Free Math Worksheets

- Find the least common denominator. - Rewrite both fractions with that denominator. - Distribute the subtraction sign carefully.

1. The least common denominator is \(a(a + 2)\). 2. Rewrite both fractions: \(\frac{2a}{a(a + 2)} - \frac{a + 2}{a(a + 2)}\). 3. Subtract the numerators: \(2a - (a + 2) = a - 2\). 4. Therefore, the difference is \(\frac{a - 2}{a(a + 2)} = \frac{a - 2}{a^2 + 2a}\). 5. The original denominators require \(a \ne 0\) and \(a \ne -2\).

c) \(\frac{a - 2}{a^2 + 2a}\)

- Can fractions with unlike denominators be added directly? - Find the least common multiple of \(2\) and \(5\). - Rewrite both fractions with the same denominator.

1. The student incorrectly added the numerators and denominators. Fractions with different denominators cannot be added that way. 2. The least common denominator of \(2\) and \(5\) is \(10\). 3. Rewrite each fraction: \(\frac{x}{2} = \frac{5x}{10}\) and \(\frac{x}{5} = \frac{2x}{10}\). 4. Add the numerators while keeping the common denominator: \(\frac{5x}{10} + \frac{2x}{10} = \frac{7x}{10}\).

The student added the numerators and denominators directly instead of first using a common denominator. The correct sum is \(\frac{7x}{10}\).

- Compare the two denominators. - Recall the rule for subtracting fractions with unlike denominators. - Determine what factor changes \(x\) into \(2x\).

1. Approach A is incorrect. When subtracting fractions, the denominators are not subtracted, and the numerators cannot be combined until the fractions have a common denominator. 2. Approach B is correct. The least common denominator is \(2x\). 3. Rewrite the second fraction: \(\frac{2}{x} = \frac{4}{2x}\). 4. Subtract: \(T = \frac{5}{2x} - \frac{4}{2x} = \frac{1}{2x}\). 5. The original expression requires \(x \ne 0\).

Approach A is incorrect, and Approach B is correct. \(T = \frac{1}{2x}\), with \(x \ne 0\).

- Add numerators directly only when the denominators are equal. - For unlike denominators, first find a common denominator. - Look for a common factor in the numerator.

1. For part a, the denominators are already equal, so add the numerators: \(\frac{7 + 2}{x} = \frac{9}{x}\), with \(x \ne 0\). 2. For part b, use the common denominator \(2y\): \(\frac{5}{2y} - \frac{2}{2y} = \frac{3}{2y}\), with \(y \ne 0\). 3. For part c, factor the numerator: \(\frac{3(x + 2)}{3} = x + 2\). The denominator is the nonzero constant \(3\), so there is no variable restriction.

a) \(\frac{9}{x}\), with \(x \ne 0\). b) \(\frac{3}{2y}\), with \(y \ne 0\). c) \(x + 2\); no restriction.

- Determine the factor that changes the given numerator or denominator into the target. - Multiply the entire numerator and denominator by the same nonzero factor.

1. For part a, the missing factor is \(\frac{12y^2}{3y} = 4y\). Multiply the numerator and denominator by \(4y\): \(\frac{5 \cdot 4y}{3y \cdot 4y} = \frac{20y}{12y^2}\). 2. For part b, the missing factor is \(\frac{10y^3}{5} = 2y^3\). Multiply the numerator and denominator by \(2y^3\): \(\frac{5 \cdot 2y^3}{3y \cdot 2y^3} = \frac{10y^3}{6y^4}\). 3. The assumption \(y \ne 0\) ensures that the original denominator and both variable multipliers are nonzero.

a) \(\frac{20y}{12y^2}\). b) \(\frac{10y^3}{6y^4}\). Both are valid for \(y \ne 0\).

- Multiplying the two denominators always produces a common denominator. - For each expression, identify the factor missing from its denominator. - Use parentheses when multiplying an entire binomial numerator or denominator.

1. Multiply the denominators: \(x(x + 2) = x^2 + 2x\). 2. Multiply the first expression by \(\frac{x}{x}\): \(\frac{3}{x + 2} = \frac{3x}{x(x + 2)} = \frac{3x}{x^2 + 2x}\). 3. Multiply the second expression by \(\frac{x + 2}{x + 2}\): \(\frac{2}{x} = \frac{2(x + 2)}{x(x + 2)} = \frac{2x + 4}{x^2 + 2x}\).

a) A common denominator is \(x(x + 2) = x^2 + 2x\). b) \(\frac{3x}{x^2 + 2x}\) and \(\frac{2x + 4}{x^2 + 2x}\), for \(x \ne -2, 0\).

- Split a fraction by dividing each term in the numerator by the denominator. - A nonzero quantity divided by itself equals \(1\). - Isolate the missing expression as you would in an equation.

1. Split the numerator in part a: \(\frac{5x + 2}{x} = \frac{5x}{x} + \frac{2}{x} = 5 + \frac{2}{x}\). 2. Split the numerator in part b: \(\frac{a - 4}{a} = \frac{a}{a} - \frac{4}{a} = 1 - \frac{4}{a}\). 3. In part c, subtract \(\frac{3}{y}\) from both sides: \(\frac{3 + y}{y} - \frac{3}{y} = \frac{y}{y} = 1\).

a) \(\frac{2}{x}\) b) \(\frac{4}{a}\) c) \(1\)

- Rewrite \(1\) as a fraction with the needed denominator. - When the denominators match, compare or combine the numerators. - Isolate the box before simplifying.

1. In part a, write \(1 = \frac{x - 1}{x - 1}\). Comparing numerators gives \(x + 3 = x - 1 + \Box\), so \(\Box = 4\). The restriction is \(x \ne 1\). 2. In part b, isolate the box: \(\Box = \frac{5}{3a} - \frac{2}{3a} = \frac{3}{3a} = \frac{1}{a}\). The restriction is \(a \ne 0\). 3. In part c, split the fraction: \(\frac{k^2 + 1}{k^2} = \frac{k^2}{k^2} + \frac{1}{k^2} = 1 + \frac{1}{k^2}\). The restriction is \(k \ne 0\).

a) \(4\), with \(x \ne 1\). b) \(\frac{1}{a}\), with \(a \ne 0\). c) \(\frac{1}{k^2}\), with \(k \ne 0\).

- Rewrite a whole number as a fraction with the needed denominator. - Find a least common denominator before subtracting. - Use parentheses when subtracting an entire expression.

1. In part a, write \(1 = \frac{a}{a}\): \(\frac{b}{a} + \frac{a}{a} = \frac{a + b}{a}\), with \(a \ne 0\). 2. In part b, use denominator \(2x\): \(\frac{4}{2x} - \frac{1}{2x} = \frac{3}{2x}\), with \(x \ne 0\). 3. In part c, write \(1 = \frac{x - 1}{x - 1}\): \(\frac{x}{x - 1} - \frac{x - 1}{x - 1} = \frac{x - (x - 1)}{x - 1} = \frac{1}{x - 1}\), with \(x \ne 1\).

a) \(a + b\), with \(a \ne 0\). b) \(\frac{3}{2x}\), with \(x \ne 0\). c) \(\frac{1}{x - 1}\), with \(x \ne 1\).

- Use the product of the two distinct denominators as the least common denominator. - Distribute the subtraction sign to the entire second numerator. - Check whether the final numerator and denominator share a factor.

1. The least common denominator is \((x + 2)(x - 1)\). 2. Rewrite and combine: \(\frac{2(x - 1) - 3(x + 2)}{(x + 2)(x - 1)}\). 3. Simplify the numerator: \(2x - 2 - 3x - 6 = -x - 8\). 4. No factor cancels, so the result is \(-\frac{x + 8}{(x + 2)(x - 1)}\). 5. The original restrictions are \(x \ne -2\) and \(x \ne 1\).

\(-\frac{x + 8}{(x + 2)(x - 1)}\), with \(x \ne -2\) and \(x \ne 1\).

- Factor before multiplying whenever possible. - Cancel common factors across the product. - Preserve restrictions from every original denominator.

1. In part a, multiply and cancel common factors: \(\frac{90x^2y^2}{15xy} = 6xy\). The restrictions are \(x \ne 0\) and \(y \ne 0\). 2. In part b, factor \(x^2 - 16 = (x - 4)(x + 4)\). 3. Cancel \(x + 4\) and one factor of \(x\): \(\frac{(x + 4)x^2}{x(x - 4)(x + 4)} = \frac{x}{x - 4}\). 4. Preserve the original restrictions \(x \ne -4\), \(x \ne 0\), and \(x \ne 4\).

a) \(6xy\), with \(x \ne 0\) and \(y \ne 0\). b) \(\frac{x}{x - 4}\), with \(x \ne -4, 0, 4\).

- Factor the second numerator. - Write the product as one fraction. - Cancel only common factors.

1. Factor \(2x + 4 = 2(x + 2)\). 2. Substitute the factorization: \(\frac{5x \cdot 2(x + 2)}{(x + 2)10x^2}\). 3. Cancel \(x + 2\), reduce \(10\), and cancel one factor of \(x\) to obtain \(\frac{1}{x}\). 4. Preserve the original restrictions \(x \ne -2\) and \(x \ne 0\).

\(\frac{1}{x}\), with \(x \ne -2\) and \(x \ne 0\).

- Look for perfect-square trinomials and differences of squares. - Rewrite division as multiplication by a reciprocal. - Cancel only common factors, not terms in a sum.

1. For a), factor \(s^2+4s+4=(s+2)^2\) and \(s^2-4=(s-2)(s+2)\). 2. Substitute and cancel common factors: \(\frac{s-2}{s+2}\cdot\frac{(s+2)^2}{(s-2)(s+2)}=1\). 3. The original denominators require \(s\ne-2\) and \(s\ne2\). 4. For b), factor \(x^2-y^2=(x-y)(x+y)\). 5. Rewrite the division as multiplication by the reciprocal: \(\frac{(x-y)(x+y)}{xy}\cdot\frac{1}{x+y}\). 6. Cancel \(x+y\) to get \(\frac{x-y}{xy}\). The original expression requires \(x\ne0\), \(y\ne0\), and \(x\ne-y\).

a) \(1\), where \(s\ne-2,2\) b) \(\frac{x-y}{xy}\), where \(x\ne0\), \(y\ne0\), and \(x\ne-y\)

- Multiply numerators and denominators, then cancel common factors. - Rewrite division as multiplication by the reciprocal. - Factor out the greatest common factor when possible.

1. For a), multiply and cancel common factors: \(\frac{2xy}{6x^2y^2}=\frac{1}{3xy}\). 2. The original expression requires \(x\ne0\) and \(y\ne0\). 3. For b), multiply by the reciprocal: \(\frac{a+b}{c}\cdot\frac{c^2}{4a+4b}\). 4. Factor \(4a+4b=4(a+b)\), then cancel \(a+b\) and one factor of \(c\): \(\frac{(a+b)c^2}{4c(a+b)}=\frac{c}{4}\). 5. The original expression requires \(c\ne0\) and \(a+b\ne0\).

a) \(\frac{1}{3xy}\), where \(x\ne0\) and \(y\ne0\) b) \(\frac{c}{4}\), where \(c\ne0\) and \(a+b\ne0\)

- Factor the difference of squares first. - Write the product as one fraction before canceling. - Keep the restrictions from the original denominators.

1. Factor the difference of squares: \(x^2-16=(x-4)(x+4)\). 2. Substitute: \(\frac{(x-4)(x+4)}{5x}\cdot\frac{10}{x-4}\). 3. Cancel \(x-4\) and simplify \(\frac{10}{5}=2\): \(\frac{2(x+4)}{x}\). 4. The original expression requires \(x\ne0\) and \(x\ne4\).

\(\frac{2(x+4)}{x}\), where \(x\ne0,4\)

- Find the least common denominator first. - Place parentheses around the entire second numerator when subtracting. - Simplify the numerator only after both fractions have the same denominator.

1. For part a, the least common denominator is \(6x\): \(\frac{9}{6x} + \frac{10}{6x} = \frac{19}{6x}\), with \(x \ne 0\). 2. For part b, the least common denominator is \(x(x - 3)\): \(\frac{x^2 - 2(x - 3)}{x(x - 3)} = \frac{x^2 - 2x + 6}{x(x - 3)}\). 3. The numerator \(x^2 - 2x + 6\) shares no factor with \(x(x - 3)\). The restrictions are \(x \ne 0\) and \(x \ne 3\).

a) \(\frac{19}{6x}\), with \(x \ne 0\). b) \(\frac{x^2 - 2x + 6}{x(x - 3)}\), with \(x \ne 0\) and \(x \ne 3\).

- Use the product of the distinct denominator factors as the least common denominator. - Multiply each entire numerator by the missing factor. - Combine like terms and check for cancellation.

1. The least common denominator is \(2x(x - 1)\). 2. Rewrite both fractions: \(\frac{5(x - 1)}{2x(x - 1)} + \frac{4x}{2x(x - 1)}\). 3. Add the numerators: \(5x - 5 + 4x = 9x - 5\). 4. No factor cancels, so the result is \(\frac{9x - 5}{2x(x - 1)}\). 5. The original restrictions are \(x \ne 0\) and \(x \ne 1\).

\(\frac{9x - 5}{2x(x - 1)}\), with \(x \ne 0\) and \(x \ne 1\).

- Identify which denominator already contains the other as a factor. - Rewrite only the fraction that needs a larger denominator. - Once denominators match, combine the numerators.

1. In part a, the least common denominator is \(2x\): \(\frac{10}{2x} + \frac{3}{2x} = \frac{13}{2x}\). 2. In part b, the least common denominator is \(3a\): \(\frac{7}{3a} - \frac{6}{3a} = \frac{1}{3a}\). 3. In part c, the least common denominator is \(b^2\): \(\frac{1}{b^2} + \frac{b}{b^2} = \frac{b + 1}{b^2}\).

a) \(\frac{13}{2x}\) b) \(\frac{1}{3a}\) c) \(\frac{b + 1}{b^2}\)

- Cancel common numerical and variable factors. - Rewrite division as multiplication by the reciprocal. - Apply the order of operations in part c).

1. For a), multiply and simplify: \(\frac{4x\cdot15}{5\cdot2x^2}=\frac{6}{x}\). The original expression requires \(x\ne0\). 2. For b), multiply by the reciprocal: \(\frac{a^2}{b}\cdot\frac{b^2}{a}=ab\). The original expression requires \(a\ne0\) and \(b\ne0\). 3. For c), multiply before subtracting: \(\frac{3}{y}-\frac{4}{y}=-\frac{1}{y}\). The original expression requires \(y\ne0\).

a) \(\frac{6}{x}\), where \(x\ne0\) b) \(ab\), where \(a\ne0\) and \(b\ne0\) c) \(-\frac{1}{y}\), where \(y\ne0\)

- Cancel matching factors in the numerator and denominator. - Rewrite \(5-z\) by factoring out \(-1\). - Avoid expanding before you simplify.

1. For a), combine the factors: \(\frac{4x(x+3)}{8x^2(x+3)}\). 2. Cancel common factors to get \(\frac{1}{2x}\). The original expression requires \(x\ne-3,0\). 3. For b), use \(5-z=-(z-5)\): \(\frac{6(z-5)}{3(5-z)}=\frac{6(z-5)}{-3(z-5)}=-2\). 4. The original expression requires \(z\ne5\).

a) \(\frac{1}{2x}\), where \(x\ne-3,0\) b) \(-2\), where \(z\ne5\)

- Factor a common \(x\) from the first denominator. - Check whether the first fraction can be reduced before adding. - Preserve restrictions from the original denominators.

