\). 3. The common output is \(f(4)=\frac36=\frac12\), so the intersection is \(\left(4, \frac12\right)\). 4. The numerator of \(f\) is the nonzero constant 3, so \(f\) has no x-intercept. 5. For \(g\), set the numerator equal to zero: \(x-1=0\), so \(x=1\). The x-intercept is \((1, 0)\).

Intersection: \(\left(4, \frac12\right)\) x-intercept of \(f\): none x-intercept of \(g\): \((1, 0)\)

- Isolate the squared expression after clearing the denominator. - Remember both signs when taking a square root. - A rational function is zero only when its numerator is zero and its denominator is not zero. - Check that every proposed value lies in the appropriate domain.

1. The domain of \(f\) excludes \(x=1\). 2. Set the functions equal: \(\frac{4}{x-1}=x-1\). 3. Multiply by \(x-1\): \(4=(x-1)^2\). 4. Thus, \(x-1=2\) or \(x-1=-2\), so \(x=3\) or \(x=-1\). 5. Substituting into \(g\) gives the intersections \((3,2)\) and \((-1,-2)\). 6. The numerator of \(f\) is the nonzero constant \(4\), so \(f\) has no zeros. For \(g\), solve \(x-1=0\), giving the zero \(x=1\), or the x-intercept \((1,0)\).

Intersections: \((-1,-2)\) and \((3,2)\) Zeros of \(f\): none Zero of \(g\): \(x=1\), with x-intercept \((1,0)\)

- Identify the input that makes the denominator zero. - Clear the denominator and solve the resulting equation. - Check the candidate in the original domain.

1. Both functions require \(x\ne2\). 2. Set the expressions equal: \(\frac{2x}{x-2}=\frac4{x-2}+1\). 3. Multiply by \(x-2\): \(2x=4+(x-2)\). 4. Simplify: \(2x=x+2\), so \(x=2\). 5. The candidate \(x=2\) is excluded from the domain. 6. Therefore, the equation has no solution and the graphs do not intersect.

The graphs do not intersect because the only candidate, \(x=2\), is excluded from the domain.

- Clear the denominator to obtain a linear equation. - Check the resulting candidate against the original domain. - An excluded candidate is not a solution.

1. The denominator requires \(x\ne4\). 2. Multiply every term by \(x-4\): \(x+2(x-4)=4\). 3. Simplify: \(x+2x-8=4\), so \(3x=12\) and \(x=4\). 4. The candidate \(x=4\) is excluded from the original domain. 5. Therefore, the equation has no solution.

\(D=\mathbb{R}\setminus\{4\}\); \(S=\varnothing\)

- Determine the original domain before canceling. - What does an identity such as \(2=2\) mean? - A canceled factor may still represent an excluded input.

1. The denominator requires \(x\ne2\), so the domain is \(\mathbb{R}\setminus\{2\}\). 2. For any input in the domain, \(\frac{2x-4}{x-2}=\frac{2(x-2)}{x-2}=2\). 3. Thus, the equation is true for every value in its domain. 4. Ben is correct because \(x=2\) is not allowed in the original expression. 5. The solution set is \(\mathbb{R}\setminus\{2\}\).

Ben is correct. \(S=\mathbb{R}\setminus\{2\}\).

- State the domain before clearing the fractions. - Multiply every term by the least common denominator. - Solve the resulting quadratic equation. - Check each solution against the excluded values.

1. The domain is \(\mathbb{Q}\setminus\{-2,0\}\). 2. Multiply the equation by the common denominator \(4x(x+2)\): \(4(x+2)+4x=3x(x+2)\). 3. Expand and rearrange: \(8x+8=3x^2+6x\), so \(3x^2-2x-8=0\). 4. Apply the quadratic formula: \(x=\frac{2\pm\sqrt{(-2)^2-4\cdot3\cdot(-8)}}{2\cdot3}=\frac{2\pm10}{6}\). 5. The solutions are \(x=2\) and \(x=-\frac{4}{3}\). Both are in the domain.

\(x=2\) or \(x=-\frac{4}{3}\)

- Set the function expressions equal to find intersections. - Identify the value that makes the denominator zero. - Clear the denominator and factor the resulting quadratic. - Check every solution in the original domain.

1. Because \(x\) is in a denominator, the domain is \(\mathbb{R}\setminus\{0\}\). 2. Set the functions equal: \(\frac{4}{x}=x-3\). 3. Multiply by \(x\): \(4=x(x-3)\), so \(x^2-3x-4=0\). 4. Factor: \((x-4)(x+1)=0\), giving \(x=4\) or \(x=-1\). Both values are in the domain. 5. A value that makes a denominator zero is not permitted in the original equation, even if it appears during algebraic manipulation.

1. \(\mathbb{R}\setminus\{0\}\) 2. \(x=-1\) and \(x=4\) 3. Values that make a denominator zero must be excluded because the original equation is undefined there.

- A fraction equals 1 when its numerator equals its nonzero denominator. - In part c, set the original and changed fractions equal. - Factor the resulting quadratic and apply the nonzero condition.

1. a) Substitute \(x=5\): \(\frac5{5+10}=\frac5{15}=\frac13\). 2. b) Let \(k\) be the increase in the numerator. Solve \(\frac{x+k}{x+10}=1\). Since the denominator is nonzero, \(x+k=x+10\), so \(k=10\). 3. c) Set the original and changed fractions equal: \(\frac{x}{x+10}=\frac{2x}{x+25}\), with \(x\ne-10,-25\). 4. Cross-multiply: \(x(x+25)=2x(x+10)\). 5. Expand and simplify: \(x^2+25x=2x^2+20x\), so \(x^2-5x=0\). 6. Factor: \(x(x-5)=0\). Since a nonzero value is required, \(x=5\).

a) \(\frac13\) b) \(10\) c) \(x=5\)

- Factor the quadratic denominator to identify the excluded values. - Multiply by the least common denominator. - Check the candidate against the original domain.

1. Factor \(x^2-2x=x(x-2)\). The domain excludes \(x=0\) and \(x=2\). 2. Multiply every term by \(x(x-2)\): \(3x-(x-2)=2\). 3. Simplify: \(3x-x+2=2\), so \(2x=0\) and \(x=0\). 4. The candidate \(x=0\) is excluded from the original domain. 5. Therefore, the solution set is empty.

\(S=\varnothing\)

- State the excluded value before clearing the denominator. - Multiply by \(x\) to eliminate the fraction. - Rearrange the resulting quadratic equation so one side is \(0\). - Look for two integers whose sum is \(3\) and whose product is \(-4\).

1. The domain of \(g\) excludes \(x=0\). Set the functions equal: \(x+3=\frac{4}{x}\). 2. Multiply by \(x\): \(x(x+3)=4\), so \(x^2+3x-4=0\). 3. Factor: \((x+4)(x-1)=0\), giving \(x=-4\) or \(x=1\). 4. Substitute into \(f\): \(f(-4)=-1\) and \(f(1)=4\). 5. Therefore, the intersections are \((-4,-1)\) and \((1,4)\).

\((-4,-1)\) and \((1,4)\)

- Clear the denominator only after identifying the excluded value. - Rearranging produces a quadratic equation. - Find the y-value corresponding to each solution for \(x\).

1. Set the functions equal: \(\frac{12}{x}=-x+7\). 2. Since \(x\ne0\), multiply by \(x\): \(12=-x^2+7x\). 3. Rearrange: \(x^2-7x+12=0\). 4. Factor: \((x-3)(x-4)=0\), so \(x=3\) or \(x=4\). 5. Substituting into \(g\) gives \(g(3)=4\) and \(g(4)=3\). Therefore, the common points are \((3,4)\) and \((4,3)\).

\((3,4)\) and \((4,3)\)

- Determine the least common denominator. - After clearing the fractions, solve the resulting quadratic equation. - Expand each product carefully before combining like terms.

1. The denominators require \(x\ne0\) and \(x\ne2\), so \(D=\mathbb{R}\setminus\{0,2\}\). 2. Multiply by the common denominator \(4x(x-2)\): \(4(x-2)+4x=3x(x-2)\). 3. Expand and rearrange: \(8x-8=3x^2-6x\), so \(3x^2-14x+8=0\). 4. Apply the quadratic formula: \(x=\frac{14\pm\sqrt{(-14)^2-4\cdot3\cdot8}}{2\cdot3}=\frac{14\pm10}{6}\). 5. This gives \(x=4\) or \(x=\frac{2}{3}\). Both values belong to the domain.

Domain: \(D=\mathbb{R}\setminus\{0,2\}\) Solution set: \(\left\{\frac{2}{3},4\right\}\)

- Factor the denominator \(x^2-4x\) before finding a common denominator. - Track the subtraction sign when expanding the second product. - Compare every candidate solution with the domain.

1. Since \(x^2-4x=x(x-4)\), the denominators require \(x\ne0\) and \(x\ne4\). Thus, \(D=\mathbb{R}\setminus\{0,4\}\). 2. Multiply the equation by \(x(x-4)\): \(2x^2-(x+4)(x-4)=16\). 3. Simplify: \(2x^2-(x^2-16)=16\), so \(x^2+16=16\). 4. Therefore, \(x^2=0\), giving \(x=0\). 5. Because \(0\notin D\), this candidate is extraneous. The equation has no solution.

Domain: \(D=\mathbb{R}\setminus\{0,4\}\) Solution set: \(\varnothing\)

- Clear the denominator by multiplying by \(x\). - Identify \(a\), \(b\), and \(c\) before using the quadratic formula. - Consider when a square root of an integer is rational.

1. Because the equation contains a reciprocal, \(x\ne0\). 2. Multiply by \(x\): \(x^2=3x+1\). 3. Rearrange: \(x^2-3x-1=0\). 4. Apply the quadratic formula: \(x=\frac{3\pm\sqrt{(-3)^2-4\cdot1\cdot(-1)}}{2}=\frac{3\pm\sqrt{13}}{2}\). 5. Both values are nonzero and therefore valid. Since \(13\) is not a perfect square, \(\sqrt{13}\) is irrational, so both solutions are irrational.

a) \(x^2-3x-1=0\) b) \(x=\frac{3+\sqrt{13}}{2}\) or \(x=\frac{3-\sqrt{13}}{2}\) c) Since \(13\) is not a perfect square, \(\sqrt{13}\) is irrational.

- Identify the value that makes the denominator zero. - Multiply every term by the denominator. - Compare the candidate with the original domain.

1. The denominator requires \(x\ne5\), so \(D=\mathbb{R}\setminus\{5\}\). 2. Multiply every term by \(x-5\): \(2x-(x-5)=10\). 3. Simplify: \(2x-x+5=10\), so \(x=5\). 4. The candidate \(x=5\) is excluded from the original domain. 5. Therefore, the solution set is empty.

1) \(D=\mathbb{R}\setminus\{5\}\) 2) The algebraic candidate is \(x=5\). 3) \(S=\varnothing\), because \(x=5\) is excluded from the domain.

- Identify the value that makes the denominator zero. - Multiply by a common denominator to eliminate the fractions. - Solve the resulting quadratic equation and check the domain.

1. The denominator requires \(x\ne1\), so \(D=\mathbb{R}\setminus\{1\}\). 2. Multiply by \(2(x-1)\): \(2(x+2)=x(x-1)\). 3. Expand and rearrange: \(2x+4=x^2-x\), so \(x^2-3x-4=0\). 4. Factor: \((x-4)(x+1)=0\), giving \(x=4\) or \(x=-1\). 5. Both values lie in the domain.

Domain: \(D=\mathbb{R}\setminus\{1\}\) Solution set: \(\{-1,4\}\)

- Account for both denominators when stating the domain. - Use the difference-of-squares pattern to identify a common denominator. - Remember both signs when solving an equation of the form \(x^2=a\).

