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Algebra II : Solving Rational Expressions

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Example Questions

Example Question #1742 : Algebra Ii

Subtract: $\frac{3}{8x}-\frac{6}{16-x}$

Possible Answers:

$\frac{51x-48}{8x^2+128}$

$\frac{12x-6}{8x+9}$

$\frac{51x+48}{8x^2+128}$

$\frac{51x+45}{8x^2+128}$

$\frac{51x-45}{8x^2+128}$

Correct answer:

$\frac{51x-48}{8x^2+128}$

Explanation:

In order to simplify this rational expression, we will need to determine the least common denominator.

8x(16-x) = 128x-8x^2

Convert the fractions.

$\frac{3(16-x)}{8x(16-x)}-\frac{6(8x)}{(16-x)(8x)}$

Simplify the fractions.

$\frac{48-3x}{ 128x-8x^2}-\frac{48x}{ 128x-8x^2}$

Simplify by combining like terms. Combine as one fraction.

$\frac{48-51x}{ 128x-8x^2}$

To rewrite this in the order of powers, we can factor a negative one on the top and bottom.

$\frac{-(-48+51x)}{- (128x+8x^2)} = \frac{51x-48}{8x^2+128}$

The answer is: $\frac{51x-48}{8x^2+128}$

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Example Question #81 : Rational Expressions

Subtract: $\frac{1}{x}-\frac{x-3}{x^2+4x-21}$

Possible Answers:

$\frac{7}{x^2+7x}$

$\frac{x-21}{x^3+4x^2-21x}$

$-\frac{x-7}{x^3+4x^2-21x}$

$-\frac{7}{x^2+7x}$

$\frac{x^2-4}{x^3+4x^2-21x}$

Correct answer:

$\frac{7}{x^2+7x}$

Explanation:

Notice that the second fraction can be simplified. Factorize the denominator.

x^2+4x-21 = (x-3)(x+7)

Rewrite the expression.

$\frac{1}{x}-\frac{x-3}{x^2+4x-21}=\frac{1}{x}-\frac{x-3}{(x-3)(x+7)}$

Reduce the fraction.

$\frac{1}{x}-\frac{1}{x+7}$

Multiply both denominators together to determine the least common denominator.

x(x+7) =x^2+7x

Convert the fractions and solve.

$\frac{x+7}{x(x+7)}-\frac{x}{x(x+7)} = \frac{x+7}{x^2+7x}-\frac{x}{x^2+7x} =\frac{7}{x^2+7x}$

The answer is: $\frac{7}{x^2+7x}$

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Example Question #82 : Rational Expressions

Add the expression: $\frac{3x}{x+1} + \frac{7}{9x}$

Possible Answers:

$\frac{27x^2+7x+7}{9x^2+9x}$

$\frac{3x^2+7}{9x^2+9x}$

$\frac{2x+22}{3x+3}$

$\frac{3x^2+3x+7}{3x^2+9x}$

$\frac{x+22}{3x+3}$

Correct answer:

$\frac{27x^2+7x+7}{9x^2+9x}$

Explanation:

Determine the least common denominator.

9x(x+1) = 9x^2+9x

Convert the fractions.

$\frac{3x(9x)}{(x+1)(9x)} + \frac{7(x+1)}{(9x)(x+1)}$

The expression becomes:

$\frac{3x(9x)+7(x+1)}{9x^2+9x} = \frac{27x^2+7x+7}{9x^2+9x}$

The answer is: $\frac{27x^2+7x+7}{9x^2+9x}$

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Example Question #83 : Rational Expressions

Add: $\frac{9}{x}-\frac{x}{x-1}$

Possible Answers:

$-\frac{10}{x-1}$

$-\frac{x^2+9x+9}{x^2-1}$

$-\frac{x-9}{x^2-x}$

$-\frac{x^2-9x+9}{x^2-x}$

$-\frac{x^2-9}{x^2-x}$

Correct answer:

$-\frac{x^2-9x+9}{x^2-x}$

Explanation:

Determine the least common denominator.

x(x-1)= x^2-x

Convert the fractions.