1. Factor the first denominator: \(x^2 - 4x = x(x - 4)\). 2. Reduce the first fraction: \(\frac{x}{x(x - 4)} = \frac{1}{x - 4}\), while preserving \(x \ne 0\). 3. Add the like-denominator fractions: \(\frac{1}{x - 4} + \frac{2}{x - 4} = \frac{3}{x - 4}\). 4. The original restrictions are \(x \ne 0\) and \(x \ne 4\).

\(\frac{3}{x - 4}\), with \(x \ne 0\) and \(x \ne 4\).

- Rewrite division as multiplication by the reciprocal. - Factor out common factors before canceling. - Cancel factors, not terms in a sum.

1. For a), multiply and cancel common factors: \(\frac{12a^2b^2}{4ab}=3ab\). 2. The original expression requires \(a\ne0\) and \(b\ne0\). 3. For b), multiply by the reciprocal: \(\frac{x+5}{3y}\cdot\frac{6y^2}{2x+10}\). 4. Factor \(2x+10=2(x+5)\), then cancel common factors: \(\frac{6y^2(x+5)}{6y(x+5)}=y\). 5. The original expression requires \(y\ne0\) and \(x\ne-5\).

a) \(3ab\), where \(a\ne0\) and \(b\ne0\) b) \(y\), where \(y\ne0\) and \(x\ne-5\)

- Rewrite each division as multiplication by a reciprocal. - Cancel common factors in the numerator and denominator. - Perform multiplication and division from left to right.

1. For a), multiply by the reciprocal: \(\frac{4x^2}{y}\cdot\frac{y^2}{2x}\). 2. Cancel common factors to get \(2xy\). The original expression requires \(x\ne0\) and \(y\ne0\). 3. For b), first multiply the first two rational expressions: \(\frac{a}{b^2}\cdot\frac{b}{a^2}=\frac{1}{ab}\). 4. Then divide by \(\frac{1}{ab}\): \(\frac{1}{ab}\cdot ab=1\). 5. The original expression requires \(a\ne0\) and \(b\ne0\).

a) \(2xy\), where \(x\ne0\) and \(y\ne0\) b) \(1\), where \(a\ne0\) and \(b\ne0\)

- The squared denominator is already a multiple of the other denominator. - Multiply the entire first numerator by \(x - 4\). - Factor the final numerator and check for cancellation.

1. The least common denominator is \((x - 4)^2\). 2. Rewrite the first fraction: \(\frac{5(x - 4)}{(x - 4)^2}\). 3. Add the numerators: \(5(x - 4) + x + 2 = 6x - 18 = 6(x - 3)\). 4. No factor cancels with the denominator, so the result is \(\frac{6(x - 3)}{(x - 4)^2}\). 5. The original restriction is \(x \ne 4\).

\(\frac{6(x - 3)}{(x - 4)^2}\), with \(x \ne 4\).

- Find the least common multiple of the numerical coefficients in the denominators. - Multiply each entire numerator by the needed factor. - Distribute the subtraction sign carefully.

1. The least common denominator is \(6x\). 2. Combine: \(\frac{3(x + 5) - 2(x - 1)}{6x}\). 3. Simplify the numerator: \(3x + 15 - 2x + 2 = x + 17\). 4. The result is \(\frac{x + 17}{6x}\), with \(x \ne 0\).

\(\frac{x + 17}{6x}\), with \(x \ne 0\).

- Rewrite the whole number using the same denominator as the fraction. - Distribute the subtraction sign through the entire second numerator. - A fraction equals zero when its numerator is zero and its denominator is nonzero. - Check the solution against the domain.

1. Rewrite \(2\) using the denominator \(x+2\): \(f(x)=\frac{3x-1}{x+2}-\frac{2(x+2)}{x+2}\). 2. Combine the numerators: \(f(x)=\frac{3x-1-2x-4}{x+2}=\frac{x-5}{x+2}\). 3. Set the numerator equal to zero: \(x-5=0\), so \(x=5\). 4. Since \(x=5\) is in the domain, the x-intercept is \((5, 0)\).

a) \(f(x)=\frac{x-5}{x+2}\) b) \((5, 0)\)

- Factor the perfect-square trinomial and the difference of squares. - Multiply before canceling common factors. - Preserve restrictions from the original expression.

1. Factor \(y^2+6y+9=(y+3)^2\) and \(y^2-16=(y-4)(y+4)\). 2. Substitute: \(\frac{(y+3)^2(y-4)}{(y-4)(y+4)(y+3)}\). 3. Cancel \(y-4\) and one factor of \(y+3\) to get \(\frac{y+3}{y+4}\). 4. The original expression requires \(y\ne-4,-3,4\).

\(\frac{y+3}{y+4}\), where \(y\ne-4,-3,4\)

- With equal denominators, combine the numerators directly. - Look for a difference of squares. - Distribute a subtraction sign to the entire second numerator.

1. In part a, add the numerators: \(\frac{4z - 5 + 5z + 14}{9} = \frac{9z + 9}{9} = z + 1\). There is no variable restriction. 2. In part b, subtract and factor: \(\frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4} = x + 4\), with \(x \ne 4\). 3. In part c, subtract the numerators: \(\frac{3a + 7 - (a - 1)}{2a} = \frac{2a + 8}{2a} = \frac{a + 4}{a}\), with \(a \ne 0\).

a) \(z + 1\); no restriction. b) \(x + 4\), with \(x \ne 4\). c) \(\frac{a + 4}{a}\), with \(a \ne 0\).

- Reversing the order of subtraction in a denominator changes its sign. - Rewrite the denominators so they match. - Track the sign outside each fraction carefully.

1. In part a, \(v - u = -(u - v)\), so \(\frac{5z}{v - u} = -\frac{5z}{u - v}\). 2. Thus, \(\frac{8z}{u - v} - \frac{5z}{u - v} = \frac{3z}{u - v}\), with \(u \ne v\). 3. In part b, \(1 - 2x = -(2x - 1)\), so subtracting the second fraction becomes adding \(\frac{k}{2x - 1}\). 4. Therefore, \(\frac{3k}{2x - 1} + \frac{k}{2x - 1} = \frac{4k}{2x - 1}\), with \(x \ne \frac{1}{2}\).

a) \(\frac{3z}{u - v}\), with \(u \ne v\). b) \(\frac{4k}{2x - 1}\), with \(x \ne \frac{1}{2}\).

- With equal denominators, combine only the numerators. - Use parentheses around a numerator being subtracted. - Factor the combined numerator before canceling.

1. In part a, subtract the numerators: \((8a - 5) - (3a - 5) = 5a\). The result is \(\frac{5a}{x}\), with \(x \ne 0\). 2. In part b, add the numerators: \((y + 7) + (y - 3) = 2y + 4 = 2(y + 2)\). 3. Cancel \(y + 2\) to obtain \(2\), with \(y \ne -2\).

a) \(\frac{5a}{x}\), with \(x \ne 0\). b) \(2\), with \(y \ne -2\).

- Reversing the order in a subtraction denominator changes its sign. - Rewrite the denominators so they match. - Use parentheses around any numerator being subtracted.

1. In part a, \(4 - x = -(x - 4)\), so the expression becomes \(\frac{5x - 2}{x - 4} - \frac{2x + 1}{x - 4}\). 2. Combine the numerators: \(5x - 2 - (2x + 1) = 3x - 3\). Thus, the result is \(\frac{3x - 3}{x - 4}\), with \(x \ne 4\). 3. In part b, \(b - a = -(a - b)\), so subtracting the second fraction becomes adding \(\frac{2a + b}{a - b}\). 4. Add the numerators: \(a + 3b + 2a + b = 3a + 4b\). The result is \(\frac{3a + 4b}{a - b}\), with \(a \ne b\).

a) \(\frac{3x - 3}{x - 4}\), with \(x \ne 4\). b) \(\frac{3a + 4b}{a - b}\), with \(a \ne b\).

- Reverse-order subtraction expressions differ by a factor of \(-1\). - Track the sign before each fraction carefully. - Factor the combined numerator when possible.

1. In part a, \(2 - m = -(m - 2)\), so \(\frac{8}{2 - m} = -\frac{8}{m - 2}\). 2. Combine and factor: \(\frac{4m - 8}{m - 2} = \frac{4(m - 2)}{m - 2} = 4\), with \(m \ne 2\). 3. In part b, \(y - x = -(x - y)\), so \(\frac{2y}{y - x} = -\frac{2y}{x - y}\). 4. Subtracting that fraction gives \(\frac{x + y}{x - y} + \frac{2y}{x - y} = \frac{x + 3y}{x - y}\), with \(x \ne y\).

a) \(4\), with \(m \ne 2\). b) \(\frac{x + 3y}{x - y}\), with \(x \ne y\).

- Find the least common multiple of the numerical factors. - Include the highest power of each variable factor. - Multiply the entire numerator by the same factor used on the denominator.

1. The least common denominator is \(6x\): \(\frac{8}{6x} + \frac{3}{6x} = \frac{11}{6x}\), with \(x \ne 0\). 2. The least common denominator is \(12z\): \(\frac{10y}{12z} - \frac{3y}{12z} = \frac{7y}{12z}\), with \(z \ne 0\). 3. The least common denominator is \(a^2\): \(\frac{2}{a^2} + \frac{3a}{a^2} = \frac{3a + 2}{a^2}\), with \(a \ne 0\). 4. The least common denominator is \(10x\): \(\frac{2(x - 2)}{10x} + \frac{x}{10x} = \frac{3x - 4}{10x}\), with \(x \ne 0\).

1) LCD: \(6x\); \(\frac{11}{6x}\), with \(x \ne 0\). 2) LCD: \(12z\); \(\frac{7y}{12z}\), with \(z \ne 0\). 3) LCD: \(a^2\); \(\frac{3a + 2}{a^2}\), with \(a \ne 0\). 4) LCD: \(10x\); \(\frac{3x - 4}{10x}\), with \(x \ne 0\).

- Find the least common multiple of the numerical denominators. - Multiply each entire numerator by the needed factor. - Distribute the subtraction sign across the second numerator.

1. In part a, the least common denominator is \(30\): \(\frac{3(3z - 5) + 2(z + 4)}{30} = \frac{9z - 15 + 2z + 8}{30} = \frac{11z - 7}{30}\). 2. In part b, the least common denominator is \(18\): \(\frac{3(7k + 2) - 2(2k - 3)}{18} = \frac{21k + 6 - 4k + 6}{18} = \frac{17k + 12}{18}\).

a) \(\frac{11z - 7}{30}\) b) \(\frac{17k + 12}{18}\)

- Find the least common multiple of the numerical factors. - Keep the variable factors in the common denominator. - Reduce the final numerical coefficient when possible.

1. In part a, the least common denominator is \(24ab\): \(\frac{9}{24ab} + \frac{10}{24ab} = \frac{19}{24ab}\). The restrictions are \(a \ne 0\) and \(b \ne 0\). 2. In part b, use denominator \(12y\): \(\frac{5x}{12y} + \frac{3x}{12y} = \frac{8x}{12y} = \frac{2x}{3y}\), with \(y \ne 0\).

a) \(\frac{19}{24ab}\), with \(a \ne 0\) and \(b \ne 0\). b) \(\frac{2x}{3y}\), with \(y \ne 0\).

- Include the highest variable power in the least common denominator. - Use parentheses around every multi-term numerator. - Do not cancel terms across addition or subtraction.

1. In part a, the least common denominator is \(12x^2\): \(\frac{8}{12x^2} + \frac{9x}{12x^2} = \frac{9x + 8}{12x^2}\), with \(x \ne 0\). 2. In part b, the least common denominator is \(12a\): \(\frac{2(a + 3) - 3(a - 2)}{12a} = \frac{2a + 6 - 3a + 6}{12a} = \frac{12 - a}{12a}\), with \(a \ne 0\).

a) \(\frac{9x + 8}{12x^2}\), with \(x \ne 0\). b) \(\frac{12 - a}{12a}\), with \(a \ne 0\).

- Build the least common denominator from every required factor. - Include the highest power of each variable. - Multiply each numerator by the factors missing from its denominator.

1. The least common denominator is \(abc\): \(\frac{4c}{abc} + \frac{3a}{abc} = \frac{4c + 3a}{abc}\). The restrictions are \(a \ne 0\), \(b \ne 0\), and \(c \ne 0\). 2. The least common denominator is \(xy^2\): \(\frac{x^2}{xy^2} - \frac{y}{xy^2} = \frac{x^2 - y}{xy^2}\). The restrictions are \(x \ne 0\) and \(y \ne 0\). 3. The least common denominator is \(18mnp\): \(\frac{15p}{18mnp} + \frac{4n}{18mnp} = \frac{15p + 4n}{18mnp}\). The restrictions are \(m \ne 0\), \(n \ne 0\), and \(p \ne 0\).

1) \(\frac{4c + 3a}{abc}\), with \(a, b, c \ne 0\). 2) \(\frac{x^2 - y}{xy^2}\), with \(x, y \ne 0\). 3) \(\frac{15p + 4n}{18mnp}\), with \(m, n, p \ne 0\).

- Check whether one denominator is already a multiple of the other. - Multiply the entire numerator by the missing factor. - Preserve every zero of an original denominator.

1. Use denominator \(a^2\): \(\frac{2a + 1 - 2a}{a^2} = \frac{1}{a^2}\), with \(a \ne 0\). 2. Use denominator \(x(x + 1)\): \(\frac{3 + 2x}{x(x + 1)} = \frac{2x + 3}{x(x + 1)}\), with \(x \ne 0\) and \(x \ne -1\).

1) \(\frac{1}{a^2}\), with \(a \ne 0\). 2) \(\frac{2x + 3}{x(x + 1)}\), with \(x \ne 0\) and \(x \ne -1\).

- Use the highest power of each variable in the common denominator. - Multiply a numerator by the same factor used on its denominator. - Combine only after the denominators match.

1. Use the least common denominator \(2x^2\): \(\frac{5}{2x^2} + \frac{6x}{2x^2} = \frac{6x + 5}{2x^2}\), with \(x \ne 0\). 2. Use the least common denominator \(w^3\): \(\frac{4z}{w^3} - \frac{w}{w^3} = \frac{4z - w}{w^3}\), with \(w \ne 0\).

1) \(\frac{6x + 5}{2x^2}\), with \(x \ne 0\). 2) \(\frac{4z - w}{w^3}\), with \(w \ne 0\).

- Use the least common multiple of the numerical coefficients. - Include the highest power of each variable factor. - Keep a multi-term numerator in parentheses while multiplying.

1. The least common denominator is \(6b^2c^2\). Rewrite and subtract: \(\frac{4ac - 5b}{6b^2c^2}\). The restrictions are \(b \ne 0\) and \(c \ne 0\). 2. The least common denominator is \(x^2y^2\). Rewrite and add: \(\frac{y(x - 1) + 3x}{x^2y^2} = \frac{xy - y + 3x}{x^2y^2}\). The restrictions are \(x \ne 0\) and \(y \ne 0\).

1) \(\frac{4ac - 5b}{6b^2c^2}\), with \(b \ne 0\) and \(c \ne 0\). 2) \(\frac{xy - y + 3x}{x^2y^2}\), with \(x \ne 0\) and \(y \ne 0\).

- Identify the least common denominator from the variable factors. - Multiply every term in a numerator by the required factor. - Combine like terms after expanding.