1. The denominators require \(x\ne1\) and \(x\ne-1\), so \(D=\mathbb{R}\setminus\{-1,1\}\). 2. Multiply by \((x-1)(x+1)\): \(4(x+1)-4(x-1)=x^2-1\). 3. Simplify: \(8=x^2-1\), so \(x^2=9\). 4. Therefore, \(x=3\) or \(x=-3\). Both values lie in the domain.

Domain: \(D=\mathbb{R}\setminus\{-1,1\}\) Solution set: \(\{-3,3\}\)

- First evaluate both functions at \(x=2\). - Clear the denominator after stating the excluded value. - Solve the resulting quadratic equation and then find each y-coordinate.

1. Evaluate the linear function: \(h(2)=2\cdot2-5=-1\). 2. Set \(g(2)=h(2)\): \(\frac{k}{2}=-1\), so \(k=-2\). 3. Set the functions equal using \(k=-2\): \(-\frac{2}{x}=2x-5\), with \(x\ne0\). 4. Multiply by \(x\) and rearrange: \(-2=2x^2-5x\), so \(2x^2-5x+2=0\). 5. Apply the quadratic formula: \(x=\frac{5\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2}=\frac{5\pm3}{4}\). Thus, \(x=2\) or \(x=\frac12\). 6. The corresponding y-values are \(h(2)=-1\) and \(h\left(\frac12\right)=-4\). Therefore, the intersections are \((2,-1)\) and \(\left(\frac12,-4\right)\).

a) \(k=-2\) b) \(\left(\frac12,-4\right)\) and \((2,-1)\)

- Identify the excluded values before clearing the denominators. - Multiply by the least common denominator. - Remember that \(x^2=a\) can have two real solutions.

1. The denominators require \(x\ne0\) and \(x\ne-3\). 2. Multiply by \(x(x+3)\): \((x-1)(x+3)=2x\). 3. Expand and simplify: \(x^2+2x-3=2x\), so \(x^2=3\). 4. Therefore, \(x=\sqrt3\) or \(x=-\sqrt3\). Both values are allowed by the domain.

Solution set: \(\{-\sqrt3,\sqrt3\}\)

- Factor every denominator before finding the least common denominator. - Clear the fractions and solve the resulting quadratic equation. - Reject any candidate that makes an original denominator zero.

1. Factor \(x^2-4=(x+2)(x-2)\). The denominators require \(x\ne-2\) and \(x\ne2\), so \(D=\mathbb{R}\setminus\{-2,2\}\). 2. Multiply by \((x+2)(x-2)\): \(3x(x-2)+2(x+2)=8\). 3. Expand and rearrange: \(3x^2-4x-4=0\). 4. Apply the quadratic formula: \(x=\frac{4\pm\sqrt{(-4)^2-4\cdot3\cdot(-4)}}{2\cdot3}=\frac{4\pm8}{6}\). The candidates are \(x=2\) and \(x=-\frac23\). 5. The value \(x=2\) is excluded from the domain and is extraneous. The only valid solution is \(x=-\frac23\).

Solution set: \(\left\{-\frac{2}{3}\right\}\). The candidate \(x=2\) is excluded because it makes a denominator zero.

- Multiply by the product of the denominators to clear the fractions. - Expand \((x-2)^2\) carefully. - Factor the resulting quadratic equation. - Check both values against the domain.

1. Multiply by \(x(x-2)\): \((x-2)^2=x\). 2. Expand and rearrange: \(x^2-4x+4=x\), so \(x^2-5x+4=0\). 3. Factor: \((x-1)(x-4)=0\), giving \(x=1\) or \(x=4\). 4. Both values belong to the stated domain.

Solution set: \(\{1,4\}\).

- Factor the denominator \(x^2-1\). - Track the subtraction sign when clearing the second fraction. - Reject any candidate excluded by the domain. - Confirm that no allowed value remains.

1. Factor \(x^2-1=(x-1)(x+1)\), which is the least common denominator. 2. Multiply by \((x-1)(x+1)\): \(x(x+1)-(x-1)=2\). 3. Simplify: \(x^2+x-x+1=2\), so \(x^2=1\). 4. The candidates are \(x=1\) and \(x=-1\). 5. Both candidates are excluded from the domain, so there is no solution.

Solution set: \(\varnothing\).

- Divide distance by circumference to find revolutions. - The smaller wheel makes more revolutions over the same distance. - Express the difference in revolutions with an equation.

1. Let \(s\) be the course length in centimeters. 2. The front wheel makes \(\frac{s}{20}\) revolutions, and the rear wheel makes \(\frac{s}{30}\) revolutions. 3. The condition gives \(\frac{s}{20}-\frac{s}{30}=15\). 4. Multiply by \(60\): \(3s-2s=900\). 5. Thus, \(s=900\,\text{cm}=9\,\text{m}\).

\(9\,\text{m}\)

- What must you check first when variables appear in denominators? - How can you remove the fractions from a proportion? - Can you rearrange equation (II) so that one variable is isolated? - Check that the final value of \(y\) is not excluded.

1. The denominators cannot equal zero, so \(y \ne -2\) and \(y \ne -5\). 2. Cross-multiply equation (I): \((2x - 1)(y + 5) = (2x + 3)(y + 2)\). 3. Expand and simplify: \(2xy + 10x - y - 5 = 2xy + 4x + 3y + 6\), so \(6x - 4y = 11\). 4. Simplify equation (II): \(3x - 3 - y - 2 = x\), so \(2x - y = 5\). 5. From equation (II), \(y = 2x - 5\). Substitute into the simplified equation (I): \(6x - 4(2x - 5) = 11\). Then \(-2x + 20 = 11\), so \(x = \frac{9}{2}\). 6. Find \(y\): \(y = 2 \cdot \frac{9}{2} - 5 = 4\). 7. Since \(4\) is not an excluded value, the solution is valid.

a) \(y \ne -2\) and \(y \ne -5\) b) The solution is \((\frac{9}{2}, 4)\).

- Replace each reciprocal expression with a temporary variable. - Add and subtract the resulting linear equations. - Substitute back to recover \(x\) and \(y\).

1. The denominators require \(x\ne-2\) and \(y\ne3\). 2. Let \(u=\frac1{x+2}\) and \(v=\frac1{y-3}\). Then \(u+v=\frac34\) and \(u-v=\frac14\). 3. Add the equations: \(2u=1\), so \(u=\frac12\). 4. Subtract the second equation from the first: \(2v=\frac12\), so \(v=\frac14\). 5. Substitute back: \(\frac1{x+2}=\frac12\) gives \(x=0\), and \(\frac1{y-3}=\frac14\) gives \(y=7\). 6. Both values satisfy the restrictions.

\(S=\{(0, 7)\}\), with \(x\ne-2\) and \(y\ne3\).

- Divide distance by rate to express each time. - Set the difference between the two times equal to 2. - After finding the fence length, divide by the sprayer rate.

1. Let \(x\) be the fence length in feet. 2. The time difference gives \(\frac{x}{24}-\frac{x}{40}=2\). 3. Multiply by 120: \(5x-3x=240\). 4. Thus, \(2x=240\), so \(x=120\,\text{ft}\). 5. With the sprayer, the time is \(\frac{120}{40}=3\) hours.

a) \(120\,\text{ft}\) b) \(3\) hours

- Replace the reciprocal expressions with temporary variables. - Solve the resulting linear system. - Substitute back to obtain two linear equations in \(x\) and \(y\).

1. Let \(u=\frac1{x+y}\) and \(v=\frac1{x-y}\). Then \(3u+4v=1\) and \(5u-2v=0.8\). 2. Multiply the second equation by 2 and add it to the first: \(13u=2.6\), so \(u=0.2\). 3. Substitute into the first equation: \(0.6+4v=1\), so \(v=0.1\). 4. Substitute back: \(x+y=5\) and \(x-y=10\). 5. Add the equations: \(2x=15\), so \(x=7.5\). Then \(y=-2.5\). 6. Both denominator expressions are nonzero.

\(S=\{(7.5, -2.5)\}\)

- Identify the values that make either denominator zero. - Multiply every term by a common multiple of the denominators. - Check the signs carefully while expanding. - Factor the quadratic after writing it in standard form.

1. The denominators require \(x\ne2\) and \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2,2\}\). 2. Multiply by \(3(x-2)(x+2)\): \(3[2x(x+2)+x(x-2)]=10(x^2-4)\). 3. Expand: \(3(3x^2+2x)=10x^2-40\), so \(9x^2+6x=10x^2-40\). 4. The quadratic standard form is \(x^2-6x-40=0\). 5. Factor: \((x-10)(x+4)=0\), giving \(x=10\) or \(x=-4\). 6. Both values lie in the domain.

\(D=\mathbb{R}\setminus\{-2,2\}\); solution set: \(\{-4,10\}\).

- A value that makes a denominator zero cannot be a solution. - Find the least common denominator of the two fractions. - Clear the fractions and write the result as a quadratic equation. - Check each solution against the domain.

1. If \(x=-2\), the denominator \(x+2\) is zero. Division by zero is undefined, so \(-2\) cannot be a solution. 2. The denominators require \(x\ne1\) and \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2,1\}\). 3. Multiply by \((x-1)(x+2)\): \((x+1)(x+2)+(x-2)(x-1)=3(x-1)(x+2)\). 4. Expand and simplify: \(2x^2+4=3x^2+3x-6\), so \(x^2+3x-10=0\). 5. Factor: \((x+5)(x-2)=0\), giving \(x=-5\) or \(x=2\). Both values lie in the domain.

a) At \(x=-2\), the denominator \(x+2\) is zero, so the equation is undefined. b) \(D=\mathbb{R}\setminus\{-2,1\}\); solution set: \(\{-5,2\}\).

- Express the planned duration using total acreage and daily rate. - Determine the number of days spent at the increased rate. - The acreage remains fixed even though the completion time changes. - Select the positive solution for a physical rate.

1. Let \(x\) be the planned rate in acres per day and \(t\) the planned duration. Then \(xt=220\), so \(t=\frac{220}{x}\). 2. The actual duration is \(t-3\) days. After the first two days, the higher rate is used for \(t-5\) days. 3. Model the total acreage: \(2x+(t-5)(x+10)=220\). 4. Substitute \(t=\frac{220}{x}\): \(2x+\left(\frac{220}{x}-5\right)(x+10)=220\). 5. Simplify: \(-3x+\frac{2200}{x}-50=0\). Multiplying by \(x\) gives \(3x^2+50x-2200=0\). 6. Factor: \((3x+110)(x-20)=0\). The candidates are \(x=20\) and \(x=-\frac{110}{3}\). 7. The positive rate is \(20\) acres per day.

\(20\) acres per day

- Express the planned rate as total brochures divided by planned time. - Increase that rate by \(10\). - Use the reduced time in the actual production equation. - Report the actual time, not the planned time.

1. Let \(x\) be the originally planned time in hours. The planned rate is \(\frac{600}{x}\) brochures per hour. 2. The new rate is \(\frac{600}{x}+10\), and the actual time is \(x-3\). 3. Use rate times time equals total production: \((x-3)\left(\frac{600}{x}+10\right)=600\). 4. Simplifying gives \(x^2-3x-180=0\). 5. Factor: \((x-15)(x+12)=0\), so \(x=15\) or \(x=-12\). 6. The planned time was \(15\) hours, so the actual time was \(15-3=12\) hours.

\(12\) hours

- Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\). - Convert the time savings to hours. - Write a time expression for each trip. - Reject a negative speed.

1. Let \(v\) be the outbound speed in miles per hour. The return speed is \(v+5\). 2. The outbound time is \(\frac{30}{v}\) hours, and the return time is \(\frac{30}{v+5}\) hours. 3. Since \(30\) minutes is \(\frac12\) hour, write \(\frac{30}{v}-\frac{30}{v+5}=\frac12\). 4. Multiply by \(2v(v+5)\): \(60(v+5)-60v=v(v+5)\). 5. Simplify: \(v^2+5v-300=0\). 6. Factor: \((v-15)(v+20)=0\), so \(v=15\) or \(v=-20\). 7. Speed must be positive, so the outbound speed was \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Add the wind speed for the tailwind flight and subtract it for the headwind flight. - Use distance divided by ground speed for each flight time. - Express “one hour longer” as a difference of the two times. - Check the solution against the domain.