$\frac{9}{x}-\frac{x}{x-1}=\frac{9(x-1)}{x^2-x}-\frac{x(x)}{x^2-x}$

Simplify the numerator.

$\frac{9(x-1)}{x^2-x}-\frac{x(x)}{x^2-x} = \frac{9x-9}{x^2-x}-\frac{x^2}{x^2-x}$

Combine as one fraction.

$\frac{-x^2+9x-9}{x^2-x}$

Pull out a common factor of negative one. This will allow us to pull the negative in front of the fraction.

$\frac{-1(x^2-9x+9)}{x^2-x}=-\frac{x^2-9x+9}{x^2-x}$

The answer is: $-\frac{x^2-9x+9}{x^2-x}$

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Example Question #84 : Rational Expressions

Add: $\frac{1}{3x+9}+\frac{2}{x}$

Possible Answers:

$\frac{5x+9}{3x^2+9x}$

$\frac{5x+18}{3x^2+9x}$

$\textup{The answer is not given.}$

$\frac{5}{3x+9}$

$\frac{7x+18}{3x^2+9x}$

Correct answer:

$\frac{7x+18}{3x^2+9x}$

Explanation:

In order to add both terms, we will need to find the least common denominator.

Multiply the denominators together.

x(3x+9) = 3x^2+9x

Convert the two fractions.

$\frac{1}{3x+9}+\frac{2}{x}=\frac{x}{3x^2+9x}+\frac{2(3x+9)}{3x^2+9x} =\frac{7x+18}{3x^2+9x}$

The answer is: $\frac{7x+18}{3x^2+9x}$

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Example Question #85 : Rational Expressions

Subtract: $\frac{1}{3x}-\frac{3}{10x}$

Possible Answers:

$-\frac{1}{30x^2}$

$\frac{1}{30x}$

$-\frac{2}{7x}$

$\frac{1}{30}x$

$\frac{1}{30x^2}$

Correct answer:

$\frac{1}{30x}$

Explanation:

In order to subtract the numerators, we will need to determine the least common denominator. Upon visualization, the denominators share an x term. This means that we will not have to change the x term.

Multiply the first denominator by ten, and the second denominator by three.

$\frac{1}{3x}-\frac{3}{10x}=\frac{1(10)}{3x(10)}-\frac{3(3)}{10x(3)}$

Simplify the numerator.

$\frac{1(10)}{3x(10)}-\frac{3(3)}{10x(3)} = \frac{10}{30x}- \frac{9}{30x}$

The answer is: $\frac{1}{30x}$

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Example Question #604 : Intermediate Single Variable Algebra

Subtract: $\frac{3}{2-x}-\frac{5}{2x}$

Possible Answers:

$-\frac{11x-10}{2x^2-2x}$

$-\frac{x+10}{2x^2-8x}$

$\frac{x+10}{2x^2-2x}$

$-\frac{4x+10}{2x^2-2x}$

$-\frac{11x-10}{2x^2-4x}$

Correct answer:

$-\frac{11x-10}{2x^2-4x}$

Explanation:

Multiply the denominators to get the least common denominator. We can then convert both fractions so that the denominators are alike.

$\frac{3}{2-x}-\frac{5}{2x}=\frac{3(2x)}{(2-x)(2x)}-\frac{5(2-x)}{2x(2-x)}$

Simplify both the top and the bottom.

$\frac{6x}{-2x^2+4x} - \frac{10-5x}{-2x^2+4x}$

Combine the numerators as one fraction. Be careful with the second fraction since the entire numerator is a quantity, which means we will need to brace 10-5x with parentheses.

$\frac{6x-(10-5x)}{-2x^2+4x} =\frac{6x-10+5x}{-2x^2+4x} = \frac{11x-10}{-2x^2+4x}$

Pull out a common factor of negative one in the denominator. This allows us to rewrite the fraction with the negative sign in front of the fraction.