1. In part a, use denominator \(x^2\): \(\frac{4x + 1 - 3x}{x^2} = \frac{x + 1}{x^2}\), with \(x \ne 0\). 2. In part b, use denominator \(ab^2\): \(\frac{b(2a - b) + a(a + 2b)}{ab^2}\). 3. Expand and combine: \(2ab - b^2 + a^2 + 2ab = a^2 + 4ab - b^2\). 4. The result is \(\frac{a^2 + 4ab - b^2}{ab^2}\), with \(a \ne 0\) and \(b \ne 0\).

a) \(\frac{x + 1}{x^2}\), with \(x \ne 0\). b) \(\frac{a^2 + 4ab - b^2}{ab^2}\), with \(a \ne 0\) and \(b \ne 0\).

- Find the least common multiple of \(15\) and \(10\). - Multiply each entire numerator by the required factor. - Distribute the subtraction sign carefully.

1. The least common denominator is \(30\). 2. Rewrite and combine: \(\frac{2(3x + 2y) - 3(x - y)}{30}\). 3. Simplify the numerator: \(6x + 4y - 3x + 3y = 3x + 7y\). 4. The result is \(\frac{3x + 7y}{30}\).

\(\frac{3x + 7y}{30}\)

- Use the least common multiple of the numerical factors. - Include the highest power of each variable in the common denominator. - Check whether the final numerator and denominator share a factor.

1. In part a, the least common denominator is \(18x^2\): \(\frac{15}{18x^2} + \frac{4x}{18x^2} = \frac{4x + 15}{18x^2}\), with \(x \ne 0\). 2. In part b, the least common denominator is \(20ab\): \(\frac{5a^2}{20ab} - \frac{2b^2}{20ab} = \frac{5a^2 - 2b^2}{20ab}\), with \(a \ne 0\) and \(b \ne 0\).

a) \(\frac{4x + 15}{18x^2}\), with \(x \ne 0\). b) \(\frac{5a^2 - 2b^2}{20ab}\), with \(a \ne 0\) and \(b \ne 0\).

- Use the highest powers of \(a\) and \(b\) in the common denominator. - Multiply each entire numerator by its missing factor. - Distribute the subtraction sign to every term in the second product.

1. The least common denominator is \(a^2b^2\). 2. Rewrite: \(\frac{b(5a - 2)}{a^2b^2} - \frac{a(3b + 4)}{a^2b^2}\). 3. Subtract and expand: \(5ab - 2b - 3ab - 4a = 2ab - 4a - 2b\). 4. The result is \(\frac{2ab - 4a - 2b}{a^2b^2}\), with \(a \ne 0\) and \(b \ne 0\).

\(\frac{2ab - 4a - 2b}{a^2b^2}\), with \(a \ne 0\) and \(b \ne 0\).

- Rewrite a whole number or variable as a fraction with the needed denominator. - Use parentheses around a multi-term numerator being subtracted. - Preserve restrictions from the original denominators.

1. In part a, write \(3 = \frac{3n}{n}\): \(\frac{3n + 2}{n}\), with \(n \ne 0\). 2. In part b, write \(y = \frac{y^2}{y}\): \(\frac{y^2 + 4 - y^2}{y} = \frac{4}{y}\), with \(y \ne 0\). 3. In part c, write \(a = \frac{a^2}{a}\): \(\frac{a^2 - (a^2 - a)}{a} = \frac{a}{a} = 1\), with \(a \ne 0\).

a) \(\frac{3n + 2}{n}\), with \(n \ne 0\). b) \(\frac{4}{y}\), with \(y \ne 0\). c) \(1\), with \(a \ne 0\).

- Use the higher variable power in the least common denominator. - Multiply every term of a numerator by the missing factor. - Put the second numerator in parentheses before subtracting.

1. In part a, use denominator \(4x^2\): \(\frac{6x - (x + 1)}{4x^2} = \frac{5x - 1}{4x^2}\), with \(x \ne 0\). 2. In part b, use denominator \(a^2\): \(\frac{a(a - 2) + 2}{a^2} = \frac{a^2 - 2a + 2}{a^2}\), with \(a \ne 0\).

a) \(\frac{5x - 1}{4x^2}\), with \(x \ne 0\). b) \(\frac{a^2 - 2a + 2}{a^2}\), with \(a \ne 0\).

- Find a common numerical denominator. - Rewrite whole-variable terms as fractions. - Distribute subtraction signs to all terms inside parentheses.

1. Use denominator \(4\): \(\frac{2x - 3 + 2(x + 1)}{4} = \frac{4x - 1}{4}\). 2. Use denominator \(3\): \(\frac{3a - (2a - b)}{3} = \frac{a + b}{3}\). 3. Use denominator \(10\): \(\frac{2(s + t) - 5(s - t) + 10t}{10} = \frac{-3s + 17t}{10}\).

1) \(\frac{4x - 1}{4}\) 2) \(\frac{a + b}{3}\) 3) \(\frac{-3s + 17t}{10}\)

- Use \(12\) as the least common denominator. - Multiply every term in each numerator by the needed factor. - Distribute the subtraction sign across \(a + 2\).

1. The least common denominator is \(12\). 2. Combine: \(\frac{3(3a - 1) - 4(a + 2) + 6a}{12}\). 3. Simplify the numerator: \(9a - 3 - 4a - 8 + 6a = 11a - 11\). 4. The result is \(\frac{11(a - 1)}{12}\).

\(\frac{11(a - 1)}{12}\)

- Use the higher power \(x^2\) as the common denominator. - Multiply the entire first numerator by \(x\). - Put the second numerator in parentheses before subtracting.

1. The least common denominator is \(x^2\). 2. Rewrite the first fraction: \(\frac{x(x + 2)}{x^2} = \frac{x^2 + 2x}{x^2}\). 3. Subtract: \(\frac{x^2 + 2x - (x^2 + 2x - 1)}{x^2}\). 4. Distribute the subtraction sign; all terms cancel except \(1\), giving \(\frac{1}{x^2}\).

\(\frac{1}{x^2}\)

- Use the higher variable power in the common denominator. - Multiply the entire second numerator by \(2\). - Combine like terms and check for common factors.

1. The least common denominator is \(2a^2\). 2. Rewrite: \(\frac{3a}{2a^2} + \frac{2(a - 2)}{2a^2}\). 3. Add the numerators: \(3a + 2a - 4 = 5a - 4\). 4. The result is \(\frac{5a - 4}{2a^2}\), with \(a \ne 0\).

\(\frac{5a - 4}{2a^2}\), with \(a \ne 0\).

- Use the least common multiple of \(6\), \(4\), and \(8\). - Multiply each entire numerator by its required factor. - Distribute the subtraction sign through the third numerator.

1. The least common denominator is \(24\). 2. Combine: \(\frac{4(2x - 5) + 6(x + 1) - 3(3x - 2)}{24}\). 3. Simplify the numerator: \(8x - 20 + 6x + 6 - 9x + 6 = 5x - 8\). 4. The result is \(\frac{5x - 8}{24}\).

\(\frac{5x - 8}{24}\)

- Identify the value that makes a denominator zero. - Use \(x^2\) as the least common denominator. - Substitute \(2\) into every occurrence of \(x\).

1. Both denominators are zero when \(x = 0\), so the domain is all real numbers except \(0\). 2. The least common denominator is \(x^2\). Rewrite the first fraction as \(\frac{2x}{x^2}\). 3. Add the numerators: \(\frac{2x + x + 2}{x^2} = \frac{3x + 2}{x^2}\). 4. Evaluate at \(x = 2\): \(T(2) = \frac{3(2) + 2}{2^2} = \frac{8}{4} = 2\).

a) \(\{x \in \mathbb{R} \mid x \ne 0\}\) b) \(\frac{3x + 2}{x^2}\) c) \(2\)

- Factor the target and original denominators first. - Multiply both numerator and denominator by the same nonzero expression. - Determine the missing factor between each original denominator and the target denominator. - Distribute carefully when expanding each product.

1. For part a, multiply the numerator and denominator by \(4(a - 2)\): \(\frac{5 \cdot 4(a - 2)}{3a \cdot 4(a - 2)} = \frac{20a - 40}{12a^2 - 24a}\). 2. For part b, factor the denominator as \(4(a - 2)\). Multiply the numerator and denominator by \(3a\): \(\frac{3a(a + 2)}{12a(a - 2)} = \frac{3a^2 + 6a}{12a^2 - 24a}\). 3. For part c, write \(2a = \frac{2a}{1}\) and multiply the numerator and denominator by \(12a(a - 2)\): \(\frac{2a \cdot 12a(a - 2)}{12a(a - 2)} = \frac{24a^3 - 48a^2}{12a^2 - 24a}\). 4. The assumptions \(a \ne 0\) and \(a \ne 2\) ensure that the target denominator and each multiplier used in forming the equivalent expressions are nonzero.

a) \(\frac{20a - 40}{12a^2 - 24a}\). b) \(\frac{3a^2 + 6a}{12a^2 - 24a}\). c) \(\frac{24a^3 - 48a^2}{12a^2 - 24a}\). All expressions are written for \(a \ne 0, 2\).

- Factor the target denominator to compare it with \(2x\). - Multiply the numerator and denominator by the same missing factor. - Expand \((x - 3)(x + 5)\) using distribution.

1. Factor the target denominator: \(2x^2 + 10x = 2x(x + 5)\). 2. The original denominator is \(2x\), so multiply the numerator and denominator by \(x + 5\). 3. Expand the new numerator: \((x - 3)(x + 5) = x^2 + 5x - 3x - 15 = x^2 + 2x - 15\). 4. The equivalent rational expression is \(\frac{x^2 + 2x - 15}{2x^2 + 10x}\), valid under the stated assumptions.

\(\frac{x^2 + 2x - 15}{2x^2 + 10x}\), for \(x \ne 0\) and \(x \ne -5\).

- Factor out the greatest common factor from each denominator. - The least common denominator contains every distinct factor at its greatest needed power. - Divide the common denominator by each original denominator to find the missing factor. - Multiply the entire numerator by that missing factor.

1. Factor the denominators: \(x^2 - 4x = x(x - 4)\) and \(2x - 8 = 2(x - 4)\). 2. The least common denominator must contain the factors \(2\), \(x\), and \(x - 4\), so it is \(2x(x - 4) = 2x^2 - 8x\). 3. Multiply \(A\) by \(\frac{2}{2}\): \(A = \frac{6}{2x^2 - 8x}\). 4. Multiply \(B\) by \(\frac{x}{x}\): \(B = \frac{x(x + 1)}{2x(x - 4)} = \frac{x^2 + x}{2x^2 - 8x}\).

1. Denominators: \(x(x - 4)\) and \(2(x - 4)\). 2. Least common denominator: \(2x(x - 4) = 2x^2 - 8x\). 3. \(A = \frac{6}{2x^2 - 8x}\) and \(B = \frac{x^2 + x}{2x^2 - 8x}\), for \(x \ne 0, 4\).

- When the denominators have no common factor, their product is a common denominator. - Expand each product in the numerator carefully. - Factor the final numerator and check whether anything cancels.

1. The denominators have no common factor, so the least common denominator is \((x - 3)(x + 1)\). 2. Rewrite and combine: \(\frac{(x + 2)(x + 1) + x(x - 3)}{(x - 3)(x + 1)}\). 3. Expand the numerator: \((x + 2)(x + 1) = x^2 + 3x + 2\) and \(x(x - 3) = x^2 - 3x\). 4. Combine like terms: \(x^2 + 3x + 2 + x^2 - 3x = 2x^2 + 2 = 2(x^2 + 1)\). 5. No factor cancels, so the result is \(\frac{2(x^2 + 1)}{(x - 3)(x + 1)}\), with \(x \ne 3\) and \(x \ne -1\).

\(\frac{2(x^2 + 1)}{(x - 3)(x + 1)}\), with \(x \ne 3\) and \(x \ne -1\).

- Factor every denominator first. - Find the least common denominator for all three fractions. - After combining the numerators, check for a common factor.

1. Factor the third denominator: \(x^2 - 2x = x(x - 2)\). 2. The least common denominator is \(x(x - 2)\). 3. Combine the fractions: \(\frac{2(x - 2) - x + 4}{x(x - 2)}\). 4. Simplify the numerator: \(2x - 4 - x + 4 = x\). 5. Cancel the factor \(x\): \(\frac{x}{x(x - 2)} = \frac{1}{x - 2}\). 6. Preserve the original restrictions \(x \ne 0\) and \(x \ne 2\).

\(\frac{1}{x - 2}\), with \(x \ne 0\) and \(x \ne 2\).

- Factor the known expanded expression to identify the multiplier. - Whatever multiplies one part of a fraction must also multiply the other part.

1. In part a, factor the new denominator: \(x^2 - 3x = x(x - 3)\). The numerator and denominator were multiplied by \(x\), so the missing numerator is \(x(x + 2) = x^2 + 2x\). 2. In part b, factor the new numerator: \(z^2 - z = z(z - 1)\). The numerator was multiplied by \(z\), so multiply the denominator by \(z\) as well: \(2z \cdot z = 2z^2\).

a) \(x^2 + 2x\), for \(x \ne 0, 3\). b) \(2z^2\), for \(z \ne 0\).

- Find the least common denominator of \(2x\) and \(x^2\). - Multiply each entire numerator by the required factor. - Combine like terms and then check for common factors.

1. The least common denominator of \(2x\) and \(x^2\) is \(2x^2\). 2. Rewrite the first fraction: \(\frac{x(x + 1)}{2x^2} = \frac{x^2 + x}{2x^2}\). 3. Rewrite the second fraction: \(\frac{2(x - 2)}{2x^2} = \frac{2x - 4}{2x^2}\). 4. Add the numerators: \(x^2 + x + 2x - 4 = x^2 + 3x - 4\). 5. Factor the numerator as \((x + 4)(x - 1)\); no factor cancels with \(2x^2\). The result is \(\frac{x^2 + 3x - 4}{2x^2}\), with \(x \ne 0\).

\(\frac{x^2 + 3x - 4}{2x^2}\), with \(x \ne 0\).

- Factor the denominator of the first fraction before combining. - After reducing, the fractions have the same denominator. - Distribute the subtraction sign to the entire second numerator.

1. Factor the first denominator and reduce: \(\frac{4x}{2(x - 1)} = \frac{2x}{x - 1}\). 2. Subtract using the common denominator: \(\frac{2x}{x - 1} - \frac{x + 1}{x - 1} = \frac{2x - (x + 1)}{x - 1}\). 3. Simplify the numerator: \(2x - x - 1 = x - 1\). 4. Thus, \(\frac{x - 1}{x - 1} = 1\) for every allowed value \(x \ne 1\). 5. Therefore, the expression is constant on its domain.

Yes. The value is always \(1\) for \(x \ne 1\).

- Factor the denominator \(x^2 - 1\) as a difference of squares. - Can you simplify the second fraction before subtracting? - Record all restrictions from the original denominators before canceling any factors.

1. Factor the second denominator: \(x^2 - 1 = (x - 1)(x + 1)\). 2. In the original expression, the denominators are zero at \(x = 1\) and \(x = -1\). Thus, \(D_{\text{original}} = \mathbb{R} \setminus \{-1, 1\}\). 3. For values in the original domain, \(\frac{x + 1}{(x - 1)(x + 1)} = \frac{1}{x - 1}\). 4. Subtract: \(\frac{2}{x - 1} - \frac{1}{x - 1} = \frac{1}{x - 1}\). 5. The simplified expression is undefined only at \(x = 1\), so \(D_{\text{simplified}} = \mathbb{R} \setminus \{1\}\). This domain is larger because it includes \(x = -1\), although the simplified expression equals the original only on the original domain.