1. Let \(v\) be the airplane's airspeed in still air, in miles per hour. Its ground speeds are \(v+50\) with the tailwind and \(v-50\) with the headwind. 2. The return flight takes one hour longer, so \(\frac{600}{v-50}-\frac{600}{v+50}=1\). The domain requires \(v>50\). 3. Multiply by \((v-50)(v+50)\): \(600(v+50)-600(v-50)=v^2-2500\). 4. Simplify: \(60{,}000=v^2-2500\), so \(v^2=62{,}500\). 5. Therefore, \(v=\pm250\). Only \(v=250\) satisfies the physical context and the domain.

\(250\,\text{mph}\)

- Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) for each traveler. - Determine how the two travel times differ. - The car's speed is \(15\,\text{mph}\) greater than the bicycle's speed. - Reject a negative speed.

1. Let \(v\) be the bicycle's average speed in miles per hour. The car's average speed is \(v+15\). 2. The bicycle's travel time is \(\frac{30}{v}\) hours, and the car's travel time is \(\frac{30}{v+15}\) hours. 3. Because the car leaves one hour later but arrives at the same time, \(\frac{30}{v}-\frac{30}{v+15}=1\). 4. Multiply by \(v(v+15)\): \(30(v+15)-30v=v(v+15)\). 5. Simplify: \(v^2+15v-450=0\). 6. Factor: \((v-15)(v+30)=0\), so \(v=15\) or \(v=-30\). 7. Speed must be positive, so the bicycle's average speed was \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Express each person's hourly work rate as the reciprocal of the time needed alone. - Add the two individual rates. - The combined rate is one job divided by \(2\) hours. - Check each algebraic solution against the context.

1. Let \(x\) be the number of hours Marcus would need alone. Elena would need \(x+3\) hours. 2. Marcus completes \(\frac{1}{x}\) of the lawn per hour, and Elena completes \(\frac{1}{x+3}\) per hour. Together, they complete \(\frac12\) of the lawn per hour. 3. Write the work-rate equation: \(\frac{1}{x}+\frac{1}{x+3}=\frac12\). 4. Multiply by \(2x(x+3)\): \(2(x+3)+2x=x(x+3)\). 5. Simplify: \(x^2-x-6=0\). 6. Factor: \((x-3)(x+2)=0\), so \(x=3\) or \(x=-2\). 7. A work time cannot be negative, so \(x=-2\) is not meaningful. Marcus would take \(3\) hours alone.

Marcus would take \(3\) hours alone. The solution \(x=-2\) is not meaningful because a work time cannot be negative.

- Add the current speed when traveling downstream and subtract it when traveling upstream. - Write each travel time as distance divided by speed. - Express the three-hour difference with an equation. - Check the solution against the physical domain.

1. Let \(c\) be the current's speed in miles per hour. The downstream speed is \(12+c\), and the upstream speed is \(12-c\). The domain requires \(0<c<12\). 2. The upstream trip takes \(3\) hours longer, so \(\frac{48}{12-c}-\frac{48}{12+c}=3\). 3. Multiply by \((12-c)(12+c)\): \(48(12+c)-48(12-c)=3(144-c^2)\). 4. Simplify: \(96c=432-3c^2\), so \(c^2+32c-144=0\). 5. Factor: \((c-4)(c+36)=0\), so \(c=4\) or \(c=-36\). 6. Only \(c=4\) satisfies the domain and represents a current speed.

\(4\,\text{mph}\)

- Determine the distance remaining after the delay. - Convert the delay to hours. - Compare the planned and actual travel times for the remaining distance. - Reject a negative speed.

1. Let \(v\) be the original speed in miles per hour. The increased speed is \(v+15\). 2. The driver must save \(15\) minutes, or \(\frac14\) hour, over the remaining \(45\) miles. 3. Write the time-savings equation: \(\frac{45}{v}-\frac{45}{v+15}=\frac14\). 4. Multiply by \(4v(v+15)\): \(180(v+15)-180v=v(v+15)\). 5. Simplify: \(v^2+15v-2700=0\). 6. Factor: \((v-45)(v+60)=0\), so \(v=45\) or \(v=-60\). 7. Speed must be positive, so the original speed was \(45\,\text{mph}\).

\(45\,\text{mph}\)

- Subtract the current speed when traveling upstream. - Write the lake and upstream travel times. - Use the one-hour difference to form an equation. - For the downstream trip, add the current speed.

1. For part a), let \(v\) be the boat's speed in still water, in miles per hour. The upstream ground speed is \(v-2\), so the domain requires \(v>2\). 2. The upstream trip takes one hour longer: \(\frac{40}{v-2}=\frac{40}{v}+1\). 3. Multiply by \(v(v-2)\): \(40v=40(v-2)+v(v-2)\). 4. Simplify: \(v^2-2v-80=0\). 5. Factor: \((v-10)(v+8)=0\), so \(v=10\) or \(v=-8\). 6. Only \(v=10\) satisfies the domain, so the still-water speed is \(10\,\text{mph}\). 7. For part b), the downstream speed is \(10+2=12\,\text{mph}\). 8. The downstream time is \(\frac{40}{12}=\frac{10}{3}\) hours, or \(3\) hours \(20\) minutes.

a) \(10\,\text{mph}\) b) \(3\) hours \(20\) minutes

- Substitute the expression for \(y\) into the linear equation. - Clear the denominator after noting that \(x\ne0\). - Interpret a repeated root as one intersection x-coordinate. - Connect one repeated intersection to tangency.

1. Substitute \(y=\frac{8}{x}\) into the second equation: \(2x+\frac{8}{x}=8\), with \(x\ne0\). 2. Multiply by \(x\): \(2x^2+8=8x\). 3. Rearrange and divide by \(2\): \(x^2-4x+4=0\). 4. Factor: \((x-2)^2=0\), so \(x=2\). Then \(y=\frac82=4\). 5. The system has exactly one solution, \((2,4)\). The repeated root means the line touches the rational curve at that point, so the line is tangent to the curve.

a) Solution set: \(\{(2,4)\}\). b) There is one intersection. The line is tangent to the rational curve at \((2,4)\).

- State the nonzero restrictions caused by the denominators. - Combine the two reciprocal terms into one fraction. - Substitute the known value of \(x+y\). - Use the sum and product to form a quadratic equation.

1. The denominators require \(x\ne0\) and \(y\ne0\). 2. Combine the fractions: \(\frac{x+y}{xy}=\frac54\). 3. Since \(x+y=5\), substitute to obtain \(\frac{5}{xy}=\frac54\), so \(xy=4\). 4. Numbers with sum \(5\) and product \(4\) are the roots of \(t^2-5t+4=0\). 5. Factor: \((t-1)(t-4)=0\). Therefore, the ordered solutions are \((1,4)\) and \((4,1)\). Both satisfy the domain restrictions.

Solution set: \(\{(1,4),(4,1)\}\).

- Use the simpler equation to express one variable in terms of the other. - Combine the rational expressions into one fraction. - Simplify the resulting quadratic equation. - Check that neither coordinate is zero.

1. Solve the second equation for \(y\): \(y=5-x\). 2. Substitute into the first equation: \(\frac{x}{5-x}+\frac{5-x}{x}=\frac{13}{6}\). 3. Combine the fractions: \(\frac{x^2+(5-x)^2}{x(5-x)}=\frac{13}{6}\). 4. Cross-multiply and simplify: \(6(2x^2-10x+25)=13x(5-x)\), which reduces to \(x^2-5x+6=0\). 5. Factor: \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\). 6. Using \(y=5-x\), the corresponding values are \(y=3\) and \(y=2\). Both pairs satisfy the nonzero restrictions.

Solution set: \(\{(2,3),(3,2)\}\).

- Isolate one variable in the linear equation. - Substitute into the rational equation and identify excluded values. - Clear the denominators to obtain a quadratic equation. - Check both coordinates against the original denominators.

1. Solve the second equation for \(y\): \(y=x+1\). 2. Substitute into the first equation: \(\frac{3}{x}-\frac{4}{x+1}=1\), with \(x\ne0\) and \(x\ne-1\). 3. Multiply by \(x(x+1)\): \(3(x+1)-4x=x(x+1)\). 4. Simplify: \(-x+3=x^2+x\), so \(x^2+2x-3=0\). 5. Factor: \((x-1)(x+3)=0\), giving \(x=1\) or \(x=-3\). 6. Using \(y=x+1\), the corresponding values are \(y=2\) and \(y=-2\). Both pairs satisfy the domain restrictions.

Solution set: \(\{(-3,-2),(1,2)\}\).

- State the two excluded values of \(x\). - Factor \(x^2-a^2\) to identify the common denominator. - Treat \(a\) as a constant while simplifying. - Factor the resulting quadratic expression in \(x\).

1. The denominators require \(x\ne a\) and \(x\ne-a\). 2. Multiply by \((x-a)(x+a)=x^2-a^2\): \(x(x+a)-2a(x-a)=8a^2\). 3. Expand and rearrange: \(x^2-ax-6a^2=0\). 4. Factor: \((x-3a)(x+2a)=0\), so \(x=3a\) or \(x=-2a\). 5. Since \(a\ne0\), neither solution equals \(a\) or \(-a\). Both are valid.

Solution set: \(\{-2a,3a\}\).

- Express each pumping rate as volume divided by time. - Relate the two individual times. - Set the difference between the rates equal to \(5\,\text{m}^3/\text{h}\). - Reject a negative time.

1. Let \(t\) be the number of hours the faster pump needs. The slower pump needs \(t+2\) hours. 2. Their pumping rates are \(\frac{120}{t}\) and \(\frac{120}{t+2}\) cubic meters per hour. 3. The rate difference is \(5\): \(\frac{120}{t}-\frac{120}{t+2}=5\). 4. Divide by \(5\) and clear the denominators: \(24(t+2)-24t=t(t+2)\). 5. Simplify: \(t^2+2t-48=0\). 6. Factor: \((t-6)(t+8)=0\), so \(t=6\) or \(t=-8\). 7. Time must be positive, so the faster pump takes \(6\) hours and the slower pump takes \(8\) hours.

The slower pump takes \(8\) hours, and the faster pump takes \(6\) hours.

- Express the original price as total budget divided by number of tablets. - Reduce that price by \(\$50\) and increase the quantity by \(8\). - Set the new total cost equal to the original budget. - Reject a negative quantity.

1. Let \(x\) be the original number of tablets. Measured in dollars, the original price per tablet is \(\frac{12{,}000}{x}\). 2. After the discount, the price in dollars is \(\frac{12{,}000}{x}-50\), and the school buys \(x+8\) tablets. 3. Use the unchanged total cost: \(\left(\frac{12{,}000}{x}-50\right)(x+8)=12{,}000\). 4. Clear the denominator and simplify: \(x^2+8x-1920=0\). 5. Factor: \((x-40)(x+48)=0\), so \(x=40\) or \(x=-48\). 6. The number of tablets must be positive, so \(x=40\). 7. The original price was \(\frac{12{,}000}{40}=300\), so it was \(\$300\) per tablet.

The school originally planned to buy \(40\) tablets at \(\$300\) each.

- Express average time per mile as total time divided by distance. - Convert the seconds to hours. - Subtract the two time-per-mile expressions. - Reject a negative distance.

