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) \(x=-8\)

- 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.

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

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

- 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.

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

- 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.

- 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 \(x=-2\)

- Use \(\text{time}=\frac{\text{distance}}{\text{speed}}\). - Express Ethan’s speed in terms of Henry’s speed. - Set the two running times equal.

1. Let Henry’s speed be \(v\,\text{mph}\). Then Ethan’s speed is \((v+1)\,\text{mph}\). 2. Equal running times give \(\frac{7}{v+1}=\frac{6}{v}\). 3. Cross-multiply: \(7v=6(v+1)\). 4. Solve: \(7v=6v+6\), so \(v=6\). 5. Both speeds are positive, and neither denominator is zero, so the solution is valid.

b) \(6\,\text{mph}\)

- 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 \((2,-1)\)

- 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: \(\{-5,2\}\).

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

- 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

- 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 \(\varnothing\).

- 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. - Check the graph intersections.

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. 5. The graph shows intersections at \((-7,-2)\) and \((0,5)\), confirming the solutions.

a) \(D=\mathbb{R}\setminus\{-2\}\) b) Solution set: \(\{-7,0\}\). c) The intersections \((-7,-2)\) and \((0,5)\) confirm the algebraic solutions.

- 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 \(x=\frac43\). 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 the student claim is false.

a) \(x=\frac43\) 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.

- 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

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

- 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.

- 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 \((0,1)\)

- 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 \(\left\{\frac{a+b}{2},a+b\right\}\).