The simplified expression is \(\frac{1}{x - 1}\). \(D_{\text{original}} = \mathbb{R} \setminus \{-1, 1\}\). \(D_{\text{simplified}} = \mathbb{R} \setminus \{1\}\), so the simplified expression has a larger domain.

- Replace division by multiplication by the reciprocal. - Factor both differences of squares. - Exclude values that make the divisor zero as well as values that make any denominator zero.

1. Multiply by the reciprocal: \(\frac{z^2 - 4}{z + 1} \cdot \frac{z^2 - 1}{z - 2}\). 2. Factor: \(z^2 - 4 = (z - 2)(z + 2)\) and \(z^2 - 1 = (z - 1)(z + 1)\). 3. Cancel \(z - 2\) and \(z + 1\), leaving \((z + 2)(z - 1) = z^2 + z - 2\). 4. The original denominators require \(z \ne -1\) and \(z \ne 1\); the divisor must also be nonzero, so \(z \ne 2\).

\((z + 2)(z - 1)\), with \(z \ne -1, 1, 2\).

- Factor the first numerator. - Cancel common factors across the product. - Determine restrictions from the original denominators before canceling.

1. Factor \(x^2 + 2x = x(x + 2)\). 2. Cancel \(x + 2\) and \(x\): \(A(x) = \frac{x(x + 2)}{x + 2} \cdot \frac{5}{x} = 5\). 3. The original denominators require \(x \ne -2\) and \(x \ne 0\). 4. Since \(100\) is in the domain and the simplified expression is constant, \(A(100) = 5\).

a) \(5\) b) \(\{x \in \mathbb{R} \mid x \ne -2, 0\}\) c) \(5\)

- First simplify the expression in parentheses. - Factor the quadratic expression. - Cancel common factors only after multiplication is written as products.

1. Combine the expression in parentheses: \(1 - \frac{x - 1}{x + 1} = \frac{x + 1 - (x - 1)}{x + 1} = \frac{2}{x + 1}\). 2. Factor \(x^2 + x = x(x + 1)\). 3. Multiply and cancel: \(\frac{2}{x + 1} \cdot \frac{x(x + 1)}{2} = x\). 4. The original restriction is \(x \ne -1\).

\(x\), with \(x \ne -1\).

- Factor every polynomial first. - Replace division by multiplication by the reciprocal. - Include values that make the divisor zero among the restrictions.

1. Factor: \(y^2 - 4y + 4 = (y - 2)^2\), \(2y + 2 = 2(y + 1)\), and \(y^2 - 1 = (y - 1)(y + 1)\). 2. Multiply by the reciprocal: \(\frac{(y - 2)^2}{2(y + 1)} \cdot \frac{(y - 1)(y + 1)}{y - 2}\). 3. Cancel \(y + 1\) and one factor of \(y - 2\), leaving \(\frac{(y - 2)(y - 1)}{2}\). 4. The original denominators require \(y \ne -1\) and \(y \ne 1\); the divisor must be nonzero, so \(y \ne 2\).

\(\frac{(y - 2)(y - 1)}{2}\), with \(y \ne -1, 1, 2\).

- For direct substitution, evaluate each fraction before multiplying. - In the general expression, cancel common numerical and variable factors. - Compare the amount of arithmetic required by the two methods.

1. Direct substitution gives \(A = \frac{12 \cdot (2.5)^2}{5 \cdot 13} \cdot \frac{10 \cdot 13}{3 \cdot 2.5} = \frac{75}{65} \cdot \frac{130}{7.5} = 20\). 2. Simplify generally: \(A = \frac{120a^2b}{15ab} = 8a\), with \(a \ne 0\) and \(b \ne 0\). 3. Substitute \(a = 2.5\): \(8 \cdot 2.5 = 20\). 4. Both methods give the same value, but simplifying first requires less arithmetic.

a) \(20\) b) \(8a\), with \(a \ne 0\) and \(b \ne 0\). c) \(8 \cdot 2.5 = 20\), the same result as in part a.

- Factor every numerator and denominator before multiplying. - For division, multiply by the reciprocal of the second rational expression. - Keep all restrictions from the original expression.

1. For a), factor \(5a+10=5(a+2)\) and \(a^2-1=(a-1)(a+1)\). 2. Cancel the common factors \(a+2\) and \(a+1\): \(\frac{5(a+2)(a+1)}{(a-1)(a+1)(a+2)}=\frac{5}{a-1}\). 3. The original expression requires \(a\ne-2,-1,1\). 4. For b), factor \(u^2-2uv+v^2=(u-v)^2\) and \(u^2-v^2=(u-v)(u+v)\). 5. Multiply by the reciprocal: \(\frac{(u-v)^2}{u+v}\cdot\frac{(u-v)(u+v)}{u-v}\). 6. Cancel common factors to get \((u-v)^2\). The original expression requires \(u\ne v\) and \(u\ne-v\).

a) \(\frac{5}{a-1}\), where \(a\ne-2,-1,1\) b) \((u-v)^2\), where \(u\ne v\) and \(u\ne-v\)

- How do you divide by a fraction? - Factor \(a^2 - 25\) as a difference of squares. - In the original expression, neither a denominator nor the divisor may equal zero.

1. Rewrite division by a fraction as multiplication by its reciprocal: \(\frac{a^2 - 25}{5a} \cdot \frac{10}{a - 5}\). 2. Factor the difference of squares: \(a^2 - 25 = (a - 5)(a + 5)\). 3. Substitute and cancel common factors: \(\frac{(a - 5)(a + 5) \cdot 10}{5a(a - 5)} = \frac{2(a + 5)}{a}\). 4. In the original expression, \(a \ne 0\) because of the first denominator, and \(a \ne 5\) because the divisor cannot equal zero. Thus, \(D_{\text{original}} = \mathbb{R} \setminus \{0, 5\}\). 5. The simplified expression \(\frac{2(a + 5)}{a}\) is undefined only at \(a = 0\), so \(D_{\text{simplified}} = \mathbb{R} \setminus \{0\}\). The domains are not the same.

The simplified expression is \(\frac{2(a + 5)}{a}\). \(D_{\text{original}} = \mathbb{R} \setminus \{0, 5\}\), and \(D_{\text{simplified}} = \mathbb{R} \setminus \{0\}\). The domains are not the same.

- Simplify each fraction separately before subtracting. - Which values make the original denominators equal to zero? - A canceled factor must be nonzero. What does that mean for the original domain?

1. Factor the first numerator: \(\frac{b(b + 2)}{b} = b + 2\) for \(b \ne 0\). 2. Factor the second numerator as a difference of squares: \(\frac{(b - 2)(b + 2)}{b - 2} = b + 2\) for \(b \ne 2\). 3. Subtract the simplified parts: \((b + 2) - (b + 2) = 0\). 4. The original expression has denominators \(b\) and \(b - 2\), so it is undefined at \(b = 0\) and \(b = 2\). 5. Canceling factors does not add those inputs to the domain of the original expression. Therefore, \(T(b) = 0\) only for \(b \in \mathbb{R} \setminus \{0, 2\}\).

The expression simplifies to \(0\) for \(b \in \mathbb{R} \setminus \{0, 2\}\). The original expression is undefined at \(b = 0\) and \(b = 2\) because at least one denominator equals zero.

- Factor the difference of squares in the second denominator. - Use the factored form to identify the least common denominator. - After adding, check for a common factor.

1. Factor the difference of squares: \(x^2 - 4 = (x - 2)(x + 2)\). 2. The least common denominator is \((x - 2)(x + 2)\). 3. Rewrite and add: \(\frac{x - 2}{(x - 2)(x + 2)} + \frac{4}{(x - 2)(x + 2)} = \frac{x + 2}{(x - 2)(x + 2)}\). 4. Cancel \(x + 2\) to obtain \(\frac{1}{x - 2}\). 5. Preserve the original restrictions \(x \ne -2\) and \(x \ne 2\).

\(\frac{1}{x - 2}\), with \(x \ne -2\) and \(x \ne 2\).

- Rewrite \(1\) as a fraction with denominator \(x - 1\). - Find the input that makes the denominator zero. - Once equivalence is established, evaluate either expression.

1. Rewrite \(1\) with denominator \(x - 1\): \(T_1(x) = \frac{x + 1}{x - 1} - \frac{x - 1}{x - 1}\). 2. Subtract the numerators: \(T_1(x) = \frac{x + 1 - (x - 1)}{x - 1} = \frac{2}{x - 1} = T_2(x)\). 3. Both expressions are undefined at \(x = 1\), so their domain is all real numbers except \(1\). 4. Substitute \(x = 0\): \(\frac{2}{0 - 1} = -2\).

a) \(T_1(x) = \frac{x + 1 - (x - 1)}{x - 1} = \frac{2}{x - 1} = T_2(x)\). b) \(\{x \in \mathbb{R} \mid x \ne 1\}\) c) \(-2\)

- Rewrite division as multiplication by the reciprocal. - Factor differences of squares before canceling. - Keep all restrictions from the original expression.

1. For a), multiply by the reciprocal: \(\frac{8x^3}{y^2}\cdot\frac{y}{4x}\). 2. Cancel common factors to get \(\frac{2x^2}{y}\). The original expression requires \(x\ne0\) and \(y\ne0\). 3. For b), factor \(a^2-4=(a-2)(a+2)\). 4. Substitute and cancel common factors: \(\frac{a+2}{3a^2}\cdot\frac{6a}{(a-2)(a+2)}=\frac{2}{a(a-2)}\). 5. The original expression requires \(a\ne-2,0,2\).

a) \(\frac{2x^2}{y}\), where \(x\ne0\) and \(y\ne0\) b) \(\frac{2}{a(a-2)}\), where \(a\ne-2,0,2\)

- Factor the quadratic denominator first. - Use the factored denominator to find the least common denominator. - Combine and simplify the numerator before checking for cancellation.

1. Factor the difference of squares: \(x^2 - 25 = (x - 5)(x + 5)\). 2. Use \((x - 5)(x + 5)\) as the least common denominator. 3. Rewrite and subtract: \(\frac{2x(x + 5) - 10x}{(x - 5)(x + 5)}\). 4. Simplify the numerator: \(2x(x + 5) - 10x = 2x^2 + 10x - 10x = 2x^2\). 5. No factor cancels, so the result is \(\frac{2x^2}{x^2 - 25}\), with \(x \ne -5\) and \(x \ne 5\).

\(\frac{2x^2}{x^2 - 25}\), with \(x \ne -5\) and \(x \ne 5\).

- Factor the quadratic denominator. - Use a common denominator to combine the right-hand side. - Compare the two denominators in the general decomposition.

1. Factor \(x^2 + x = x(x + 1)\). The least common denominator on the right is \(x(x + 1)\). 2. Combine: \(\frac{1}{x + 1} + \frac{1}{x(x + 1)} = \frac{x + 1}{x(x + 1)} = \frac{1}{x}\), using \(x \ne -1\). 3. Substitute \(x = 10\): \(\frac{1}{10} = \frac{1}{11} + \frac{1}{110}\). 4. The two denominators are \(x + 1\) and \(x(x + 1)\). For \(x > 0\), they are equal only when \(x = 1\). Thus, when \(x \ne 1\), the unit fractions are different.

a) \(\frac{1}{x + 1} + \frac{1}{x(x + 1)} = \frac{x + 1}{x(x + 1)} = \frac{1}{x}\). b) \(\frac{1}{10} = \frac{1}{11} + \frac{1}{110}\) c) For \(x > 0\), the denominators \(x + 1\) and \(x(x + 1)\) are equal only when \(x = 1\), so they are different under the stated condition.

- Factor the quadratic denominator first. - Distribute the subtraction sign to the entire second numerator. - Factor the combined numerator before canceling.

1. Factor the second denominator: \(x^2 + x = x(x + 1)\). 2. The least common denominator is \(x(x + 1)\). 3. Rewrite and subtract: \(\frac{3x - (x - 2)}{x(x + 1)} = \frac{2x + 2}{x(x + 1)}\). 4. Factor the numerator: \(\frac{2(x + 1)}{x(x + 1)}\). 5. Cancel \(x + 1\) to obtain \(\frac{2}{x}\). 6. Preserve the original restrictions \(x \ne -1\) and \(x \ne 0\).

\(\frac{2}{x}\), with \(x \ne -1\) and \(x \ne 0\).

- Use a common denominator when subtracting rational expressions. - Factor before multiplying in part b. - Record all values that make an original denominator zero before canceling.

1. For part a, use the common denominator \(2x\): \(\frac{3}{x} - \frac{1}{2x} = \frac{6}{2x} - \frac{1}{2x} = \frac{5}{2x}\). 2. Both the original and simplified expressions in part a require \(x \ne 0\). Thus, \(D_{\text{original}} = D_{\text{simplified}} = \mathbb{R} \setminus \{0\}\). 3. For part b, factor \(x^2 - 1 = (x - 1)(x + 1)\): \(\frac{x - 1}{x^2} \cdot \frac{3x}{(x - 1)(x + 1)}\). 4. Cancel \(x - 1\) and one factor of \(x\) to obtain \(\frac{3}{x(x + 1)}\). 5. The original expression is undefined at \(x = 0\), \(x = 1\), and \(x = -1\), so \(D_{\text{original}} = \mathbb{R} \setminus \{-1, 0, 1\}\). 6. The simplified expression is undefined at \(x = 0\) and \(x = -1\), so \(D_{\text{simplified}} = \mathbb{R} \setminus \{-1, 0\}\).

a) \(\frac{5}{2x}\); \(D_{\text{original}} = \mathbb{R} \setminus \{0\}\); \(D_{\text{simplified}} = \mathbb{R} \setminus \{0\}\). b) \(\frac{3}{x(x + 1)}\); \(D_{\text{original}} = \mathbb{R} \setminus \{-1, 0, 1\}\); \(D_{\text{simplified}} = \mathbb{R} \setminus \{-1, 0\}\).

- Exclude values that make a denominator zero. - Find the least common multiple of the numerical coefficients in the denominators. - Use parentheses around an entire numerator when subtracting.

1. In part a, \(x \ne 0\). The least common denominator is \(12x\): \(\frac{15}{12x} + \frac{8}{12x} - \frac{2}{12x} = \frac{21}{12x} = \frac{7}{4x}\). 2. In part b, \(a \ne 0\). The least common denominator is \(6a\): \(\frac{2a + 3}{6a} - \frac{2(a - 1)}{6a} = \frac{2a + 3 - 2a + 2}{6a} = \frac{5}{6a}\).

a) Domain: \(\{x \in \mathbb{R} \mid x \ne 0\}\); \(\frac{7}{4x}\) b) Domain: \(\{a \in \mathbb{R} \mid a \ne 0\}\); \(\frac{5}{6a}\)

- Factor the target denominator as a difference of squares. - Identify the factor missing from the original denominator. - Multiply the entire original numerator by that factor.

1. Factor the target denominator as a difference of squares: \(4a^2 - 9b^2 = (2a + 3b)(2a - 3b)\). 2. The original denominator must be multiplied by \(2a - 3b\), so multiply the numerator by the same factor. 3. Expand: \((a - b)(2a - 3b) = 2a^2 - 3ab - 2ab + 3b^2 = 2a^2 - 5ab + 3b^2\). 4. The stated conditions ensure that both the original denominator and the multiplier are nonzero.