1. Driver A's average time per mile is \(\frac{2}{x}\) hours, and driver B's is \(\frac{2}{x+10}\) hours. 2. Convert \(36\) seconds to hours: \(\frac{36}{3600}=\frac{1}{100}\) hour. 3. Write the time-per-mile difference: \(\frac{2}{x}-\frac{2}{x+10}=\frac{1}{100}\). 4. Multiply by \(100x(x+10)\): \(200(x+10)-200x=x(x+10)\). 5. Simplify: \(x^2+10x-2000=0\). 6. Factor: \((x-40)(x+50)=0\), so \(x=40\) or \(x=-50\). 7. Distance must be positive, so driver A's route is \(40\) miles.

\(40\) miles

- Express each person's daily work rate as the reciprocal of the individual completion time. - Multiply each rate by the number of days that person works. - The two completed portions add to one whole project. - Reject a negative time.

1. Let \(x\) be the number of days the designer would need alone. The assistant would need \(x+10\) days. 2. The designer completes \(\frac{4}{x}\) of the project, and the assistant completes \(\frac{9}{x+10}\). 3. Their completed fractions sum to one: \(\frac{4}{x}+\frac{9}{x+10}=1\). 4. Multiply by \(x(x+10)\): \(4(x+10)+9x=x(x+10)\). 5. Simplify: \(x^2-3x-40=0\). 6. Factor: \((x-8)(x+5)=0\), so \(x=8\) or \(x=-5\). 7. Time must be positive, so the designer needs \(8\) days and the assistant needs \(18\) days.

The designer needs \(8\) days, and the assistant needs \(18\) days.

- Add the two hourly work rates. - Express the apprentice's time using the landscaper's time. - For part b), multiply the landscaper's hourly rate by \(6\) hours.

1. Let \(g\) be the number of hours the landscaper would need alone. The apprentice would need \(g+9\) hours. 2. Add their work rates: \(\frac{1}{g}+\frac{1}{g+9}=\frac16\). 3. Multiply by \(6g(g+9)\): \(6(g+9)+6g=g(g+9)\). 4. Simplify: \(g^2-3g-54=0\). 5. Factor: \((g-9)(g+6)=0\), so \(g=9\) or \(g=-6\). 6. The positive solution gives \(9\) hours for the landscaper and \(18\) hours for the apprentice. 7. During \(6\) hours, the landscaper completes \(6\left(\frac19\right)=\frac23\) of the job.

a) The landscaper needs \(9\) hours, and the apprentice needs \(18\) hours. b) The landscaper completes \(\frac23\) of the job.

- Express each work rate as the reciprocal of its completion time. - Write each individual time in terms of the shared time. - Set the shared rate equal to the sum of the individual rates. - Use the positive square root for a time.

1. Let \(x\) be the number of hours the two mowers need together. Their individual times are \(x+16\) and \(x+9\). 2. The combined rate equals the sum of the individual rates: \(\frac{1}{x}=\frac{1}{x+16}+\frac{1}{x+9}\). 3. Multiply by \(x(x+16)(x+9)\): \((x+16)(x+9)=x(x+9)+x(x+16)\). 4. Expand and simplify: \(x^2+25x+144=2x^2+25x\), so \(x^2=144\). 5. Since time is positive, \(x=12\) hours. 6. The individual times are \(12+16=28\) hours and \(12+9=21\) hours.

Together, the mowers take \(12\) hours. Mower 1 takes \(28\) hours alone, and mower 2 takes \(21\) hours alone.

- Solve the numerator equation first. - Check each candidate in the original denominator. - A value that makes both numerator and denominator zero is not in the domain.

1. A rational expression equals zero only when its numerator is zero and its denominator is nonzero. 2. For \(\frac{x^2+1}{x-1}\), the equation \(x^2+1=0\) has no real solution, so the expression has no real zero. 3. For \(\frac{(x-4)(x+2)}{x-4}\), the numerator is zero at \(x=4\) and \(x=-2\). The value \(x=4\) is excluded by the denominator, so the only zero is \(x=-2\). 4. For \(\frac{0.5x-2}{x^2-16}\), the numerator is zero at \(x=4\). However, \(x=4\) also makes the denominator zero, so the expression has no zero.

1) No real zero 2) \(x=-2\) 3) No real zero

- Add the wind speed for the tailwind flight and subtract it for the headwind flight. - Write a travel-time expression for each flight. - Set their sum equal to the total time. - Check the result against the requirement that the airplane can fly into the wind.

1. Let \(v\) be the airspeed in still air, in miles per hour. The domain requires \(v>30\). 2. The outbound and return ground speeds are \(v+30\) and \(v-30\). 3. The total-time equation is \(\frac{180}{v+30}+\frac{180}{v-30}=2.5\). 4. Multiply by \((v+30)(v-30)\) and simplify: \(360v=2.5(v^2-900)\). 5. Multiply by \(2\) and rearrange: \(v^2-144v-900=0\). 6. Factor: \((v-150)(v+6)=0\), so \(v=150\) or \(v=-6\). 7. Only \(v=150\) satisfies the domain, so the airspeed is \(150\,\text{mph}\).

\(150\,\text{mph}\)

- Cross-multiply after noting that \(x\ne0\). - Determine when \(k^3\) is positive or negative. - A real square cannot be negative. - Include both signs when taking a square root.

1. Because \(x\) is in a denominator, \(x\ne0\). 2. Cross-multiply: \(x^2=k^3\). 3. If \(k<0\), then \(k^3<0\), so the equation has no real solutions. 4. If \(k>0\), then \(x=\pm\sqrt{k^3}=\pm k\sqrt{k}\). Both values are nonzero and therefore valid.

For \(k>0\), the solution set is \(\{-k\sqrt{k},k\sqrt{k}\}\). For \(k<0\), the solution set is \(\varnothing\).

- Convert both time differences to hours. - Add the starting advantage and arrival advantage to obtain the travel-time difference. - Use distance divided by speed for each train. - Reject a negative speed.

1. Let \(v\) be the freight train's speed in miles per hour. The express train's speed is \(v+20\). 2. Their travel times are \(\frac{300}{v}\) and \(\frac{300}{v+20}\) hours. 3. The freight train's travel time exceeds the express train's by the \(30\)-minute head start plus the \(45\)-minute earlier arrival: \(\frac12+\frac34=\frac54\) hours. 4. Write the equation \(\frac{300}{v}-\frac{300}{v+20}=\frac54\). 5. Clear the denominators and simplify: \(v^2+20v-4800=0\). 6. Factor: \((v-60)(v+80)=0\), so \(v=60\) or \(v=-80\). 7. Speed must be positive, so the freight train travels at \(60\,\text{mph}\).

\(60\,\text{mph}\)

- Exclude the value that makes the denominator zero. - Equation solutions are the x-coordinates of graph intersections. - Simplify before multiplying by the denominator. - Factor out \(x\) from the final quadratic expression.

1. The denominator requires \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2\}\). 2. The graph shows intersections at \((-4,3)\) and \((0,-1)\), suggesting \(x=-4\) and \(x=0\). 3. Subtract \(1\): \(\frac{-4}{x+2}=-x-2\). 4. Multiply by \(x+2\): \(-4=(-x-2)(x+2)\). 5. Expand and simplify: \(-4=-x^2-4x-4\), so \(x^2+4x=0\). 6. Factor: \(x(x+4)=0\), giving \(x=0\) or \(x=-4\). Both values are in the domain, confirming the graph.

a) \(D=\mathbb{R}\setminus\{-2\}\) b) Intersections: \((-4,3)\) and \((0,-1)\); predicted solutions: \(x=-4\) and \(x=0\) c) Solution set: \(\{-4,0\}\), confirming the prediction.

- Use the overall shape of each graph to identify its function family. - Solutions of \(f(x)=g(x)\) are x-coordinates of intersection points. - Set the formulas for \(g\) and \(h\) equal. - State the domain restriction before clearing the denominator.

1. The graph of \(f\) is a line, so \(f\) is linear. The graph of \(g\) is a downward-opening parabola, so \(g\) is quadratic. The graph of \(h\) has two branches and is undefined at \(x=0\), so \(h\) is rational. 2. The graphs of \(f\) and \(g\) intersect at approximately \(x=-2\) and \(x=1.5\). 3. From the graph labels, \(g(x)=-x^2+4\) and \(h(x)=\frac{2}{x}\). Their intersections satisfy \(-x^2+4=\frac{2}{x}\). 4. Since \(x\ne0\), multiply both sides by \(x\): \(-x^3+4x=2\). Moving all terms to one side gives \(x^3-4x+2=0\).

a) \(f\) is linear, \(g\) is quadratic, and \(h\) is rational. b) \(x\approx-2\) and \(x\approx1.5\) c) \(-x^2+4=\frac{2}{x}\), which is equivalent to \(x^3-4x+2=0\) for \(x\ne0\).

- Locate the points where the line and rational curve meet. - Rewrite \(f(x)=g(x)\) to compare it with the choices. - Multiply by \(x\) and factor the resulting quadratic.

1. The graph shows intersections at \((1, 3)\) and \((3, 1)\). 2. Intersections satisfy \(f(x)=g(x)\), so \(\frac{3}{x}=-x+4\). Adding \(x\) to both sides gives \(\frac{3}{x}+x=4\), which is equation (A). 3. Solve \(\frac{3}{x}=-x+4\), with \(x\ne0\). Multiplying by \(x\) gives \(3=-x^2+4x\), so \(x^2-4x+3=0\). 4. Factor: \((x-1)(x-3)=0\), so \(x=1\) or \(x=3\).

1. \((1, 3)\) and \((3, 1)\) 2. (A), because it is equivalent to \(f(x)=g(x)\) 3. Solution set: \(\{1, 3\}\).

- Exclude the value that makes the denominator zero. - Multiply by the denominator to clear the fraction. - Factor out the common variable from the quadratic expression.

1. The denominator requires \(x\ne-2\), so \(D=\mathbb{R}\setminus\{-2\}\). 2. Multiply by \(x+2\): \(10=(x+5)(x+2)\). 3. Expand: \(10=x^2+7x+10\), so \(x^2+7x=0\). 4. Factor: \(x(x+7)=0\), giving \(x=0\) or \(x=-7\). Both values are in the domain.

a) \(D=\mathbb{R}\setminus\{-2\}\) b) Solution set: \(\{-7,0\}\).

- Use the slope formula with the two given points. - Set the function expressions equal. - Clear the denominator after noting the excluded x-value. - Factor the resulting quadratic expression. - Substitute each x-value to find its y-coordinate.

1. The slope of the line is \(m=\frac{2-0}{1-(-1)}=1\). Substituting \((-1, 0)\) into \(g(x)=x+n\) gives \(n=1\), so \(g(x)=x+1\). 2. Set the functions equal: \(\frac{4}{x+2}-1=x+1\), with \(x\ne-2\). 3. Add \(1\): \(\frac{4}{x+2}=x+2\). 4. Multiply by \(x+2\): \(4=(x+2)^2\). 5. Expand and simplify: \(x^2+4x=0\). Factor: \(x(x+4)=0\), so \(x=0\) or \(x=-4\). 6. Substituting into \(g\) gives the intersections \((0, 1)\) and \((-4, -3)\).

a) \(g(x)=x+1\) b) \((-4, -3)\) and \((0, 1)\)

- A reciprocal is undefined when its denominator is zero. - Read \(f(2)\) before taking the reciprocal. - Set the two function values equal at an intersection. - Solve for the possible values of \(f(x)\), then read the corresponding inputs from the graph.

1. Since \(f(0)=0\), the expression \(g(0)=\frac{1}{f(0)}\) would require division by zero. Therefore, \(0\) is excluded from the domain of \(g\). 2. The graph shows \(f(2)=3\), so \(g(2)=\frac{1}{3}\). 3. At an intersection, \(f(x)=\frac{1}{f(x)}\). Because \(f(x)\ne0\) there, multiply by \(f(x)\) to obtain \(f(x)^2=1\). 4. Thus, \(f(x)=1\) or \(f(x)=-1\). From the graph, these values occur at \(x=1\) and \(x=-1\), respectively.

a) \(0\) is excluded because \(f(0)=0\), and \(g(2)=\frac{1}{3}\). b) \(x=-1\) and \(x=1\)

- Set the two function values equal. - Use the intersection point in the equation of the line. - Examine the equation that remains after clearing the denominator.