$\frac{11x-10}{-2x^2+4x}= \frac{11x-10}{-(2x^2-4x)}= -\frac{11x-10}{2x^2-4x}$

The answer is: $-\frac{11x-10}{2x^2-4x}$

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Example Question #611 : Intermediate Single Variable Algebra

Solve: $\frac{7}{5x}+\frac{9}{10x}$

Possible Answers:

$\frac{23}{10}x$

$\frac{25}{10}x$

$\frac{63}{50x}$

$\frac{25}{10x}$

$\frac{23}{10x}$

Correct answer:

$\frac{23}{10x}$

Explanation:

In order to solve this expression, we will need to determine the least common denominator. Notice that the denominators both share an x term. We do not need to change that.

Multiply the denominator of the first fraction by two to get the least common denominator, which is 10x .

$\frac{7(2)}{5x(2)}+\frac{9}{10x} = \frac{14}{10x}+\frac{9}{10x}$

Add the numerators.

The answer is: $\frac{23}{10x}$

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Example Question #88 : Rational Expressions

Solve: $\frac{9x}{x+1}-\frac{9}{x}$

Possible Answers:

$\frac{9x^2-9x+9}{x^2+x}$

$\frac{9-x}{x}$

$\frac{9x-9}{x}$

$\frac{9x^2-9x-9}{x^2+x}$

$\frac{9x}{x+1}$

Correct answer:

$\frac{9x^2-9x-9}{x^2+x}$

Explanation:

Determine the least common denominator by multiplying the denominators together.

x(x+1) = x^2+x

Convert the fractions given.

$\frac{9x}{x+1}-\frac{9}{x}=\frac{(x)(9x)}{(x)(x+1)}-\frac{9(x+1)}{x(x+1)}$

Simplify the numerators and denominators.

$\frac{9x^2}{x^2+x}-\frac{9x+9}{x^2+x}$

Combine the terms as one fraction. Make sure to brace the 9x+9 term since this is a quantity.

$\frac{9x^2}{x^2+x}-\frac{9x+9}{x^2+x} = \frac{9x^2-(9x+9)}{x^2+x}$

The answer is: $\frac{9x^2-9x-9}{x^2+x}$

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Example Question #1751 : Algebra Ii

Add: $\frac{4x}{x-1}-\frac{10}{x}$

Possible Answers:

$\frac{4x^2-10}{x^2-x-10}$

$\frac{4x+10}{x-1}$

$\frac{4x^2-10x+10}{x^2-x}$

$\frac{4x-10}{x-1}$

$\frac{4x^2-10x-10}{x^2-x}$

Correct answer:

$\frac{4x^2-10x+10}{x^2-x}$

Explanation:

Identify the least common denominator by multiplying the denominators together.

Convert the fractions.

$\frac{4x}{x-1}-\frac{10}{x} = \frac{4x(x)}{x(x-1)}-\frac{10(x-1)}{x(x-1)}$

Simplify the numerator and denominator.

$\frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}$

Combine both fractions as one. Make sure to enclose the second number in parentheses since the negative sign is distributive.

$\frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}=\frac{4x^2-(10x-10)}{x^2-x} =\frac{4x^2-10x+10}{x^2-x}$

The answer is: $\frac{4x^2-10x+10}{x^2-x}$

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Algebra II : Solving Rational Expressions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1742 : Algebra Ii

Example Question #81 : Rational Expressions

Example Question #82 : Rational Expressions

Example Question #83 : Rational Expressions

Example Question #84 : Rational Expressions

Example Question #85 : Rational Expressions

Example Question #604 : Intermediate Single Variable Algebra

Example Question #611 : Intermediate Single Variable Algebra

Example Question #88 : Rational Expressions

Example Question #1751 : Algebra Ii

All Algebra II Resources

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