The new numerator is \(2a^2 - 5ab + 3b^2\), valid when \(2a + 3b \ne 0\) and \(2a - 3b \ne 0\).

- Factor each numerator before canceling. - Determine excluded values from the original denominators. - A canceled factor still creates a restriction in the original expression.

1. Factor the numerators: \(x^2-4x=x(x-4)\) and \(3x+6=3(x+2)\). 2. Substitute: \(T(x)=\frac{x(x-4)}{x+2}\cdot\frac{3(x+2)}{x-4}\). 3. Cancel \(x-4\) and \(x+2\) to get \(T(x)=3x\). 4. In the original expression, \(x+2=0\) when \(x=-2\), and \(x-4=0\) when \(x=4\). These values remain excluded after simplification.

a) \(3x\), where \(x\ne-2,4\) b) \(x=-2\) and \(x=4\)

- Factor the quadratic denominator as a difference of squares. - Rewrite the second fraction with the full factored denominator. - Distribute the subtraction sign carefully.

1. Factor the difference of squares: \(x^2 - 25 = (x - 5)(x + 5)\). 2. Rewrite the second fraction using the least common denominator: \(\frac{x - 5}{(x - 5)(x + 5)}\). 3. Subtract the numerators: \(2x - (x - 5) = x + 5\). 4. Cancel \(x + 5\): \(\frac{x + 5}{(x - 5)(x + 5)} = \frac{1}{x - 5}\). 5. Preserve the original restrictions \(x \ne -5\) and \(x \ne 5\).

\(\frac{1}{x - 5}\), with \(x \ne -5\) and \(x \ne 5\).

- Factor differences of squares and perfect-square trinomials. - Rewrite division as multiplication by the reciprocal. - Keep restrictions from every denominator in the original expression.

1. For a), factor \(x^2-1=(x-1)(x+1)\) and \(3x+6=3(x+2)\). 2. Cancel \(x-1\) and \(x+2\): \(\frac{3(x-1)(x+1)(x+2)}{(x+2)(x-1)}=3(x+1)\). 3. The original expression requires \(x\ne-2,1\). 4. For b), multiply by the reciprocal, then factor: \(a^2-10a+25=(a-5)^2\) and \(a^2-1=(a-1)(a+1)\). 5. Cancel common factors: \(\frac{(a-5)^2(a-1)(a+1)}{(a+1)(a-5)}=(a-5)(a-1)\). 6. The original expression requires \(a\ne-1,1,5\).

a) \(3(x+1)\), where \(x\ne-2,1\) b) \((a-5)(a-1)\), where \(a\ne-1,1,5\)

- Treat \(4x^2\) as \((2x)^2\) when factoring. - Look for a perfect-square trinomial in part b). - Rewrite division as multiplication by the reciprocal.

1. For a), factor \(4x^2-9=(2x-3)(2x+3)\). 2. Cancel \(2x+3\) and one factor of \(x\): \(\frac{(2x-3)(2x+3)x^2}{2x(2x+3)}=\frac{x(2x-3)}{2}\). 3. The original expression requires \(x\ne0\) and \(x\ne-\frac{3}{2}\). 4. For b), multiply by the reciprocal and factor \(y^2-16=(y-4)(y+4)\) and \(y^2-8y+16=(y-4)^2\). 5. The product becomes \(\frac{(y-4)(y+4)(y-4)}{(y-4)^2(y+4)}=1\). 6. The original expression requires \(y\ne-4,4\).

a) \(\frac{x(2x-3)}{2}\), where \(x\ne0,-\frac{3}{2}\) b) \(1\), where \(y\ne-4,4\)

- Combine the fractions inside the parentheses first. - Use the simplified expression to evaluate the given value. - Check the original denominators to find excluded values.

1. Use the common denominator \(2k\): \(\frac{2}{k}-\frac{1}{2k}=\frac{4}{2k}-\frac{1}{2k}=\frac{3}{2k}\). 2. Multiply and simplify: \(\frac{3}{2k}\cdot\frac{4k^2}{3}=2k\). 3. For \(k=1.5\), \(2k=2\cdot1.5=3\). 4. The denominator is zero when \(k=0\), so the original expression is undefined there.

a) \(2k\), where \(k\ne0\) b) \(3\) c) \(k=0\)

- Factor the difference of squares. - Rewrite division as multiplication by the reciprocal. - Simplify completely, then compare with the right side.

1. Rewrite the division as multiplication by the reciprocal: \(\frac{x^2-y^2}{x}\cdot\frac{x^2}{x+y}\). 2. Factor the difference of squares: \(x^2-y^2=(x-y)(x+y)\). 3. Substitute and cancel common factors: \(\frac{(x-y)(x+y)}{x}\cdot\frac{x^2}{x+y}=x(x-y)\). 4. Distribute: \(x(x-y)=x^2-xy\). 5. This matches the right side for every allowed pair satisfying \(x\ne0\) and \(x+y\ne0\).

Yes. For \(x\ne0\) and \(x+y\ne0\), the left side simplifies to \(x^2-xy\).

- Factor the difference of squares. - Rewrite the first fraction with the full factored denominator. - Combine the numerators and check for cancellation.

1. Write the difference: \(B - A = \frac{1}{x - 2} - \frac{2}{x^2 - 4}\). 2. Factor \(x^2 - 4 = (x - 2)(x + 2)\). 3. Use the common denominator \((x - 2)(x + 2)\): \(\frac{x + 2 - 2}{(x - 2)(x + 2)} = \frac{x}{(x - 2)(x + 2)}\). 4. No factor cancels, so the result is \(\frac{x}{x^2 - 4}\), with \(x \ne -2\) and \(x \ne 2\).

\(\frac{x}{x^2 - 4}\), with \(x \ne -2\) and \(x \ne 2\).

- Factor both denominators first. - Include every distinct factor in the least common denominator. - Distribute the subtraction sign to the entire second numerator.

1. Factor the denominators: \(2x - 4 = 2(x - 2)\) and \(x^2 - 4 = (x - 2)(x + 2)\). 2. The least common denominator is \(2(x - 2)(x + 2)\). 3. Combine: \(\frac{3(x + 2) - 2(x + 1)}{2(x - 2)(x + 2)}\). 4. Simplify the numerator: \(3x + 6 - 2x - 2 = x + 4\). 5. No factor cancels, so the result is \(\frac{x + 4}{2(x - 2)(x + 2)}\), with \(x \ne -2\) and \(x \ne 2\).

\(\frac{x + 4}{2(x - 2)(x + 2)}\), with \(x \ne -2\) and \(x \ne 2\).

- Apply the order of operations. - Factor the denominator in the product. - Simplify the product before subtracting.

1. Perform the multiplication before the subtraction. Factor \(b^2 + 2b = b(b + 2)\). 2. Simplify the product: \(\frac{b}{3} \cdot \frac{6}{b(b + 2)} = \frac{6b}{3b(b + 2)} = \frac{2}{b + 2}\). 3. Subtract the like-denominator fractions: \(\frac{4}{b + 2} - \frac{2}{b + 2} = \frac{2}{b + 2}\). 4. Preserve the original restrictions \(b \ne -2\) and \(b \ne 0\).

\(\frac{2}{b + 2}\), with \(b \ne -2\) and \(b \ne 0\).

- Simplify the product before subtracting. - Factor \(2x - 6\). - Distribute the subtraction sign to the entire second numerator.

1. Simplify the product: \(\frac{x(x - 1)}{x(x - 3)} = \frac{x - 1}{x - 3}\), while preserving \(x \ne 0\). 2. Factor \(2x - 6 = 2(x - 3)\). 3. Use the common denominator \(2(x - 3)\): \(\frac{2(x - 1) - (x + 2)}{2(x - 3)}\). 4. Simplify the numerator: \(2x - 2 - x - 2 = x - 4\). 5. The result is \(\frac{x - 4}{2(x - 3)}\), with \(x \ne 0\) and \(x \ne 3\).

\(\frac{x - 4}{2(x - 3)}\), with \(x \ne 0\) and \(x \ne 3\).

- Factor the difference of squares. - Rewrite the first fraction with the complete factored denominator. - Use parentheses around the second numerator when subtracting.

1. Factor \(x^2 - 4 = (x - 2)(x + 2)\). 2. Use the common denominator \((x - 2)(x + 2)\): \(\frac{3(x + 2) - (x + 4)}{(x - 2)(x + 2)}\). 3. Simplify the numerator: \(3x + 6 - x - 4 = 2x + 2 = 2(x + 1)\). 4. No factor cancels, so the result is \(\frac{2(x + 1)}{(x - 2)(x + 2)}\), with \(x \ne -2\) and \(x \ne 2\).

\(\frac{2(x + 1)}{(x - 2)(x + 2)}\), with \(x \ne -2\) and \(x \ne 2\).

- Factor both quadratic denominators. - Include each distinct factor in the least common denominator. - Relate \(1 - x\) to \(x - 1\) before canceling.

1. Factor the denominators: \(x^2 - x = x(x - 1)\) and \(x^2 - 1 = (x - 1)(x + 1)\). 2. The least common denominator is \(x(x - 1)(x + 1)\). 3. Combine: \(\frac{x + 1 - 2x}{x(x - 1)(x + 1)} = \frac{1 - x}{x(x - 1)(x + 1)}\). 4. Since \(1 - x = -(x - 1)\), cancel \(x - 1\) to obtain \(-\frac{1}{x(x + 1)}\). 5. Preserve the original restrictions \(x \ne -1\), \(x \ne 0\), and \(x \ne 1\).

\(-\frac{1}{x(x + 1)}\), with \(x \ne -1, 0, 1\).

- Check the step where a negative sign is distributed across parentheses. - Every sign inside a subtracted expression changes. - Confirm that the common denominator itself was formed correctly.

1. The common-denominator setup is correct. 2. The error occurs when distributing the negative sign: \(-(x^2 - 2x) = -x^2 + 2x\), not \(-x^2 - 2x\). 3. The correct numerator is \(5x + 5 - x^2 + 2x = -x^2 + 7x + 5\). 4. Therefore, the correct result is \(\frac{-x^2 + 7x + 5}{x(x + 1)}\). 5. The original restrictions are \(x \ne 0\) and \(x \ne -1\).

The student failed to change the sign of \(-2x\) when distributing the subtraction. The correct result is \(\frac{-x^2 + 7x + 5}{x(x + 1)}\), with \(x \ne 0\) and \(x \ne -1\).

- Rewrite the polynomial part using the denominator \(x-3\). - Both terms must have the same denominator before they can be combined. - A rational expression is zero when its numerator is zero and its denominator is nonzero. - Check each solution against the domain.

1. Use the common denominator \(x-3\): \(f(x)=\frac{(x+1)(x-3)}{x-3}+\frac{4}{x-3}\). 2. Expand and combine the numerator: \((x+1)(x-3)+4=x^2-2x-3+4=x^2-2x+1\). 3. Thus, \(f(x)=\frac{x^2-2x+1}{x-3}=\frac{(x-1)^2}{x-3}\). 4. The numerator is zero only at \(x=1\). Since \(x=1\) is in the domain, the x-intercept is \((1, 0)\).

\(f(x)=\frac{x^2-2x+1}{x-3}\), and the only x-intercept is \((1, 0)\).

- Find a common denominator for the two terms. - When subtracting the second fraction, distribute the negative sign through its entire numerator. - A rational expression equals zero only when its numerator is zero and its denominator is nonzero. - Factor or solve the resulting quadratic equation.

1. A common denominator is \(2(x-2)\). 2. Rewrite each term: \(g(x)=\frac{8}{2(x-2)}-\frac{x(x-2)}{2(x-2)}\). 3. Combine the numerators: \(g(x)=\frac{8-x^2+2x}{2x-4}=\frac{-x^2+2x+8}{2x-4}\). 4. Set the numerator equal to zero: \(-x^2+2x+8=0\), or \(x^2-2x-8=0\). 5. Factor: \((x-4)(x+2)=0\), so \(x=4\) or \(x=-2\). Both values are in the domain \(\mathbb{R}\setminus\{2\}\).

\(g(x)=\frac{-x^2+2x+8}{2x-4}\). The x-intercepts are \((4, 0)\) and \((-2, 0)\).

- Rewrite the polynomial term using the denominator \(x-2\). - Which value is excluded by the denominator? - A rational expression equals zero when its numerator is zero and its denominator is nonzero.

1. Rewrite \(f\) with the common denominator \(x-2\): \(f(x)=\frac{(x-2)^2}{x-2}-\frac{1}{x-2}=\frac{(x-2)^2-1}{x-2}=\frac{x^2-4x+3}{x-2}=g(x)\). 2. Both expressions are undefined at \(x=2\), so the common domain is \(D=\mathbb{R}\setminus\{2\}\). 3. Factor the numerator: \(x^2-4x+3=(x-1)(x-3)\). Its zeros are \(x=1\) and \(x=3\), and both are in the domain. Therefore, the x-intercepts are \((1, 0)\) and \((3, 0)\).

1. \(f(x)=\frac{(x-2)^2-1}{x-2}=\frac{x^2-4x+3}{x-2}=g(x)\) 2. \(D=\mathbb{R}\setminus\{2\}\) 3. \((1, 0)\) and \((3, 0)\)

- Rewrite every term using the common denominator \(x\). - A fraction equals zero when its numerator is zero and its denominator is nonzero. - Factor the resulting quadratic numerator. - Which input is excluded from the original function?

1. Use the common denominator \(x\): \(f(x)=\frac{8}{x}+\frac{x^2}{x}-\frac{6x}{x}\). 2. Combine the numerators: \(f(x)=\frac{x^2-6x+8}{x}\). 3. Factor the numerator: \(x^2-6x+8=(x-2)(x-4)\). 4. The numerator is zero at \(x=2\) and \(x=4\). Both values are in the domain \(\mathbb{R}\setminus\{0\}\), so both are x-intercepts.

a) \(f(x)=\frac{x^2-6x+8}{x}\) b) \((2, 0)\) and \((4, 0)\)

- Set the denominator equal to zero. - Fractions with equal denominators are combined by adding their numerators. - Factor the difference of squares before canceling.

1. The denominator is zero when \(x^2 - 16 = 0\), so \(x = -4\) and \(x = 4\) are excluded. 2. Add the numerators: \(A = \frac{x + 4}{x^2 - 16}\). 3. Factor the denominator: \(x^2 - 16 = (x - 4)(x + 4)\). 4. Cancel \(x + 4\) to obtain \(A = \frac{1}{x - 4}\), while preserving both original restrictions. 5. At \(x = 6\), \(A = \frac{1}{6 - 4} = \frac{1}{2}\).

a) Undefined at \(x = -4\) and \(x = 4\). b) \(\frac{1}{x - 4}\), with \(x \ne -4\) and \(x \ne 4\). c) \(\frac{1}{2}\)

- Rewrite a reversed subtraction denominator by factoring out \(-1\). - Factor a difference of squares in part a. - In part b, distribute the subtraction sign across the entire second numerator.

1. Since \(4 - x = -(x - 4)\), \(A(x) = \frac{x^2 - 16}{x - 4}\). 2. Factor \(x^2 - 16 = (x - 4)(x + 4)\), then cancel \(x - 4\) to obtain \(A(x) = x + 4\), with \(x \ne 4\). 3. Since \(1 - y = -(y - 1)\), \(B(y) = \frac{2y + 3}{y - 1} - \frac{y + 4}{y - 1}\). 4. Combine the numerators: \(2y + 3 - (y + 4) = y - 1\). 5. Thus, \(B(y) = \frac{y - 1}{y - 1} = 1\), with \(y \ne 1\).

a) \(A(x) = x + 4\), with \(x \ne 4\). b) \(B(y) = 1\), with \(y \ne 1\).