1. Set the functions equal: \(\frac{12}{x}=\frac{3}{x}+1.5\). 2. Subtract \(\frac{3}{x}\): \(\frac{9}{x}=1.5\). 3. Since \(x\ne0\), multiply by \(x\): \(9=1.5x\), so \(x=6\). 4. Then \(f(6)=2\), so the intersection is \((6, 2)\). 5. Substitute into \(h(x)=mx-1\): \(2=6m-1\), so \(m=0.5\). 6. For \(x>0\), the intersection equation reduces to \(9=1.5x\), a linear equation with exactly one positive solution.

a) \((6, 2)\) b) \(m=0.5\) c) The intersection equation has exactly one positive solution, \(x=6\).

- Rewrite the decimal as a fraction before clearing denominators. - State the excluded values before multiplying by a common denominator. - After solving the resulting quadratic equation, check whether each value belongs to \(\mathbb{Q}\).

1. The denominators are zero at \(x=1\) and \(x=-1\), so the domain is \(\mathbb{Q}\setminus\{-1,1\}\). 2. Rewrite \(2.5\) as \(\frac{5}{2}\), and multiply the equation by \(2(x-1)(x+1)\): \(2x(x+1)+2x(x-1)=5(x-1)(x+1)\). 3. Expand and simplify: \(4x^2=5x^2-5\), so \(x^2=5\). 4. The real solutions of \(x^2=5\) are \(x=\pm\sqrt5\), but neither value is rational. 5. Therefore, the equation has no solutions in \(\mathbb{Q}\).

\(\varnothing\)

- Intersecting graphs have the same output at the stated input. - Use the function without the parameter to find that output first. - Substitute the intersection coordinates into the second function.

1. At the intersection input, \(f(2)=\frac62=3\). 2. Therefore, \(g(2)=3\), so \(\frac3{2+a}=3\), with \(a\ne-2\). 3. Multiply by \(2+a\): \(3=3(2+a)\). 4. Divide by 3: \(1=2+a\), so \(a=-1\). 5. This value keeps the denominator nonzero at \(x=2\).

\(a=-1\)

- Identify the value excluded by the denominator. - Intersection points satisfy an equation formed by setting the two function values equal. - When does a quadratic equation have exactly one real solution? - A solution outside the original domain does not represent an intersection point.

1. The denominator excludes \(x=1\), so \(D_{f_k}=\mathbb{R}\setminus\{1\}\). The x-intercept of \(g\) is \((1, 0)\). For \(f_k\), the numerator is zero at \(x=-\frac{k}{2}\). This gives an x-intercept unless \(-\frac{k}{2}=1\), or \(k=-2\). Thus, for \(k\neq -2\), the x-intercept is \(\left(-\frac{k}{2}, 0\right)\); for \(k=-2\), there is none. 2. For \(k=6\), solve \(\frac{2x+6}{x-1}=x-1\). Multiplying by \(x-1\), with \(x\neq 1\), gives \(x^2-4x-5=0\), so \(x=5\) or \(x=-1\). Using \(g(x)=x-1\), the intersection points are \((5, 4)\) and \((-1, -2)\). 3. The equation \(f_k(x)=g(x)\) is equivalent to \(f_k(x)-g(x)=0\). Therefore, the x-coordinates of valid intersection points are exactly the zeros of \(d_k\). 4. In general, the intersection equation becomes \(x^2-4x+1-k=0\), with \(x\neq 1\). 5. A repeated solution occurs when the discriminant is zero: \(16-4(1-k)=0\), so \(k=-3\). The repeated root is \(x=2\), which is valid. 6. Exactly one valid point also occurs when one of two roots is the excluded value \(x=1\). Substitution gives \(k=-2\); then the roots are \(1\) and \(3\), leaving only \(x=3\) as a valid intersection. 7. Therefore, \(k=-3\) or \(k=-2\).

a) \(D_{f_k}=\mathbb{R}\setminus\{1\}\). The x-intercept of \(g\) is \((1, 0)\). For \(k\neq -2\), the x-intercept of \(f_k\) is \(\left(-\frac{k}{2}, 0\right)\); for \(k=-2\), \(f_k\) has no x-intercept. b) \((5, 4)\) and \((-1, -2)\) c) Their x-coordinates are the zeros of \(d_k(x)=f_k(x)-g(x)\). d) \(k=-3\) or \(k=-2\)

- Set the two function expressions equal to find intersection points. - A quadratic equation has no real solutions when its discriminant is negative. - Check whether any algebraic solution is excluded from the domain of \(h_a\).

1. The denominator excludes \(x=1\), so \(D_{h_a}=\mathbb{R}\setminus\{1\}\). For \(a=-6\), solve \(3x-6=0\), giving the x-intercept \((2, 0)\). 2. For \(a=-5\), solve \(\frac{3x-5}{x-1}=x-1\). Multiplying by \(x-1\), with \(x\neq 1\), gives \(x^2-5x+6=0\). Thus, \(x=2\) or \(x=3\), and the intersection points are \((2, 1)\) and \((3, 2)\). 3. In general, the intersection equation is \(x^2-5x+1-a=0\). It has no real solutions when its discriminant is negative: \(25-4(1-a)<0\). 4. Simplifying gives \(21+4a<0\), so \(a<-\frac{21}{4}\). 5. A root at the excluded value \(x=1\) occurs when \(a=-3\), but the other root is \(x=4\), so that case still has one valid intersection. Therefore, the graphs have no common point exactly when \(a<-\frac{21}{4}\).

a) \(D_{h_a}=\mathbb{R}\setminus\{1\}\), and the x-intercept of \(h_{-6}\) is \((2, 0)\). b) \((2, 1)\) and \((3, 2)\) c) \(a<-\frac{21}{4}\)

- Clear denominators carefully while retaining domain restrictions. - When can a common factor of \(x\) or \(x^2\) be factored from every term? - Check every solution in the original equation’s domain. - An equation in \(x^4\) and \(x^2\) may become quadratic after a substitution.

1. In equation 1, factoring out \(x^2\) is useful: \(x^2(x^2-9)=0\). The solutions are \(x=0\) and \(x=\pm 3\). None is excluded by \(x+1\neq 0\), so the solution set is \(\{-3, 0, 3\}\). 2. In equation 2, first multiply by \(x^2\), with \(x\neq 0\): \(x^4-5x^2-36=0\). Substitute \(u=x^2\): \(u^2-5u-36=0\), so \(u=9\) or \(u=-4\). Only \(u=9\) gives real values, so the solution set is \(\{-3, 3\}\). Factoring out a power of \(x\) is not the useful first step. 3. In equation 3, multiply by \(x\), with \(x\neq 0\): \(x^2+x-12=0\). Factoring gives \((x+4)(x-3)=0\), so the solution set is \(\{-4, 3\}\). Factoring out a power of \(x\) is not the useful first step. 4. In equation 4, factoring out \(x^2\) is useful: \(x^2(x+2)=0\). Since \(x^2+4\) is never zero, the solution set is \(\{-2, 0\}\).

Factoring out a power of \(x\) is a useful first step for equations 1 and 4. 1) \(\{-3, 0, 3\}\) 2) \(\{-3, 3\}\) 3) \(\{-4, 3\}\) 4) \(\{-2, 0\}\)

- Factor the difference of squares in the denominator. - Combine the left side over a common denominator. - Consider \(a=0\) separately before dividing by \(a\).

1. The denominators require \(x\ne a\) and \(x\ne-a\). 2. Since \(x^2-a^2=(x-a)(x+a)\), combine the left side: \(\frac{a(x+a)+a(x-a)}{x^2-a^2}=\frac{2ax}{x^2-a^2}\). 3. On the domain, equate numerators: \(2ax=a^2\). 4. If \(a\ne0\), divide by \(2a\): \(x=\frac a2\). This value is not \(a\) or \(-a\) when \(a\ne0\). 5. If \(a=0\), the original equation is \(0=0\), but its denominators require \(x\ne0\). Thus, every nonzero real \(x\) is a solution.

If \(a\ne0\), \(x=\frac a2\). If \(a=0\), \(S=\mathbb{R}\setminus\{0\}\).

- Treat each reciprocal denominator expression as a new variable. - Solve the resulting linear system first. - Substitute back and solve for \(x\) and \(y\).

1. Let \(u=\frac1{x+2y}\) and \(v=\frac1{2x-y}\). The system becomes \(10u+5v=3\) and \(5u-10v=-1\). 2. Multiply the first equation by 2 and add the second: \(25u=5\), so \(u=\frac15\). 3. Substitute into \(10u+5v=3\): \(2+5v=3\), so \(v=\frac15\). 4. Substitute back: \(x+2y=5\) and \(2x-y=5\). 5. Solve the linear system to obtain \(x=3\) and \(y=1\). 6. Both original denominators equal 5, so the solution satisfies the restrictions.

\(S=\{(3, 1)\}\)

- Use time equals distance divided by speed. - Downstream speed is the boat speed plus the current; upstream speed is the difference. - Substitute variables for the reciprocals of the two travel speeds.

1. Let \(d\) be the downstream speed and \(u\) the upstream speed, in miles per hour. 2. Using \(\text{time}=\frac{\text{distance}}{\text{speed}}\), write \(\frac{60}{d}+\frac{36}{u}=6\) and \(\frac{40}{d}+\frac{60}{u}=7\). 3. Let \(p=\frac1d\) and \(q=\frac1u\). Then \(60p+36q=6\) and \(40p+60q=7\). 4. Solving the linear system gives \(p=\frac1{20}\) and \(q=\frac1{12}\), so \(d=20\) and \(u=12\). 5. If \(b\) is the speed in still water and \(c\) is the current speed, then \(b+c=20\) and \(b-c=12\). 6. Thus, \(b=16\) and \(c=4\).

The boat’s speed in still water is \(16\,\text{mph}\), and the current’s speed is \(4\,\text{mph}\).

- Use \(V=\frac{m}{\rho}\) for each liquid. - Add the component volumes for each mixture. - Substitute variables for the reciprocal densities.

1. Since \(V=\frac{m}{\rho}\), the mixtures give \(\frac{100}{\rho_1}+\frac{100}{\rho_2}=150\) and \(\frac{120}{\rho_1}+\frac{40}{\rho_2}=100\). 2. Let \(u=\frac1{\rho_1}\) and \(v=\frac1{\rho_2}\). Then \(u+v=1.5\) and \(3u+v=2.5\). 3. Subtract the first equation from the second: \(2u=1\), so \(u=0.5\). 4. Then \(v=1\). 5. Therefore, \(\rho_1=\frac1u=2\) and \(\rho_2=\frac1v=1\).

\(\rho_1=2\,\text{g/cm}^3\) and \(\rho_2=1\,\text{g/cm}^3\)

- Replace each reciprocal denominator expression with a temporary variable. - Eliminate one temporary variable from the resulting linear system. - Substitute back and solve the final linear system in \(a\) and \(b\).

1. Let \(u=\frac1{3a+b}\) and \(v=\frac1{a+2b}\). Then \(21u+8v=5\) and \(14u-12v=-1\). 2. Multiply the first equation by 2 and the second by 3: \(42u+16v=10\) and \(42u-36v=-3\). 3. Subtract the second equation from the first: \(52v=13\), so \(v=\frac14\). 4. Substitute into the first equation: \(21u+2=5\), so \(u=\frac17\). 5. Substitute back: \(3a+b=7\) and \(a+2b=4\). 6. Solving this linear system gives \(a=2\) and \(b=1\). 7. Both original denominator expressions are nonzero.

\(S=\{(2, 1)\}\), where the ordered pair is \((a, b)\).

- Identify the values of \(x\) that make either denominator zero. - Multiply by the product of the denominators to clear the fractions. - Use the square-of-a-sum and square-of-a-difference identities. - Check the resulting x-values against the domain.

1. The denominators require \(x\ne a\) and \(x\ne-a\), so \(D=\mathbb{R}\setminus\{-a,a\}\). 2. Multiply by \(5(x-a)(x+a)\): \(5(x+a)^2+5(x-a)^2=26(x^2-a^2)\). 3. Expand and simplify: \(10x^2+10a^2=26x^2-26a^2\), so \(36a^2=16x^2\). 4. Therefore, \(x^2=\frac94a^2\), giving \(x=\frac32a\) or \(x=-\frac32a\). Since \(a\ne0\), neither solution equals \(a\) or \(-a\), so both are valid. 5. For \(x=3\), solve \(3=\frac32a\) or \(3=-\frac32a\). This gives \(a=2\) or \(a=-2\).

a) \(D=\mathbb{R}\setminus\{-a,a\}\) b) \(x=\frac32a\) or \(x=-\frac32a\) c) \(a=-2\) or \(a=2\)

- Cross-multiply only after identifying the excluded values. - Check whether the variable remains after simplifying each case. - State the domain before accepting a candidate solution. - A value that makes a denominator zero cannot solve the original equation.

1. For \(a=0\), the domain excludes \(x=0\) and \(x=2\). The equation is \(\frac{x}{x-2}=\frac{x-4}{x}\). Cross-multiplying gives \(x^2=(x-2)(x-4)=x^2-6x+8\), so \(x=\frac{4}{3}\). This value is in the domain. 2. For \(a=3\), the equation is \(\frac{x-3}{x-2}=\frac{x-4}{x-3}\). Cross-multiplying gives \((x-3)^2=(x-2)(x-4)\), which simplifies to \(9=8\). This contradiction means there is no solution. 3. For \(a=4\), the domain is \(\mathbb{R}\setminus\{2,4\}\). Cross-multiplication produces \((x-4)^2=(x-2)(x-4)\), which simplifies to \(x=4\). 4. However, \(x=4\) is excluded because it makes the denominator \(x-a\) equal to zero. Therefore, the equation has no solution, and Noah's claim is false.

a) \(x=\frac{4}{3}\) b) The algebra reduces to the contradiction \(9=8\), so there is no solution. c) The claim is false. The candidate \(x=4\) is excluded from the domain, so there is no solution.

- Relate daily production, number of days, and total production. - Express the planned number of days in terms of the daily quota. - Determine how many days were worked at each production rate. - Add the production from the two phases.

1. Let \(x\) be the planned number of components per day and \(t\) the planned number of days. Then \(xt=400\), so \(t=\frac{400}{x}\). 2. The team works \(t-1\) days in total. After the first two days, it works \(t-3\) days at \(x+10\) components per day. 3. Model the actual production: \(2x+(t-3)(x+10)=430\). 4. Substitute \(t=\frac{400}{x}\): \(2x+\left(\frac{400}{x}-3\right)(x+10)=430\). 5. Simplifying gives \(x^2+60x-4000=0\). 6. Apply the quadratic formula: \(x=\frac{-60\pm\sqrt{60^2-4\cdot1\cdot(-4000)}}{2}=\frac{-60\pm140}{2}\). 7. The solutions are \(x=40\) and \(x=-100\). A production rate must be positive, so the planned quota was \(40\) components per day.

\(40\) components per day

- a) Write the per-person contribution before and after the additional students join. - b) Test the proposed doubled group directly; you do not need to solve a second quadratic equation. - Compare the two per-person contributions for the \(\$144\) gift.

1. For part a), let \(x\) be the original number of students. The original contribution is \(\frac{72}{x}\), and the new contribution is \(\frac{72}{x+3}\). 2. The contribution decreases by \(\$2\), so \(\frac{72}{x}-\frac{72}{x+3}=2\). 3. Multiply by \(x(x+3)\): \(72(x+3)-72x=2x(x+3)\). 4. Simplify: \(x^2+3x-108=0\). 5. Factor: \((x-9)(x+12)=0\). The candidates are \(x=9\) and \(x=-12\), so \(9\) students originally planned to participate. 6. For part b), if \(18\) students divide \(\$144\), each pays \(\$8\). If \(3\) more students join, each pays \(\frac{144}{21}=\frac{48}{7}\) dollars. 7. The decrease is \(8-\frac{48}{7}=\frac{8}{7}\approx1.14\) dollars, not \(\$2\). Therefore, doubling the original group does not preserve the stated decrease.

a) \(9\) students b) No. With \(18\) students, adding \(3\) more would reduce each contribution by only \(\frac{8}{7}\) dollars, or about \(\$1.14\), not \(\$2\).

- Account for both the change in motor speed and the river current. - Write one time expression for the upstream trip and one for the downstream trip. - Set the sum of the times equal to \(6\) hours. - Check the solutions against the requirement that the boat move upstream.

1. Let \(v\) be the boat's original speed in still water, in miles per hour. The domain requires \(v>3\). 2. The upstream ground speed is \(v-3\). 3. On the return trip, the boat's still-water speed is \(v+3\), so its downstream ground speed is \(v+6\). 4. The total-time equation is \(\frac{36}{v-3}+\frac{36}{v+6}=6\). 5. Divide by \(6\): \(\frac{6}{v-3}+\frac{6}{v+6}=1\). 6. Multiply by \((v-3)(v+6)\): \(6(v+6)+6(v-3)=(v-3)(v+6)\). 7. Simplify: \(v^2-9v-36=0\). 8. Factor: \((v-12)(v+3)=0\), so \(v=12\) or \(v=-3\). 9. Only \(v=12\) satisfies \(v>3\), so the original still-water speed was \(12\,\text{mph}\).

\(12\,\text{mph}\)

- First subtract the break from the total elapsed time. - Add the wind speed on one trip and subtract it on the other. - Write a time expression for each direction. - Check the solution against the requirement that the cyclist can ride into the wind.

1. Subtract the break from the total time. The actual riding time is \(5\) hours. 2. Let \(v\) be the cyclist's speed in still air, in miles per hour. The domain requires \(v>3\). 3. The outbound speed is \(v+3\), and the return speed is \(v-3\). Therefore, \(\frac{36}{v+3}+\frac{36}{v-3}=5\). 4. Multiply by \((v+3)(v-3)\): \(36(v-3)+36(v+3)=5(v^2-9)\). 5. Simplify: \(5v^2-72v-45=0\). 6. Factor: \((5v+3)(v-15)=0\), so \(v=-\frac35\) or \(v=15\). 7. Only \(v=15\) satisfies the domain, so the cyclist's speed in still air is \(15\,\text{mph}\).

\(15\,\text{mph}\)

- Express each pump's hourly rate as the reciprocal of its individual filling time. - Determine how many hours each pump runs. - Add the fractions of the tank filled by the two pumps. - Check each solution against the relationship between the two individual times.

1. Let \(x\) be the number of hours the first pump would need alone. The second pump would need \(x-4\) hours, so the domain requires \(x>4\). 2. The first pump runs for \(5\) hours in all, and the second pump runs for \(3\) hours. 3. Their contributions fill one tank: \(\frac{5}{x}+\frac{3}{x-4}=1\). 4. Multiply by \(x(x-4)\): \(5(x-4)+3x=x(x-4)\). 5. Simplify: \(x^2-12x+20=0\). 6. Factor: \((x-10)(x-2)=0\), so \(x=10\) or \(x=2\). 7. Only \(x=10\) satisfies \(x>4\). Thus, the first pump takes \(10\) hours alone, and the second takes \(10-4=6\) hours alone.

The first pump takes \(10\) hours, and the second pump takes \(6\) hours.

- a) Add the machines' hourly work rates. - Express machine B's time in terms of machine A's time. - b) Multiply each hourly operating cost by the number of hours that machine runs. - When both machines run, include both hourly costs.

1. For part a), let \(x\) be the number of hours machine A would need alone. Machine B would need \(x+5\) hours. 2. Add their work rates: \(\frac{1}{x}+\frac{1}{x+5}=\frac16\). 3. Multiply by \(6x(x+5)\): \(6(x+5)+6x=x(x+5)\). 4. Simplify: \(x^2-7x-30=0\). 5. Factor: \((x-10)(x+3)=0\), so \(x=10\) or \(x=-3\). Time must be positive, so machine A takes \(10\) hours and machine B takes \(15\) hours. 6. For part b), operating both machines for \(6\) hours gives \((110+70)\cdot6=1080\), so the cost is \(\$1080\). 7. Operating machine B alone for \(15\) hours gives \(70\cdot15=1050\), so the cost is \(\$1050\). 8. Because \(1050<1080\), using machine B alone costs \(\$30\) less.

a) Machine A takes \(10\) hours, and machine B takes \(15\) hours. b) Using machine B alone costs less: \(\$1050\), compared with \(\$1080\) for using both machines.

- Express the first-year profit as a fraction of the starting capital. - Determine the capital available at the beginning of the second year. - Apply the same percentage return to that new amount. - Clear the denominator before solving the quadratic equation.

1. The first-year rate of return, written as a decimal, is \(\frac{500}{x}\). 2. At the beginning of the second year, the capital is \(x+500\). 3. Apply the same rate of return during the second year: \((x+500)\left(1+\frac{500}{x}\right)=3600\). 4. Rewrite and clear the denominator: \(\frac{(x+500)^2}{x}=3600\), so \((x+500)^2=3600x\). 5. Expand and rearrange: \(x^2-2600x+250{,}000=0\). 6. Factor: \((x-100)(x-2500)=0\). 7. Therefore, the mathematically possible starting amounts are \(x=100\) and \(x=2500\). Both are positive and satisfy the model.

The original starting capital could have been \(\$100\) or \(\$2500\).

- Write the concentrate fraction after one dilution. - Two identical dilution steps multiply the fraction by itself. - Convert the percentage to a fraction. - Check each algebraic solution against the amount removed.

1. The container must satisfy \(V>2\). After the first replacement, the concentrate fraction is \(\frac{V-2}{V}\). 2. The same fraction remains after each identical dilution step, so the final fraction is \(\left(\frac{V-2}{V}\right)^2\). 3. Since \(56.25\%=\frac{9}{16}\), write \(\left(1-\frac{2}{V}\right)^2=\frac{9}{16}\). 4. Taking square roots gives \(1-\frac{2}{V}=\frac34\) or \(1-\frac{2}{V}=-\frac34\). 5. The positive branch gives \(\frac{2}{V}=\frac14\), so \(V=8\). 6. The negative branch gives \(V=\frac87\), which violates \(V>2\) because the process removes \(2\) gallons. Therefore, the capacity is \(8\) gallons.

\(8\) gallons

- Think of the route as two segments separated by the meeting point. - Each hiker's pre-meeting segment is traveled by the other hiker after the meeting. - Use distance equals rate times time for those matching segments. - Equate two expressions for the speed ratio.

1. Let \(t\) be the number of hours before the hikers meet, and let their speeds be \(v_A\) and \(v_B\). 2. Avery's pre-meeting distance is the distance Blake covers in \(9\) hours after the meeting, so \(v_A t=9v_B\). 3. Blake's pre-meeting distance is the distance Avery covers in \(4\) hours after the meeting, so \(v_B t=4v_A\). 4. The first equation gives \(\frac{v_A}{v_B}=\frac{9}{t}\), while the second gives \(\frac{v_A}{v_B}=\frac{t}{4}\). 5. Set the ratios equal: \(\frac{9}{t}=\frac{t}{4}\), so \(t^2=36\). Since time is positive, \(t=6\) hours. 6. Avery's total time is \(6+4=10\) hours, and Blake's total time is \(6+9=15\) hours.

Avery's total travel time was \(10\) hours, and Blake's was \(15\) hours.