- Combine the numerators first. - Look for a difference of squares or a perfect-square trinomial. - Preserve the zeros of the original denominators.

1. In part a, combine the numerators: \(\frac{k^2 - 25}{k - 5}\). 2. Factor the difference of squares: \(k^2 - 25 = (k - 5)(k + 5)\). Cancel \(k - 5\) to obtain \(k + 5\), with \(k \ne 5\). 3. In part b, combine and reorder the numerator: \(\frac{x^2 - 4x + 4}{x - 2}\). 4. Factor the perfect-square trinomial: \(x^2 - 4x + 4 = (x - 2)^2\). Cancel one factor of \(x - 2\) to obtain \(x - 2\), with \(x \ne 2\).

a) \(k + 5\), with \(k \ne 5\). b) \(x - 2\), with \(x \ne 2\).

- Factor out \(-1\) from a reversed subtraction denominator. - Look for a perfect-square trinomial after combining. - Factor the final numerator before canceling.

1. In part a, \(5 - y = -(y - 5)\), so the expression becomes \(\frac{y^2 + 25 - 10y}{y - 5}\). 2. Factor the numerator: \(y^2 - 10y + 25 = (y - 5)^2\). Cancel one factor of \(y - 5\) to obtain \(y - 5\), with \(y \ne 5\). 3. In part b, \(3 - z = -(z - 3)\), so subtracting the second fraction becomes adding \(\frac{z - 4}{z - 3}\). 4. Add the numerators: \(2z - 5 + z - 4 = 3z - 9 = 3(z - 3)\). Cancel \(z - 3\) to obtain \(3\), with \(z \ne 3\).

a) \(y - 5\), with \(y \ne 5\). b) \(3\), with \(z \ne 3\).

- Rewrite \(9 - a^2\) using a difference of squares and a negative sign. - Use the factored denominator to find the least common denominator. - Cancel only after combining the fractions.

1. Rewrite \(9 - a^2 = -(a^2 - 9) = -(a - 3)(a + 3)\). 2. Thus, \(T(a) = \frac{1}{a - 3} - \frac{6}{(a - 3)(a + 3)}\). 3. Use the common denominator \((a - 3)(a + 3)\): \(T(a) = \frac{a + 3 - 6}{(a - 3)(a + 3)} = \frac{a - 3}{(a - 3)(a + 3)}\). 4. Cancel \(a - 3\) to obtain \(T(a) = \frac{1}{a + 3}\). 5. Preserve the original restrictions \(a \ne -3\) and \(a \ne 3\).

\(\frac{1}{a + 3}\), with \(a \ne -3\) and \(a \ne 3\).

- Test the claim with simple nonzero values. - Use the product \(xy\) as a common denominator. - Isolate \(T\) before combining the fractions.

1. For part a, let \(x = 1\) and \(y = 1\). The left side is \(1 + 1 = 2\), while the claimed right side is \(\frac{2}{2} = 1\). This counterexample disproves the rule. 2. For part b, use the common denominator \(xy\): \(\frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy}\), where \(x \ne 0\) and \(y \ne 0\). 3. For part c, isolate \(T\): \(T = \frac{3}{x} - \frac{5}{2x} = \frac{6 - 5}{2x} = \frac{1}{2x}\), where \(x \ne 0\).

a) For \(x = y = 1\), the two sides are \(2\) and \(1\), so the rule is false. b) \(\frac{x + y}{xy}\), with \(x \ne 0\) and \(y \ne 0\). c) \(T = \frac{1}{2x}\), with \(x \ne 0\).

- Check the common-denominator step separately from the subtraction step. - Treat the second numerator as a parenthesized expression. - Distribute a leading negative sign to every term.

1. The least common denominator \(3x\) and the rewrite \(\frac{5}{x} = \frac{15}{3x}\) are correct. 2. The subtraction sign must apply to the entire numerator \(x - 2\): \(15 - (x - 2) = 15 - x + 2\). 3. Therefore, \(T = \frac{17 - x}{3x}\), with \(x \ne 0\).

The work is incorrect because the subtraction sign was not distributed to both terms in \(x - 2\). The correct result is \(\frac{17 - x}{3x}\), with \(x \ne 0\).

- Use \(x^2y\) as the least common denominator. - Multiply each entire numerator by its missing factor. - Enclose the second numerator in parentheses before subtracting.

1. The least common denominator of \(xy\) and \(x^2\) is \(x^2y\). 2. Rewrite: \(A = \frac{x(3x + y)}{x^2y}\) and \(B = \frac{y(2x - y)}{x^2y}\). 3. Subtract: \(A - B = \frac{3x^2 + xy - (2xy - y^2)}{x^2y}\). 4. Distribute the subtraction sign and combine: \(3x^2 + xy - 2xy + y^2 = 3x^2 - xy + y^2\). 5. The result is \(\frac{3x^2 - xy + y^2}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\). The minus sign changes the sign of every term in the second numerator.

\(A - B = \frac{3x^2 - xy + y^2}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\). The entire second numerator is subtracted, so both of its terms change sign.

- Include the highest power of \(x\) and the factor \(y\) in the common denominator. - Multiply the entire second numerator by \(x\). - Distribute the subtraction sign after expanding.

1. The least common denominator is \(x^2y\). 2. Rewrite the fractions: \(\frac{2y}{x^2y} - \frac{x(x - 3)}{x^2y}\). 3. Subtract and expand: \(2y - x(x - 3) = 2y - x^2 + 3x\). 4. The result is \(\frac{2y - x^2 + 3x}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\).

\(\frac{2y - x^2 + 3x}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\).

- Build one least common denominator for all terms. - Keep a multi-term numerator in parentheses when multiplying. - After combining, factor out any common numerical factor.

1. In part a, the least common denominator is \(12x^2y^2\). 2. Rewrite and combine: \(\frac{9xy - 8x^2 + 2y^2}{12x^2y^2}\). The restrictions are \(x \ne 0\) and \(y \ne 0\). 3. In part b, the least common denominator is \(2x^2\). 4. Combine the numerators: \(x(x - 2) + 6 - x^2 = x^2 - 2x + 6 - x^2 = -2x + 6\). 5. Reduce: \(\frac{-2x + 6}{2x^2} = \frac{3 - x}{x^2}\), with \(x \ne 0\).

a) \(\frac{9xy - 8x^2 + 2y^2}{12x^2y^2}\), with \(x \ne 0\) and \(y \ne 0\). b) \(\frac{3 - x}{x^2}\), with \(x \ne 0\).

- Use \(x^2y\) as the least common denominator. - Keep \(x - 3\) and \(y + 1\) in parentheses while multiplying. - Track the subtraction sign before combining like terms.

1. The least common denominator is \(x^2y\). 2. Rewrite the terms: \(\frac{2xy}{x^2y} - \frac{y(x - 3)}{x^2y} + \frac{x(y + 1)}{x^2y}\). 3. Combine and expand: \(2xy - (xy - 3y) + xy + x = 2xy + x + 3y\). 4. The result is \(\frac{2xy + x + 3y}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\).

\(\frac{2xy + x + 3y}{x^2y}\), with \(x \ne 0\) and \(y \ne 0\).

- Factor the difference of squares in the second denominator. - Use the factored form to find the common denominator. - Factor the combined numerator before canceling.

1. Factor \(x^2 - y^2 = (x - y)(x + y)\). 2. Rewrite the first fraction: \(\frac{x(x + y)}{(x - y)(x + y)}\). 3. Subtract the numerators: \(x^2 + xy - (x^2 + y^2) = xy - y^2 = y(x - y)\). 4. Cancel \(x - y\) to obtain \(\frac{y}{x + y}\). 5. Preserve the original restrictions \(x \ne y\) and \(x \ne -y\).

\(\frac{y}{x + y}\), with \(x \ne y\) and \(x \ne -y\).

- Use \(6\) as the least common denominator. - Expand both squared binomials. - Group the quadratic, linear, and constant terms separately.

1. The least common denominator is \(6\). 2. Combine: \(\frac{2(a + 1)^2 + (a - 1)^2 - 3(a^2 + 1)}{6}\). 3. Expand: \(2(a^2 + 2a + 1) + (a^2 - 2a + 1) - 3a^2 - 3\). 4. Combine like terms: \(2a\). 5. Reduce \(\frac{2a}{6}\) to \(\frac{a}{3}\).

\(\frac{a}{3}\)

- Factor the quadratic denominator first. - Rewrite \(1\) with the least common denominator. - Look for a perfect-square trinomial after combining.

1. In part a, factor \(y^2 - 3y = y(y - 3)\). Using this denominator gives \(\frac{2y - 12}{y(y - 3)} = \frac{2(y - 6)}{y(y - 3)}\), with \(y \ne 0\) and \(y \ne 3\). 2. In part b, use denominator \(b^2\): \(\frac{b^2 - b - (b - 1)}{b^2} = \frac{b^2 - 2b + 1}{b^2} = \frac{(b - 1)^2}{b^2}\), with \(b \ne 0\).

a) \(\frac{2(y - 6)}{y(y - 3)}\), with \(y \ne 0\) and \(y \ne 3\). b) \(\frac{(b - 1)^2}{b^2}\), with \(b \ne 0\).

- Rewrite all terms over one common denominator. - Keep multi-term numerators in parentheses while multiplying. - Carefully distribute each subtraction sign.

1. With denominator \(4\), the numerator is \(3x - 3y - 2(2x + y) + 4x = 3x - 5y\), so the result is \(\frac{3x - 5y}{4}\). 2. With denominator \(12\), the numerator is \(4(2u + v) - 3(u - 2v) - v = 5u + 9v\), so the result is \(\frac{5u + 9v}{12}\). 3. With denominator \(6\), the numerator is \(12 - (3k - 1) - 2(k + 2) = 9 - 5k\), so the result is \(\frac{9 - 5k}{6}\).

1) \(\frac{3x - 5y}{4}\) 2) \(\frac{5u + 9v}{12}\) 3) \(\frac{9 - 5k}{6}\)

- Include all three distinct linear factors in the common denominator. - Multiply each numerator by the one factor missing from its denominator. - Group the expanded numerator into terms containing \(x\) and terms that do not.

1. The least common denominator is \((x - a)(x - b)(x - c)\). 2. The combined numerator is \(a(x - c) + b(x - a) + c(x - b)\). 3. Expand: \(ax - ac + bx - ab + cx - bc\). 4. Group the terms: \(x(a + b + c) - (ab + bc + ca)\). 5. The result is \(\frac{x(a + b + c) - (ab + bc + ca)}{(x - a)(x - b)(x - c)}\), with \(x \ne a\), \(x \ne b\), and \(x \ne c\).

\(\frac{x(a + b + c) - (ab + bc + ca)}{(x - a)(x - b)(x - c)}\), with \(x \ne a, b, c\).

- Factor the difference of squares. - Relate \(2 - x\) to \(x - 2\). - Distribute the subtraction sign across the expanded numerator.

1. Factor \(x^2 - 4 = (x - 2)(x + 2)\) and rewrite \(2 - x = -(x - 2)\). 2. The expression becomes \(\frac{x}{(x - 2)(x + 2)} - \frac{1}{x - 2}\). 3. Use the common denominator: \(\frac{x - (x + 2)}{(x - 2)(x + 2)} = \frac{-2}{x^2 - 4}\). 4. The original restrictions are \(x \ne -2\) and \(x \ne 2\).

\(-\frac{2}{x^2 - 4}\), with \(x \ne -2\) and \(x \ne 2\).

- Factor the difference of squares in part 1. - In part 2, combine the numerator into one fraction first. - Use the relationship \(b-a=-(a-b)\).

1. Factor \(x^2-4=(x-2)(x+2)\). Then \(\frac{3x^2(x-2)(x+2)}{6x(x-2)}=\frac{x(x+2)}{2}\). The original expression requires \(x\ne0,2\). 2. Combine the numerator of the complex fraction: \(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}\). 3. Divide by \(a-b\): \(\frac{b-a}{ab}\cdot\frac{1}{a-b}\). 4. Since \(b-a=-(a-b)\), the result is \(-\frac{1}{ab}\). The original expression requires \(a\ne0\), \(b\ne0\), and \(a\ne b\).

1) \(\frac{x(x+2)}{2}\), where \(x\ne0,2\) 2) \(-\frac{1}{ab}\), where \(a\ne0\), \(b\ne0\), and \(a\ne b\)

- Factor each denominator before finding a common denominator. - In the complex fraction, combine the numerator and denominator separately. - Preserve restrictions from the original expression.

1. Factor the denominators: \(x^2-x=x(x-1)\) and \(x^2-1=(x-1)(x+1)\). 2. Use the common denominator \(x(x-1)(x+1)\): \(\frac{x+1-x}{x(x-1)(x+1)}=\frac{1}{x(x^2-1)}\). The original expression requires \(x\ne-1,0,1\). 3. For part 2, rewrite the numerator and denominator: \(y-\frac{1}{y}=\frac{y^2-1}{y}\) and \(1+\frac{1}{y}=\frac{y+1}{y}\). 4. Divide and factor: \(\frac{y^2-1}{y}\cdot\frac{y}{y+1}=\frac{(y-1)(y+1)}{y+1}=y-1\). 5. The original expression requires \(y\ne0\) and \(y\ne-1\).

1) \(\frac{1}{x(x^2-1)}\), where \(x\ne-1,0,1\) 2) \(y-1\), where \(y\ne-1,0\)

- Factor the difference of squares. - Combine the terms in parentheses before dividing. - Rewrite division as multiplication by the reciprocal.

1. Factor \(a^2-b^2=(a-b)(a+b)\). 2. Combine the terms in parentheses: \(\frac{a(a+b)-2ab}{(a-b)(a+b)}=\frac{a^2-ab}{(a-b)(a+b)}\). 3. Factor and cancel: \(\frac{a(a-b)}{(a-b)(a+b)}=\frac{a}{a+b}\). 4. Divide by the second rational expression: \(\frac{a}{a+b}\cdot\frac{a+b}{a-b}=\frac{a}{a-b}\). 5. The original expression requires \(a\ne b\) and \(a\ne-b\).

\(\frac{a}{a-b}\), where \(a\ne b\) and \(a\ne-b\)

- Combine the two fractions in parentheses first. - Expand the squared binomials carefully, including the subtraction sign. - Compare the resulting factor with the second fraction.

1. Combine the fractions in parentheses using the denominator \((x-3)(x+3)=x^2-9\): \(\frac{(x+3)^2-(x-3)^2}{x^2-9}\). 2. Expand the numerator: \((x^2+6x+9)-(x^2-6x+9)=12x\). 3. The expression becomes \(\frac{12x}{x^2-9}\cdot\frac{x^2-9}{12x}=1\). 4. The original expression requires \(x\ne-3,0,3\), so the claim is true for all allowed values.

The claim is true. The expression equals \(1\) for \(x\ne-3,0,3\).

- Factor the difference of squares to find the common denominator. - Distribute the subtraction sign across the middle numerator. - Simplify the parentheses before performing the final division.