- Convert both remaining times to hours. - Express each side of the meeting point using the person who walks it after the meeting. - Use the equal pre-meeting times to find a speed ratio. - Then use the total route length.

1. Convert the remaining times to hours: Alex walks for \(\frac34\) hour after the meeting, and Brooke walks for \(\frac43\) hours. 2. Let their speeds be \(v_A\) and \(v_B\). The segment from \(A\) to the meeting point has length \(\frac43v_B\), and the segment from the meeting point to \(B\) has length \(\frac34v_A\). 3. Because they traveled for the same amount of time before meeting, \(\frac{\frac43v_B}{v_A}=\frac{\frac34v_A}{v_B}\). 4. Simplify: \(\frac{v_A^2}{v_B^2}=\frac{16}{9}\), so \(v_A=\frac43v_B\) because both speeds are positive. 5. Use the total distance: \(\frac43v_B+\frac34v_A=7\). 6. Substitute \(v_A=\frac43v_B\): \(\frac43v_B+v_B=7\), so \(\frac73v_B=7\) and \(v_B=3\). 7. Therefore, \(v_A=\frac43\cdot3=4\).

Alex walks at \(4\,\text{mph}\), and Brooke walks at \(3\,\text{mph}\).

- Match each train's pre-meeting segment with the segment traveled by the other train afterward. - Express those segments as rate times time. - First find the ratio of the squared speeds. - Equal travel times before the meeting make the distance ratio equal to the speed ratio.

1. Let \(t\) be the common travel time before the meeting. The distances traveled before the meeting are \(d_1=v_1t\) and \(d_2=v_2t\). 2. After the meeting, train 1 travels the segment \(d_2\) in time \(t_1\), so \(d_2=v_1t_1\). Train 2 travels the segment \(d_1\) in time \(t_2\), so \(d_1=v_2t_2\). 3. Substitute into the common-time relationship: \(\frac{v_2t_2}{v_1}=\frac{v_1t_1}{v_2}\). 4. Rearranging gives \(\frac{v_1^2}{v_2^2}=\frac{t_2}{t_1}\). Because speeds are positive, \(\frac{v_1}{v_2}=\sqrt{\frac{t_2}{t_1}}\). 5. For part a), \(\frac{v_1}{v_2}=\sqrt{\frac{4.5}{2}}=\sqrt{2.25}=1.5=\frac32\). Thus, \(v_1:v_2=3:2\). 6. For part b), both trains travel for the same time before meeting, so their distance ratio also is \(3:2\). The faster train travels \(\frac{3}{3+2}=\frac35=60\%\) of the total distance.

a) \(v_1:v_2=3:2\) b) \(60\%\) c) The matching route segments give \(v_1t=v_2t_2\) and \(v_2t=v_1t_1\). Eliminating \(t\) yields \(\frac{v_1^2}{v_2^2}=\frac{t_2}{t_1}\), and positive speeds give the stated square-root relationship.

- Convert the time difference to hours. - Use distance divided by speed for each travel time. - For part b), examine the time-difference expression when both speeds increase but their difference stays fixed. - A fixed speed difference has a smaller effect on travel time at higher speeds.

1. For part a), let \(v\) be the bus's speed in miles per hour. The car's speed is \(v+20\). 2. Convert \(30\) minutes to \(\frac12\) hour and write the time-difference equation: \(\frac{120}{v}-\frac{120}{v+20}=\frac12\). 3. Multiply by \(2v(v+20)\): \(240(v+20)-240v=v(v+20)\). 4. Simplify: \(v^2+20v-4800=0\). 5. Factor: \((v-60)(v+80)=0\), so \(v=60\) or \(v=-80\). 6. Speed must be positive, so the bus travels at \(60\,\text{mph}\), and the car travels at \(80\,\text{mph}\). 7. For part b), with a fixed speed difference of \(20\,\text{mph}\), the time advantage is \(\frac{120}{v}-\frac{120}{v+20}=\frac{2400}{v(v+20)}\). 8. Increasing both speeds increases the denominator while the numerator stays fixed, so the time advantage decreases.

a) The bus travels at \(60\,\text{mph}\), and the car travels at \(80\,\text{mph}\). b) The car's time advantage would decrease.

- Express each post-meeting route segment using speed times time. - The hikers traveled for equal amounts of time before meeting. - Use the \(1\,\text{mph}\) difference to relate their speeds. - Add the two route segments after finding the speeds.

1. Let \(v\) be the slower hiker's speed in miles per hour. The faster hiker's speed is \(v+1\). 2. The segment from \(P\) to the meeting point has length \(4(v+1)\), and the segment from the meeting point to \(Q\) has length \(9v\). 3. The hikers spent the same amount of time walking before they met, so \(\frac{4(v+1)}{v}=\frac{9v}{v+1}\). 4. Cross-multiply: \(4(v+1)^2=9v^2\). 5. Because speeds are positive, \(2(v+1)=3v\), so \(v=2\). The faster speed is \(3\,\text{mph}\). 6. The two route segments are \(4\cdot3=12\) miles and \(9\cdot2=18\) miles. 7. The total distance is \(12+18=30\) miles.

\(30\) miles

- Use the post-meeting times to express the two route segments. - Account for the one-hour difference in starting times. - Introduce a variable for the ratio of the two speeds. - Use the difference between the pre-meeting distances after finding the ratio.

1. Let the walking speeds be \(v_A\) and \(v_B\), in miles per hour. 2. The segment from the meeting point to \(B\) is \(v_A\) miles because Alex walks it in \(1\) hour. The segment from \(A\) to the meeting point is \(6v_B\) miles because Blake walks it in \(6\) hours. 3. Alex's pre-meeting time is \(\frac{6v_B}{v_A}\), and Blake's is \(\frac{v_A}{v_B}\). Since Alex started one hour earlier, \(\frac{6v_B}{v_A}=\frac{v_A}{v_B}+1\). 4. Let \(r=\frac{v_A}{v_B}\). Then \(\frac{6}{r}=r+1\), so \(r^2+r-6=0\). 5. Factor: \((r-2)(r+3)=0\). A speed ratio is positive, so \(r=2\), meaning \(v_A=2v_B\). 6. Alex had walked \(8\) miles farther at the meeting: \(6v_B-v_A=8\). 7. Substitute \(v_A=2v_B\): \(4v_B=8\), so \(v_B=2\,\text{mph}\) and \(v_A=4\,\text{mph}\). 8. The total distance is \(6\cdot2+1\cdot4=16\) miles.

Alex walks at \(4\,\text{mph}\), Blake walks at \(2\,\text{mph}\), and the total distance is \(16\) miles.

- Convert the time difference to hours. - Write a time expression for each segment. - The second speed is \(2\,\text{mph}\) greater than the first. - Check both positive solutions in the original equation.

1. Let \(v\) be the speed on the first segment, in miles per hour. The speed on the second segment is \(v+2\). 2. The travel times are \(\frac{12}{v}\) and \(\frac{18}{v+2}\) hours. 3. Since the second segment takes \(15\) minutes, or \(\frac14\) hour, longer, \(\frac{18}{v+2}-\frac{12}{v}=\frac14\). 4. Multiply by \(4v(v+2)\): \(72v-48(v+2)=v(v+2)\). 5. Simplify: \(v^2-22v+96=0\). 6. Factor: \((v-6)(v-16)=0\), so \(v=6\) or \(v=16\). 7. Both values are positive and satisfy the original time relationship.

The first-segment speed could be \(6\,\text{mph}\) or \(16\,\text{mph}\).

- State the restriction caused by \(y\) in the denominator. - Solve the linear equation for \(x\) and substitute. - Clear the denominators after substitution. - Factor the resulting quadratic in \(y\).

1. Since \(a\ne0\) and \(y\) is in a denominator, \(y\ne0\). 2. From \(x-y=a\), write \(x=y+a\). 3. Substitute: \(\frac{y+a}{a}+\frac{a}{y}=\frac72\), which simplifies to \(\frac{y}{a}+\frac{a}{y}=\frac52\). 4. Multiply by \(2ay\): \(2y^2+2a^2=5ay\), so \(2y^2-5ay+2a^2=0\). 5. Factor: \((2y-a)(y-2a)=0\). Thus, \(y=\frac a2\) or \(y=2a\). 6. Using \(x=y+a\), the ordered pairs are \(\left(\frac{3a}{2},\frac a2\right)\) and \((3a,2a)\). Because \(a\ne0\), both satisfy \(y\ne0\).

Solution set: \(\left\{\left(\frac{3a}{2},\frac a2\right),(3a,2a)\right\}\).

- Express the planned total travel time using the unknown speed. - Find the distance remaining after the first \(2\) hours. - Convert the stop time to hours. - Set the planned time equal to the actual travel-and-stop time.

1. Let \(v\) be the planned speed in miles per hour. The planned total travel time is \(\frac{180}{v}\) hours. 2. In the first \(2\) hours, the train travels \(2v\) miles, leaving \(180-2v\) miles. 3. Convert \(12\) minutes to \(\frac15\) hour. The actual time is \(2+\frac15+\frac{180-2v}{v+15}\). 4. Since the train arrives on time, \(\frac{180}{v}=\frac{11}{5}+\frac{180-2v}{v+15}\). 5. Clear the denominators and simplify: \(v^2+165v-13{,}500=0\). 6. Factor: \((v-60)(v+225)=0\), so \(v=60\) or \(v=-225\). 7. Speed must be positive, so the planned speed was \(60\,\text{mph}\).

\(60\,\text{mph}\)

- Express the total distance using both speeds and the time before the meeting. - The distance hiker B covers before the meeting is the distance hiker A covers afterward. - Use the \(1\,\text{mph}\) speed difference to eliminate one variable. - Check both algebraic solutions against the stated speed restriction.

1. Let the speeds be \(v_A\) and \(v_B\), and let \(t\) be the time before they meet. Then \((v_A+v_B)t=22.5\) and \(v_A=v_B+1\). 2. After the meeting, hiker A covers in \(2\) hours the distance hiker B covered before the meeting. Thus, \(v_Bt=2v_A\), so \(t=\frac{2v_A}{v_B}\). 3. Substitute \(v_B=v_A-1\) into the distance equation: \((2v_A-1)\frac{2v_A}{v_A-1}=22.5\). 4. Clear the denominator and simplify: \(4v_A^2-24.5v_A+22.5=0\). 5. The solutions are \(v_A=5\) and \(v_A=1.125\). 6. Since \(v_B=v_A-1\), the corresponding pairs are \((v_A,v_B)=(5,4)\) and \((1.125,0.125)\). 7. The second pair violates the condition that both speeds exceed \(1\,\text{mph}\). Therefore, \(v_A=5\,\text{mph}\) and \(v_B=4\,\text{mph}\).

Hiker A walks at \(5\,\text{mph}\), and hiker B walks at \(4\,\text{mph}\).

- Express each individual time using the shared time \(x\). - Add the individual filling rates. - Identify the slowest pipe from the individual times. - After removing that pipe, add the two remaining rates and take the reciprocal.

1. The individual filling times are \(x+4\), \(x+10\), and \(2x\) hours. 2. Add the individual rates: \(\frac{1}{x}=\frac{1}{x+4}+\frac{1}{x+10}+\frac{1}{2x}\). 3. Subtract \(\frac{1}{2x}\): \(\frac{1}{2x}=\frac{1}{x+4}+\frac{1}{x+10}\). 4. Multiply by \(2x(x+4)(x+10)\): \((x+4)(x+10)=2x(x+10)+2x(x+4)\). 5. Simplify: \(3x^2+14x-40=0\). 6. Factor: \((3x+20)(x-2)=0\), so \(x=2\) or \(x=-\frac{20}{3}\). 7. Time must be positive, so all three pipes together take \(2\) hours. The individual times are \(6\) hours for A, \(12\) hours for B, and \(4\) hours for C. 8. Pipe B is the slowest. Without B, the combined rate is \(\frac16+\frac14=\frac{5}{12}\) tank per hour. 9. The new filling time is \(\frac{12}{5}\) hours, or \(2\) hours \(24\) minutes.

All three pipes together take \(2\) hours. Pipe A takes \(6\) hours, pipe B takes \(12\) hours, and pipe C takes \(4\) hours alone. Without pipe B, the tank takes \(\frac{12}{5}\) hours, or \(2\) hours \(24\) minutes, to fill.

- Add the current speed downstream and subtract it upstream. - Add the two travel times for part a). - For part b), compare the effect of equal speed increases and decreases on travel time. - The slower leg lasts longer than the faster leg.

1. For part a), let \(v\) be the boat's speed in still water, in miles per hour. The domain requires \(v>4\). 2. The downstream and upstream speeds are \(v+4\) and \(v-4\), so \(\frac{32}{v+4}+\frac{32}{v-4}=6\). 3. Multiply by \((v+4)(v-4)\): \(32(v-4)+32(v+4)=6(v^2-16)\). 4. Simplify: \(3v^2-32v-48=0\). 5. Factor: \((3v+4)(v-12)=0\), so \(v=-\frac43\) or \(v=12\). 6. Only \(v=12\) satisfies the domain, so the still-water speed is \(12\,\text{mph}\). 7. For part b), the boat on the calm lake returns sooner. The speed lost while traveling upstream causes more added time than the equal speed gained downstream removes, because the boat spends longer traveling at the lower speed.

a) \(12\,\text{mph}\) b) The boat on the calm lake returns sooner.

- Identify both denominator restrictions. - Use the difference-of-squares identity after cross-multiplying. - Separate the cases in which the radicand is positive, negative, or zero. - Check what \(x=0\) means in the original equation.

1. The denominators require \(a\ne-5\) and \(x\ne0\). 2. Cross-multiply: \(x^2=(a+5)(a-5)=a^2-25\). 3. If \(a<-5\) or \(a>5\), then \(a^2-25>0\), giving \(x=\pm\sqrt{a^2-25}\). Both values are nonzero. 4. If \(-5<a<5\), then \(a^2-25<0\), so there are no real solutions. 5. If \(a=5\), the only candidate is \(x=0\), which is excluded. The value \(a=-5\) is outside the given parameter domain. 6. Therefore, within the allowed parameter values, there are no real solutions for \(-5<a\le5\).

a) For \(a<-5\) or \(a>5\), the solution set is \(\{-\sqrt{a^2-25},\sqrt{a^2-25}\}\). For \(-5<a\le5\), the solution set is \(\varnothing\). b) The equation has no real solutions for \(-5<a\le5\). The parameter value \(a=-5\) is not allowed.

- Write one equation for the combined hourly rate. - Write another equation for the time spent filling one-third and two-thirds of the tank. - Solve one equation for one individual time and substitute. - Check both positive solution pairs.

1. Let \(t_1\) and \(t_2\) be the individual filling times, in hours. 2. Their combined-rate equation is \(\frac{1}{t_1}+\frac{1}{t_2}=\frac16\). 3. The sequential procedure gives \(\frac13t_1+\frac23t_2=15\), so \(t_1=45-2t_2\). 4. Substitute into the rate equation: \(\frac{1}{45-2t_2}+\frac{1}{t_2}=\frac16\). 5. Clear the denominators and simplify: \(2t_2^2-51t_2+270=0\). 6. Factor: \((2t_2-15)(t_2-18)=0\), so \(t_2=\frac{15}{2}\) or \(t_2=18\). 7. If \(t_2=18\), then \(t_1=9\). If \(t_2=\frac{15}{2}\), then \(t_1=30\). 8. Both pairs satisfy the two original conditions.

There are two possibilities: - Pump 1 takes \(9\) hours and pump 2 takes \(18\) hours. - Pump 1 takes \(30\) hours and pump 2 takes \(7.5\) hours.

- Convert the sequential time to a fraction of an hour. - Use the sequential half-job condition to express one individual time in terms of the other. - Substitute that expression into the combined-rate equation. - Factor the resulting quadratic equation.

1. Convert \(4\) hours \(10\) minutes to \(\frac{25}{6}\) hours. 2. Let \(x\) and \(y\) be the individual times, in hours. The combined-rate equation is \(\frac1x+\frac1y=\frac12\). 3. Printing half the job on each copier sequentially gives \(\frac{x}{2}+\frac{y}{2}=\frac{25}{6}\), so \(x+y=\frac{25}{3}\). Thus, \(y=\frac{25}{3}-x\). 4. Substitute into the combined-rate equation: \(\frac1x+\frac{1}{\frac{25}{3}-x}=\frac12\). 5. Clear the denominators and simplify: \(3x^2-25x+50=0\). 6. Factor: \((3x-10)(x-5)=0\), so \(x=\frac{10}{3}\) or \(x=5\). 7. The corresponding value of \(y\) is \(5\) or \(\frac{10}{3}\). Therefore, one copier takes \(5\) hours, and the other takes \(\frac{10}{3}\) hours, or \(3\) hours \(20\) minutes.

One copier takes \(5\) hours, and the other takes \(3\) hours \(20\) minutes.

- Use the midpoint condition to compare the two travel times for \(40\) miles. - Use the remaining distance after \(2\) hours to find the sum of the speeds. - Substitute one speed into the other equation. - Reject a negative speed.

1. Let the speeds be \(v_A\) and \(v_B\), in miles per hour. 2. To meet at the midpoint, each cyclist travels \(40\) miles. Since cyclist B starts \(2\) hours earlier, \(\frac{40}{v_B}=\frac{40}{v_A}+2\). 3. If they start together, they cover \(80-20=60\) miles in \(2\) hours, so \(2(v_A+v_B)=60\). Thus, \(v_B=30-v_A\). 4. Substitute into the midpoint equation: \(\frac{40}{30-v_A}=\frac{40}{v_A}+2\). 5. Clear the denominators and simplify: \(v_A^2+10v_A-600=0\). 6. Factor: \((v_A-20)(v_A+30)=0\), so \(v_A=20\) or \(v_A=-30\). 7. Speed must be positive, so \(v_A=20\,\text{mph}\). Then \(v_B=30-20=10\,\text{mph}\).

The cyclist from city A rides at \(20\,\text{mph}\), and the cyclist from city B rides at \(10\,\text{mph}\).

- Express each pace as total minutes divided by miles traveled. - Use the one-minute-per-mile difference to form an equation. - After finding both distances to \(Z\), consider the possible orders of the three points on the line. - The starting points may be on the same side or opposite sides of the destination.

1. Let \(x\) be the distance from \(A\) to \(Z\), in miles. Then the distance from \(B\) to \(Z\) is \(x+5\). 2. The cyclists' average paces are \(\frac{30}{x}\) and \(\frac{30}{x+5}\) minutes per mile. 3. The pace difference is \(1\) minute per mile, so \(\frac{30}{x}-\frac{30}{x+5}=1\). 4. Clear the denominators: \(30(x+5)-30x=x(x+5)\). 5. Simplify: \(x^2+5x-150=0\). 6. Factor: \((x-10)(x+15)=0\), so \(x=10\) or \(x=-15\). 7. Distance must be positive, so \(AZ=10\) miles and \(BZ=15\) miles. 8. If \(A\) and \(B\) are on the same side of \(Z\), then \(AB=15-10=5\) miles. If they are on opposite sides of \(Z\), then \(AB=15+10=25\) miles.

The distance between \(A\) and \(B\) could be \(5\) miles or \(25\) miles.

- Their speed sum equals the total distance because they meet after one hour. - Each cyclist's remaining distance equals the distance the other cyclist traveled before the meeting. - Use a variable for the ratio of their speeds. - Convert \(50\) minutes to a fraction of an hour.

1. Because they meet after \(1\) hour, their speeds satisfy \(v_J+v_M=25\). 2. At the meeting point, Julia has traveled \(v_J\) miles and Marcus has traveled \(v_M\) miles. Julia's remaining distance is \(v_M\), while Marcus's remaining distance is \(v_J\). 3. Their remaining travel times are \(\frac{v_M}{v_J}\) and \(\frac{v_J}{v_M}\) hours. 4. Julia arrives \(50\) minutes, or \(\frac56\) hour, later: \(\frac{v_M}{v_J}-\frac{v_J}{v_M}=\frac56\). 5. Let \(r=\frac{v_J}{v_M}\). Then \(\frac1r-r=\frac56\), which gives \(6r^2+5r-6=0\). 6. Factor: \((3r-2)(2r+3)=0\). A speed ratio is positive, so \(r=\frac23\). 7. Thus, \(v_J=\frac23v_M\). Combining this with \(v_J+v_M=25\) gives \(v_M=15\,\text{mph}\) and \(v_J=10\,\text{mph}\).

Julia cycles at \(10\,\text{mph}\), and Marcus cycles at \(15\,\text{mph}\).

- Use the meeting time to find the sum of the speeds. - Each boat's remaining distance is the distance the other boat traveled before the meeting. - Use a speed ratio to simplify the remaining-time difference. - Multiply the slower speed by \(2\) hours for part b).

1. Because the boats cover \(60\) miles together in \(2\) hours, their speeds satisfy \(v_s+v_f=30\), where \(v_s\) is the slower speed and \(v_f\) is the faster speed. 2. Before the meeting, the slower boat travels \(2v_s\) miles and the faster boat travels \(2v_f\) miles. 3. After the meeting, the faster boat must travel \(2v_s\) miles, while the slower boat must travel \(2v_f\) miles. 4. The slower boat's remaining time exceeds the faster boat's by \(3\) hours: \(\frac{2v_f}{v_s}-\frac{2v_s}{v_f}=3\). 5. Let \(r=\frac{v_s}{v_f}\). Then \(\frac{2}{r}-2r=3\), so \(2r^2+3r-2=0\). 6. Factor: \((2r-1)(r+2)=0\). Since the ratio is positive, \(r=\frac12\), so \(v_f=2v_s\). 7. From \(v_s+v_f=30\), \(3v_s=30\), giving \(v_s=10\,\text{mph}\) and \(v_f=20\,\text{mph}\). 8. The slower boat travels \(2\cdot10=20\) miles before the meeting.

a) The boats travel at \(10\,\text{mph}\) and \(20\,\text{mph}\). b) The slower boat has traveled \(20\) miles when they meet.

- State the values excluded by the two denominators. - Clear the denominators and collect the result as a quadratic in \(x\). - Factor the quadratic using \(a+b\) as a repeated expression. - Check when a candidate is excluded or when the two candidates coincide.

1. The denominators require \(x\ne a\) and \(x\ne b\). 2. Multiply by \((x-a)(x-b)\): \(a(x-a)+b(x-b)=2(x-a)(x-b)\). 3. Expand and rearrange: \(2x^2-3(a+b)x+(a+b)^2=0\). 4. Factor: \([2x-(a+b)][x-(a+b)]=0\). The candidates are \(x=a+b\) and \(x=\frac{a+b}{2}\). 5. Since \(a\ne0\) and \(b\ne0\), the value \(x=a+b\) cannot equal \(a\) or \(b\), so it is always valid. 6. The value \(x=\frac{a+b}{2}\) equals an excluded value exactly when \(a=b\). In that case, only \(x=2a\) remains. 7. If \(a\ne b\) and \(a+b=0\), the two candidates coincide at \(x=0\), giving one valid solution. 8. If \(a\ne b\) and \(a+b\ne0\), the two candidates are distinct and valid.

Restrictions: \(x\ne a\) and \(x\ne b\). If \(a=b\), the solution set is \(\{2a\}\). If \(a\ne b\) and \(a+b=0\), the solution set is \(\{0\}\). If \(a\ne b\) and \(a+b\ne0\), the solution set is \(\left\{\frac{a+b}{2},a+b\right\}\).