1. Factor \(x^2-25=(x-5)(x+5)\), the common denominator inside the parentheses. 2. Combine the numerators: \(x(x+5)-(x^2+25)+5(x-5)=10x-50=10(x-5)\). 3. Simplify the expression in parentheses: \(\frac{10(x-5)}{(x-5)(x+5)}=\frac{10}{x+5}\). 4. Divide: \(\frac{10}{x+5}\div\frac{10}{x+5}=1\). 5. The original expression requires \(x\ne-5,5\).

\(1\), where \(x\ne-5,5\)

- Factor out \(2\), then use the difference of squares. - Recognize the perfect-square trinomial in the denominator. - Simplify before substituting the numerical values.

1. Factor \(2a^2-2b^2=2(a-b)(a+b)\) and \(a^2-2ab+b^2=(a-b)^2\). 2. Substitute and cancel: \(\frac{2(a-b)(a+b)}{(a-b)^2}\cdot\frac{a-b}{2}=a+b\). 3. The original expression requires \(a\ne b\). 4. Convert \(5\frac{3}{8}=5.375\), then evaluate: \(5.375+(-2.375)=3\).

a) \(a+b\), where \(a\ne b\) b) \(3\)

- Factor the difference of squares to find a common denominator. - Factor the combined numerator before multiplying. - Determine the domain from the original expression.

1. Factor \(x^2-9=(x-3)(x+3)\), the common denominator inside the parentheses. 2. Combine the numerators: \(x(x+3)-2(x-3)-12=x^2+x-6=(x+3)(x-2)\). 3. The parentheses simplify to \(\frac{(x+3)(x-2)}{(x-3)(x+3)}=\frac{x-2}{x-3}\). 4. Multiply by \(\frac{x-3}{x+2}\) and cancel \(x-3\): \(\frac{x-2}{x+2}\). 5. The original denominators require \(x\ne-3,-2,3\).

\(\frac{x-2}{x+2}\), with domain \(\mathbb{R}\setminus\{-3,-2,3\}\)

- Factor the denominator \(a^2+3a\). - Combine the fractions in parentheses first. - Rewrite division as multiplication by the reciprocal.

1. Factor \(a^2+3a=a(a+3)\). 2. Combine the terms in parentheses: \(\frac{a^2}{a(a+3)}+\frac{9}{a(a+3)}=\frac{a^2+9}{a(a+3)}\). 3. Multiply by the reciprocal: \(\frac{a^2+9}{a(a+3)}\cdot\frac{a}{a^2+9}\). 4. Cancel common factors to get \(\frac{1}{a+3}\). 5. The original expression requires \(a\ne0,-3\).

\(\frac{1}{a+3}\), where \(a\ne-3,0\)

- Factor the difference of squares. - Rewrite \(3-x\) as \(-(x-3)\). - Combine the terms in parentheses before dividing.

1. Factor \(x^2-9=(x-3)(x+3)\) and use \(3-x=-(x-3)\). 2. Combine the fractions in parentheses: \(\frac{2x-(x+3)}{(x-3)(x+3)}=\frac{x-3}{(x-3)(x+3)}=\frac{1}{x+3}\). 3. Divide by the second rational expression: \(\frac{1}{x+3}\cdot\frac{x+3}{x-3}=\frac{1}{x-3}\). 4. The original expression requires \(x\ne-3,3\).

\(\frac{1}{x-3}\), with domain \(\mathbb{R}\setminus\{-3,3\}\)

- Factor the difference of squares to find the common denominator. - Factor the combined numerator before canceling. - Rewrite the final division as multiplication by the reciprocal.

1. Factor \(x^2-1=(x-1)(x+1)\), the common denominator inside the parentheses. 2. Combine the numerators: \((x+1)+(x-1)-2=2x-2=2(x-1)\). 3. The parentheses simplify to \(\frac{2(x-1)}{(x-1)(x+1)}=\frac{2}{x+1}\). 4. Divide by the second rational expression: \(\frac{2}{x+1}\cdot\frac{x+1}{x-1}=\frac{2}{x-1}\). 5. The original expression requires \(x\ne-1,1\).

\(\frac{2}{x-1}\), where \(x\ne-1,1\)

- Use the difference of squares as the common denominator. - Expand the two squared binomials carefully. - Cancel only after writing the product in factored form.

1. Combine the fractions in parentheses over \((x-y)(x+y)=x^2-y^2\). 2. The numerator is \((x+y)^2-(x-y)^2=4xy\), so the parentheses simplify to \(\frac{4xy}{x^2-y^2}\). 3. Multiply and cancel: \(\frac{4xy}{x^2-y^2}\cdot\frac{x^2-y^2}{x}=4y\). 4. The original expression requires \(x\ne0\), \(x\ne y\), and \(x\ne-y\).

\(4y\), where \(x\ne0\), \(x\ne y\), and \(x\ne-y\)

- Check every original denominator and the numerator of the divisor. - Rewrite \(1\) using the denominator \(x+2\). - Multiply by the reciprocal and cancel common factors.

1. The original denominators require \(x+2\ne0\) and \(x^2+2x=x(x+2)\ne0\), so \(x\ne-2,0\). 2. Because the second rational expression is the divisor, its numerator must also be nonzero: \(x-2\ne0\), so \(x\ne2\). 3. Simplify the parentheses: \(\frac{2x}{x+2}-1=\frac{2x-(x+2)}{x+2}=\frac{x-2}{x+2}\). 4. Divide: \(\frac{x-2}{x+2}\cdot\frac{x(x+2)}{x-2}=x\).

a) \(x=-2,0,2\) b) \(x\), where \(x\ne-2,0,2\)

- Use the product of the two distinct linear denominators. - Expand both squared binomials. - Distribute the subtraction sign to every term in the second square.

1. The least common denominator is \((x - 2)(x + 2) = x^2 - 4\). 2. Combine: \(\frac{(x + 2)^2 - (x - 2)^2}{x^2 - 4}\). 3. Expand the numerator: \(x^2 + 4x + 4 - (x^2 - 4x + 4) = 8x\). 4. The result is \(\frac{8x}{x^2 - 4}\), with \(x \ne -2\) and \(x \ne 2\).

\(\frac{8x}{x^2 - 4}\), with \(x \ne -2\) and \(x \ne 2\).

- Factor the difference of squares. - Combine the fractions inside the parentheses first. - Keep restrictions from all original denominators.

1. Factor \(x^2-4=(x-2)(x+2)\), the common denominator inside the parentheses. 2. Combine the fractions: \(\frac{x(x+2)-2x}{(x-2)(x+2)}=\frac{x^2}{(x-2)(x+2)}\). 3. Multiply by \(\frac{x+2}{x^2}\) and cancel common factors to get \(\frac{1}{x-2}\). 4. The original expression requires \(x\ne-2,0,2\).

\(\frac{1}{x-2}\), where \(x\ne-2,0,2\)

- Rewrite \(1\) with denominator \(x-3\). - Factor the difference of squares. - Cancel common factors only after factoring.

1. Combine the terms in parentheses: \(\frac{x}{x-3}-1=\frac{x-(x-3)}{x-3}=\frac{3}{x-3}\). 2. Factor \(x^2-9=(x-3)(x+3)\). 3. Multiply and cancel: \(\frac{3}{x-3}\cdot\frac{(x-3)(x+3)}{3}=x+3\). 4. The original expression requires \(x\ne3\).

\(x+3\), where \(x\ne3\)

- Find a common denominator inside the parentheses. - Factor the perfect-square trinomial. - Remember that the divisor itself cannot equal zero.

1. Combine the fractions in parentheses: \(\frac{(a+b)-(a-b)}{(a-b)(a+b)}=\frac{2b}{(a-b)(a+b)}\). 2. Factor \(a^2+2ab+b^2=(a+b)^2\). 3. Multiply by the reciprocal: \(\frac{2b}{(a-b)(a+b)}\cdot\frac{(a+b)^2}{2b}=\frac{a+b}{a-b}\). 4. The original expression requires \(a\ne b\), \(a\ne-b\), and \(b\ne0\).

\(\frac{a+b}{a-b}\), where \(a\ne b\), \(a\ne-b\), and \(b\ne0\)

- Factor the numerator and denominator using binomial identities. - Simplify before substituting \(x=8\). - Preserve the restrictions from the original expression.

1. Factor \(x^2-16=(x-4)(x+4)\) and \(x^2-8x+16=(x-4)^2\). 2. Substitute and cancel: \(\frac{(x-4)(x+4)}{(x-4)^2}\cdot\frac{x-4}{x}=\frac{x+4}{x}\). 3. The original expression requires \(x\ne0,4\). 4. For \(x=8\), \(\frac{x+4}{x}=\frac{12}{8}=1.5\).

The simplified expression is \(\frac{x+4}{x}\), where \(x\ne0,4\). Its value at \(x=8\) is \(1.5\).

- Factor all three denominators. - Use one denominator containing each distinct factor. - Expand the combined numerator, then look for a factor to cancel.

1. Factor the denominators: \(a(a - b)\), \(a(a + b)\), and \((a - b)(a + b)\). 2. The least common denominator is \(a(a - b)(a + b)\). 3. The combined numerator is \((a + 2b)(a + b) - (a - 2b)(a - b) - 4ab\). 4. Expand and simplify: \((a^2 + 3ab + 2b^2) - (a^2 - 3ab + 2b^2) - 4ab = 2ab\). 5. Cancel \(a\): \(\frac{2ab}{a(a - b)(a + b)} = \frac{2b}{a^2 - b^2}\). 6. Evaluate: \(\frac{2(1)}{3^2 - 1^2} = \frac{2}{8} = 0.25\). 7. The original restrictions are \(a \ne 0\), \(a \ne b\), and \(a \ne -b\).

\(\frac{2b}{a^2 - b^2}\), with \(a \ne 0\) and \(a \ne \pm b\); value: \(0.25\).

- Factor a common \(x\) from the second denominator. - Look for a difference of squares after combining. - Domain restrictions come from the original expression.

1. Factor \(x^2 - 3x = x(x - 3)\). 2. Use the common denominator \(x(x - 3)\): \(\frac{x^2 - 9}{x(x - 3)}\). 3. Factor the numerator: \(x^2 - 9 = (x - 3)(x + 3)\). 4. Cancel \(x - 3\) to obtain \(\frac{x + 3}{x}\). 5. The original denominators require \(x \ne 0\) and \(x \ne 3\).

\(\frac{x + 3}{x}\), with domain \(\{x \in \mathbb{R} \mid x \ne 0, 3\}\).

- Use the least common multiple of the numerical denominator factors in part a. - In part b, use the product of the two distinct linear denominators. - Put the second numerator in parentheses before subtracting.

1. In part a, use denominator \(6x\): \(\frac{10 - 12 + 3}{6x} = \frac{1}{6x}\), with \(x \ne 0\). 2. In part b, use denominator \((x + 3)(x - 3) = x^2 - 9\). 3. Combine: \(\frac{2(x - 3) - (x + 3)}{x^2 - 9} = \frac{x - 9}{x^2 - 9}\), with \(x \ne -3\) and \(x \ne 3\).

a) \(\frac{1}{6x}\), with \(x \ne 0\). b) \(\frac{x - 9}{x^2 - 9}\), with \(x \ne -3\) and \(x \ne 3\).

- Write each combined function as one rational expression. - Zeros occur where the numerator is zero, provided the denominator is not zero. - For large positive \(x\), compare each function with its polynomial term. - Multiplying an entire function by \(-1\) reflects its graph across the x-axis.

1. Combine the terms in the sum: \(s(x)=x+\frac{2}{x}=\frac{x^2+2}{x}\). The numerator \(x^2+2\) is never zero for real \(x\), so \(s\) has no real zeros. This matches panel a). 2. For the first difference, \(d_1(x)=x-\frac{2}{x}=\frac{x^2-2}{x}\). Its zeros satisfy \(x^2-2=0\), so \(x=\pm\sqrt{2}\). As \(x\to\infty\), \(d_1(x)\) behaves like \(x\) and is positive. This matches panel b). 3. For the second difference, \(d_2(x)=\frac{2}{x}-x=-d_1(x)\). It has the same zeros, but its graph is the reflection of the graph of \(d_1\) across the x-axis. As \(x\to\infty\), it behaves like \(-x\). This matches panel c).

a) \(s(x)=f(x)+g(x)\) b) \(d_1(x)=f(x)-g(x)\) c) \(d_2(x)=g(x)-f(x)\)

- Treat the box as an unknown divisor. - Rewrite division by a fraction as multiplication by its reciprocal. - Check your result in the original equation.

1. Solve for the box: \(\square=\frac{9k}{m^2}\div\frac{3}{m}\). 2. Multiply by the reciprocal: \(\frac{9k}{m^2}\cdot\frac{m}{3}\). 3. Simplify: \(\frac{9km}{3m^2}=\frac{3k}{m}\). 4. Substitution verifies that \(\frac{9k}{m^2}\div\frac{3k}{m}=\frac{3}{m}\).

\(\square=\frac{3k}{m}\), where \(m\ne0\) and \(k\ne0\)

- Find the values that make either denominator zero. - Multiply the two linear denominators to obtain a common denominator. - Examine how \(x\) appears in the simplified expression.

1. The denominators are zero at \(x = -2\) and \(x = 2\), so the domain is all real numbers except \(-2\) and \(2\). 2. Use the common denominator \((x + 2)(2 - x) = 4 - x^2\): \(A(x) = \frac{3(2 - x) + 3(x + 2)}{4 - x^2} = \frac{12}{4 - x^2}\). 3. Evaluate: \(A(1) = \frac{12}{4 - 1} = 4\) and \(A(-1) = \frac{12}{4 - 1} = 4\). 4. The variable occurs only as \(x^2\). Since \((-k)^2 = k^2\), replacing \(k\) by \(-k\) leaves the denominator, and therefore the function value, unchanged.

a) Domain: \(\{x \in \mathbb{R} \mid x \ne -2, 2\}\); \(A(x) = \frac{12}{4 - x^2}\). b) \(A(1) = 4\) and \(A(-1) = 4\); the values are equal. c) Because \((-k)^2 = k^2\), \(A(-k) = \frac{12}{4 - (-k)^2} = \frac{12}{4 - k^2} = A(k)\).

- Find a common denominator and combine the numerators. - Substitute each proposed value into the simplified expression. - In part c, replace every \(x\) by \(-(x + 1)\) and expand carefully.

1. The denominators require \(x \ne 0\) and \(x \ne -1\). 2. Combine the fractions: \(B(x) = \frac{x + 1 - x}{x(x + 1)} = \frac{1}{x(x + 1)} = \frac{1}{x^2 + x}\). 3. Evaluate: \(B(2) = \frac{1}{4 + 2} = \frac{1}{6}\) and \(B(-3) = \frac{1}{9 - 3} = \frac{1}{6}\). Both values satisfy the equation. 4. Substitute \(-(x + 1)\) into the simplified expression: \(B(-(x + 1)) = \frac{1}{(-(x + 1))^2 - (x + 1)}\). 5. Simplify the denominator: \((x + 1)^2 - x - 1 = x^2 + x\). Therefore, \(B(-(x + 1)) = \frac{1}{x^2 + x} = B(x)\).

a) Domain: \(\{x \in \mathbb{R} \mid x \ne 0, -1\}\); \(B(x) = \frac{1}{x^2 + x}\). b) Both \(x = 2\) and \(x = -3\) satisfy the equation. c) Substitution and simplification give \(B(-(x + 1)) = \frac{1}{x^2 + x} = B(x)\).

- Find the value that makes a denominator zero. - Use \(2x\) as the common denominator. - Dividing by a fraction is equivalent to multiplying by its reciprocal.

1. The denominator requires \(x \ne 0\), so the domain is all nonzero real numbers. 2. Use the common denominator \(2x\): \(f(x) = \frac{4}{2x} + \frac{x^2}{2x} = \frac{x^2 + 4}{2x}\). 3. Substitute \(\frac{4}{x}\) into the original form: \(f\left(\frac{4}{x}\right) = \frac{2}{4/x} + \frac{4/x}{2} = \frac{x}{2} + \frac{2}{x} = f(x)\). 4. Evaluate: \(f(1) = 2 + \frac{1}{2} = 2.5\) and \(f(4) = \frac{1}{2} + 2 = 2.5\).

a) Domain: \(\{x \in \mathbb{R} \mid x \ne 0\}\); \(f(x) = \frac{x^2 + 4}{2x}\). b) \(f\left(\frac{4}{x}\right) = \frac{x}{2} + \frac{2}{x} = f(x)\). c) \(f(1) = 2.5\) and \(f(4) = 2.5\).

- Simplify each complex fraction by multiplying by the reciprocal. - Combine the two resulting rational expressions. - Determine when \(\frac{n + 1}{2}\) is a natural number and when the two denominators can be equal.

1. For \(n = 7\), the right-hand side is \(\frac{1}{4} + \frac{1}{28} = \frac{7}{28} + \frac{1}{28} = \frac{8}{28} = \frac{2}{7}\). 2. In general, simplify the complex fractions: \(\frac{1}{(n + 1)/2} = \frac{2}{n + 1}\) and \(\frac{1}{n(n + 1)/2} = \frac{2}{n(n + 1)}\). 3. Combine: \(\frac{2}{n + 1} + \frac{2}{n(n + 1)} = \frac{2n + 2}{n(n + 1)} = \frac{2(n + 1)}{n(n + 1)} = \frac{2}{n}\). 4. If \(n\) is odd, then \(n + 1\) is even, so both displayed denominators are natural numbers. If \(n > 1\), the second denominator is \(n\) times the first and is therefore different from it.

a) \(\frac{2}{7} = \frac{1}{4} + \frac{1}{28}\) b) \(\frac{2}{n + 1} + \frac{2}{n(n + 1)} = \frac{2n + 2}{n(n + 1)} = \frac{2}{n}\). c) Odd \(n\) makes both unit-fraction denominators natural numbers, and \(n > 1\) makes the two denominators different.

- Find the least common denominator of the two expressions on the right. - Match \(a\) and \(b\) with the factors \(3\) and \(5\). - Test both directions of the claim, including a prime value and the boundary case \(n = 1\).

1. The least common denominator on the right is \(ab(a + b)\). 2. Combine: \(\frac{b}{ab(a + b)} + \frac{a}{ab(a + b)} = \frac{a + b}{ab(a + b)} = \frac{1}{ab}\). 3. For \(a = 3\) and \(b = 5\), \(\frac{1}{15} = \frac{1}{3(8)} + \frac{1}{5(8)} = \frac{1}{24} + \frac{1}{40}\). 4. The claim is false. For any natural number \(n > 1\), including a prime, choose \(a = 1\) and \(b = n\). Then \(\frac{1}{n} = \frac{1}{n + 1} + \frac{1}{n(n + 1)}\), and the unit fractions are different. 5. For \(n = 1\), the only positive factorization is \(1 = 1 \cdot 1\), which gives two equal unit fractions. Thus, this identity produces two different unit fractions for every natural number \(n > 1\), not only for composite \(n\).

a) \(\frac{b + a}{ab(a + b)} = \frac{1}{ab}\) b) \(\frac{1}{15} = \frac{1}{24} + \frac{1}{40}\) c) The claim is false. Every \(n > 1\), including prime \(n\), works by choosing \(a = 1\) and \(b = n\); for \(n = 1\), the two unit fractions produced by the identity are equal.

- Factor the difference of squares. - Rewrite \(1\) as a fraction with the least common denominator. - Domain restrictions come from the original expression, not only the simplified formula.

1. Factor \(x^2 - 9 = (x - 3)(x + 3)\), so the least common denominator is \((x - 3)(x + 3)\). 2. Rewrite all terms: \(A(x) = \frac{x(x + 3) - 18 - (x - 3)(x + 3)}{(x - 3)(x + 3)}\). 3. Simplify the numerator: \(x^2 + 3x - 18 - (x^2 - 9) = 3x - 9 = 3(x - 3)\). 4. Cancel \(x - 3\) to obtain \(A(x) = \frac{3}{x + 3}\). 5. The original restrictions remain \(x \ne -3\) and \(x \ne 3\). 6. At \(x = 3\), the original denominators \(x - 3\) and \(x^2 - 9\) are zero. Cancellation does not restore an input excluded from the original domain.

a) \((x - 3)(x + 3)\) b) \(\frac{3}{x + 3}\), with \(x \ne -3\) and \(x \ne 3\). c) At \(x = 3\), the original expression requires division by zero, so \(3\) remains excluded.

- For division by a rational expression, check both denominators and require the divisor to be nonzero. - Rewrite division as multiplication by the reciprocal. - Track each original restriction through the reciprocal and cancellation steps.

1. The first denominator requires \(x \ne -1\). The denominator of the divisor requires \(x^2 - 1 \ne 0\), so \(x \ne -1\) and \(x \ne 1\). The divisor itself must also be nonzero, so \(x \ne 0\). Therefore, the original domain is \(D_{\text{original}} = \mathbb{R} \setminus \{-1, 0, 1\}\). 2. Rewrite division as multiplication by the reciprocal: \(\frac{1}{x + 1} \cdot \frac{x^2 - 1}{x}\). 3. Factor and simplify: \(\frac{1}{x + 1} \cdot \frac{(x - 1)(x + 1)}{x} = \frac{x - 1}{x}\). 4. The simplified expression has domain \(D_{\text{simplified}} = \mathbb{R} \setminus \{0\}\). 5. The restriction \(x \ne -1\) is hidden when the factor \(x + 1\) is canceled. The restriction \(x \ne 1\) came from the denominator of the original divisor; after taking the reciprocal, the factor \(x - 1\) appears in the numerator. Both values must still be excluded when the simplified expression is used as an equivalent form of the original.

1. The original expression is undefined at \(x = -1\), \(x = 0\), and \(x = 1\). 2. The simplified expression is \(\frac{x - 1}{x}\). 3. Its own domain excludes only \(x = 0\). The restrictions \(x = -1\) and \(x = 1\) are no longer visible after taking the reciprocal and simplifying, but they remain excluded from the original domain.

- Relate \(1 - x\) to \(x - 1\). - Factor the difference of squares in the first denominator. - After combining, look for a perfect-square trinomial.

1. Factor \(x^2 - 1 = (x - 1)(x + 1)\). 2. Since \(1 - x = -(x - 1)\), rewrite \(\frac{2}{1 - x} = -\frac{2}{x - 1}\). 3. Use the common denominator \((x - 1)(x + 1)\): \(\frac{x^2 + 3 - 2(x + 1)}{(x - 1)(x + 1)}\). 4. Simplify and factor the numerator: \(x^2 - 2x + 1 = (x - 1)^2\). 5. Cancel one factor of \(x - 1\) to obtain \(\frac{x - 1}{x + 1}\). 6. Preserve the original restrictions \(x \ne -1\) and \(x \ne 1\).

\(\frac{x - 1}{x + 1}\), with \(x \ne -1\) and \(x \ne 1\).

- Rewrite division as multiplication by the reciprocal. - Factor before canceling. - Use \(a(a+1)\) as the common denominator for the final addition.

1. Rewrite the division as multiplication by the reciprocal: \(\frac{2a-4}{a+1}\cdot\frac{a}{a-2}\). 2. Factor \(2a-4=2(a-2)\) and cancel \(a-2\): \(\frac{2a}{a+1}\). 3. Factor \(a^2+a=a(a+1)\). Write the first term with this denominator: \(\frac{2a^2}{a(a+1)}\). 4. Add: \(\frac{2a^2}{a(a+1)}+\frac{3}{a(a+1)}=\frac{2a^2+3}{a(a+1)}\). 5. The original expression requires \(a\ne-1,0,2\).

\(\frac{2a^2+3}{a(a+1)}\), where \(a\ne-1,0,2\)

- Isolate \(A\) first. - Use \(x(x - 2)\) as the common denominator. - Factor the final numerator and check for cancellation.

1. Isolate \(A\): \(A = \frac{5x - 6}{x(x - 2)} - \frac{3}{x}\). 2. Rewrite the second fraction with denominator \(x(x - 2)\): \(\frac{3(x - 2)}{x(x - 2)}\). 3. Subtract: \(A = \frac{5x - 6 - 3(x - 2)}{x(x - 2)} = \frac{2x}{x(x - 2)}\). 4. Cancel \(x\) to obtain \(A = \frac{2}{x - 2}\). 5. The original equation is defined only for \(x \ne 0\) and \(x \ne 2\).

\(A = \frac{2}{x - 2}\), with \(x \ne 0\) and \(x \ne 2\).

- Treat each blank as an unknown in an equation. - Isolate the missing expression first. - Rewrite whole-number or variable terms with the needed denominator.

1. In part a, subtract \(\frac{5}{x}\) from the right side: \(\frac{5 + x^2 - 5}{x} = \frac{x^2}{x} = x\), with \(x \ne 0\). 2. In part b, add \(\frac{1}{a}\) to \(a\): \(a + \frac{1}{a} = \frac{a^2 + 1}{a}\), with \(a \ne 0\). 3. In part c, subtract \(2\) from \(\frac{2z + 4}{z}\): \(\frac{2z + 4 - 2z}{z} = \frac{4}{z}\), with \(z \ne 0\).

a) \(x\), with \(x \ne 0\). b) \(\frac{a^2 + 1}{a}\), with \(a \ne 0\). c) \(\frac{4}{z}\), with \(z \ne 0\).

- Factor every denominator completely. - Include the highest power of each factor in the least common denominator. - Distribute the subtraction sign through the entire third numerator.

1. Factor the denominators: \(a^2 - b^2 = (a - b)(a + b)\), \(a^2 + 2ab + b^2 = (a + b)^2\), and \(a^2 + ab = a(a + b)\). 2. The least common denominator is \(a(a - b)(a + b)^2\). 3. The combined numerator is \(a(a + b) + a(a - b) - 2(a - b)(a + b)\). 4. Simplify: \(a^2 + ab + a^2 - ab - 2(a^2 - b^2) = 2b^2\). 5. The result is \(\frac{2b^2}{a(a - b)(a + b)^2}\), with \(a \ne 0\), \(a \ne b\), and \(a \ne -b\).

\(\frac{2b^2}{a(a - b)(a + b)^2}\), with \(a \ne 0\), \(a \ne b\), and \(a \ne -b\).

- Use the difference of squares as the common denominator. - Expand the two squared binomials carefully. - For the domain, exclude values that make a denominator or the divisor zero.

1. Combine the fractions in parentheses over \((a-b)(a+b)=a^2-b^2\). 2. The numerator is \((a+b)^2-(a-b)^2=4ab\), so the parentheses simplify to \(\frac{4ab}{a^2-b^2}\). 3. Divide by the second rational expression: \(\frac{4ab}{a^2-b^2}\cdot\frac{a^2-b^2}{8ab}=\frac{1}{2}\). 4. The original denominators require \(a\ne b\) and \(a\ne-b\). The divisor must also be nonzero, so \(a\ne0\) and \(b\ne0\).

a) \(\frac{1}{2}\) b) \(\{(a, b)\in\mathbb{R}^2\mid a\ne b,\ a\ne-b,\ a\ne0,\ b\ne0\}\)

- Write each parenthetical expression as one fraction. - Factor both differences of squares. - Exclude values that make the divisor equal to zero.

1. Combine the first parentheses: \(\frac{y}{y+1}+1=\frac{2y+1}{y+1}\). 2. Combine the second parentheses: \(1-\frac{3y^2}{1-y^2}=\frac{1-4y^2}{1-y^2}\). 3. Factor \(1-4y^2=(1-2y)(1+2y)\) and \(1-y^2=(1-y)(1+y)\). 4. Multiply by the reciprocal: \(\frac{2y+1}{y+1}\cdot\frac{(1-y)(1+y)}{(1-2y)(1+2y)}\). 5. Cancel \(2y+1=1+2y\) and \(y+1=1+y\) to get \(\frac{1-y}{1-2y}\). 6. The original expression requires \(y\ne-1,1\), and the divisor must be nonzero, so \(y\ne-\frac{1}{2},\frac{1}{2}\).

\(\frac{1-y}{1-2y}\), where \(y\ne-1,-\frac{1}{2},\frac{1}{2},1\)

- Factor the quadratic denominators. - Combine the fractions in parentheses before dividing. - Determine excluded values from every original denominator.

1. Use the common denominator \((k-2)(k+2)=k^2-4\): \(\frac{1}{k+2}-\frac{k}{k^2-4}=\frac{k-2-k}{k^2-4}=\frac{-2}{(k-2)(k+2)}\). 2. Multiply by the reciprocal of the divisor and factor \(k^2+2k=k(k+2)\): \(\frac{-2}{(k-2)(k+2)}\cdot\frac{k(k+2)}{2}=-\frac{k}{k-2}\). 3. For \(k=6\), \(-\frac{k}{k-2}=-\frac{6}{4}=-1.5\). 4. The original denominators are zero at \(k=-2,0,2\), so those values are excluded.

a) \(-\frac{k}{k-2}\), where \(k\ne-2,0,2\) b) \(-1.5\) c) \(k=-2,0,2\)

- Factor the third denominator by grouping. - Combine all numerators over the factored common denominator. - Factor the resulting numerator by grouping before canceling.

1. Factor the third denominator by grouping: \(a^2 - ab + ac - bc = (a - b)(a + c)\). 2. Use \((a - b)(a + c)\) as the least common denominator. 3. The combined numerator is \(2a(a + c) + a(a - b) - (a^2 + 3ac + ab - bc)\). 4. Simplify: \(2a^2 + 2ac + a^2 - ab - a^2 - 3ac - ab + bc = 2a^2 - ac - 2ab + bc\). 5. Factor by grouping: \(2a(a - b) - c(a - b) = (2a - c)(a - b)\). 6. Cancel \(a - b\) to obtain \(\frac{2a - c}{a + c}\). 7. Evaluate: \(\frac{2(3) - 2}{3 + 2} = \frac{4}{5} = 0.8\). 8. The original restrictions are \(a \ne b\) and \(a \ne -c\).

\(\frac{2a - c}{a + c}\), with \(a \ne b\) and \(a \ne -c\); value: \(0.8\).

- Rewrite reversed differences by factoring out \(-1\). - Use one product containing each distinct denominator factor. - After expanding, group terms to recover factors from the denominator.

1. Rewrite \(b - a = -(a - b)\), \(c - a = -(a - c)\), and \(c - b = -(b - c)\). 2. Use the common denominator \((a - b)(a - c)(b - c)\). 3. The combined numerator is \(a^2(b - c) - b^2(a - c) + c^2(a - b)\). 4. Expand and regroup: \(a^2b - a^2c - ab^2 + b^2c + ac^2 - bc^2\). 5. Factor: \((b - c)(a^2 - a(b + c) + bc) = (b - c)(a - b)(a - c)\). 6. The numerator equals the denominator, so the expression simplifies to \(1\).

\(1\)