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Algebra II : Adding and Subtracting Rational Expressions

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Example Question #90 : Rational Expressions

Add: $\frac{7}{9x}+\frac{2(4-x)}{x}$

Possible Answers:

$\frac{9x-15}{9x}$

$\frac{9x-79}{9x}$

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

$\frac{-18x+79}{9x}$

$\frac{-18x+79}{9}$

Correct answer:

$\frac{-18x+79}{9x}$

Explanation:

Evaluate by changing the denominator of the second fraction so that both fractions have similar denominators.

$\frac{7}{9x}+\frac{2(4-x)}{x} = \frac{7}{9x}+\frac{2[9](4-x)}{[9]x}$

Use distribution to simplify the numerator.

$\frac{7}{9x}+\frac{2[9](4-x)}{[9]x} =\frac{7}{9x}+\frac{72-18x}{9x}$

Combine like-terms.

The answer is: $\frac{-18x+79}{9x}$

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Example Question #1751 : Algebra Ii

Add: $\frac{3}{10x}-\frac{10}{x}$

Possible Answers:

$\frac{7}{10-x}$

$-\frac{97}{10}x$

$-\frac{7}{10x}$

$-\frac{97}{10x}$

$-\frac{7}{10}x$

Correct answer:

$-\frac{97}{10x}$

Explanation:

In order to subtract the fractions, we will need to determine the least common denominator.

Multiply the second fraction's numerator and denominator by ten, since both fractions share an x-term.

$\frac{3}{10x}-\frac{10(10)}{x(10)} = \frac{3}{10x}-\frac{100}{10x}$

Combine the numerator to form one fraction.

The answer is: $-\frac{97}{10x}$

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Example Question #1752 : Algebra Ii

Solve: $\frac{x}{2x-7}+ \frac{7x}{x-7}$

Possible Answers:

$\frac{15x^2-42x}{2x^2-21x+49}$

$\frac{14x^2-49x}{2x^2-21x+49}$

$\frac{15x-56}{2x^2-21x+49}$

$\frac{15x^2-56x}{2x^2-21x+49}$

$\frac{15x}{2x-7}$

Correct answer:

$\frac{15x^2-56x}{2x^2-21x+49}$

Explanation:

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

Use the FOIL method to multiply the two denominators together.

(2x-7)(x-7) = 2x^2-7x-14x+49

Simplify by combining like-terms. The LCD is: 2x^2-21x+49

Change the fractions.

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

Simplify the fractions.

$\frac{x^2-7x}{2x^2-21x+49}+\frac{14x^2-49x}{2x^2-21x+49}$

Combine like-terms.

The answer is: $\frac{15x^2-56x}{2x^2-21x+49}$

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Example Question #1751 : Algebra Ii

Add: $\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x}$

Possible Answers:

$\frac{49}{12x}$

$\frac{14}{3x}$

$\frac{49}{24x}$

$\frac{49}{12}x$

$\frac{14}{3}x$

Correct answer:

$\frac{49}{12x}$

Explanation:

Determine the least common denominator in order to add the numerator.

Each denominator shares an term. The least common denominator is since it is divisible by each coefficient of the denominator.

Convert the fractions.

$\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x} = \frac{3(6)}{2x(6)}+\frac{4(4)}{3x(4)}+\frac{5(3)}{4x(3)}$

Simplify the top and bottom.

$\frac{18}{12x}+\frac{16}{12x}+\frac{15}{12x}=\frac{49}{12x}$

The answer is: $\frac{49}{12x}$

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Example Question #1754 : Algebra Ii

Solve: $\frac{2}{3x}+\frac{x}{4}-\frac{1}{x}$

Possible Answers:

$\frac{1}{4}x-\frac{1}{3x}$

$\frac{1}{12}x$

$\frac{1}{12x}$

$\frac{3}{4}x-\frac{4}{3x}$

$\frac{1}{4x}-\frac{1}{3}x$

Correct answer:

$\frac{1}{4}x-\frac{1}{3x}$

Explanation:

Determine the least common denominator. Each term will need an x-variable, and all three denominators will need a common coefficient.

The least common denominator is 12x .

Convert the fractions.

$\frac{2(4)}{3x(4)}+\frac{x(3x)}{4(3x)}-\frac{1(12)}{x(12)} = \frac{8}{12x}+\frac{3x^2}{12x}-\frac{12}{12x}$

The expression becomes:

$\frac{3x^2-4}{12x} = \frac{x}{4}-\frac{1}{3x}$

The answer is: $\frac{1}{4}x-\frac{1}{3x}$

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Example Question #1755 : Algebra Ii

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

Possible Answers:

$-\frac{x^2-3x-2}{x^2-1}$

$-\frac{x^2+3x+2}{x^2-1}$

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

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

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

Correct answer:

$-\frac{x^2-3x-2}{x^2-1}$

Explanation:

Notice that x^2-1 can be factorized by the difference of squares. This means that the second fraction will need to multiply the quantity of (x-1) in order for the denominators to be similar.

Convert the fractions.

$\frac{x+3}{x^2-1}-\frac{(x-1)(x-1)}{(x+1)(x-1)}$

Simplify the fractions and combine as one fraction.

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

The expression becomes:

$\frac{x+3-x^2+2x-1}{x^2-1}$

Combine like-terms.

$\frac{-x^2+3x+2}{x^2-1}$

Factor out a negative one from the numerator to pull the negative sign in front of the fraction.

$\frac{-(x^2-3x-2)}{x^2-1}$

The answer is: $-\frac{x^2-3x-2}{x^2-1}$

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Example Question #51 : Solving Rational Expressions

For all positive values of x, which of the following expressions is equivalent to the following:

$\frac{6}{x}+\frac{x+6}{6}$

Possible Answers:

$\frac{x+6}{x}$

$\frac{x^2+36}{6x}$

$\frac{x^2+6x+36}{6x}$

x+6

$\frac{x^2+5x+6}{6x}$

Correct answer:

$\frac{x^2+6x+36}{6x}$

Explanation:

In order to add the numerators in this expression, we will need to calculate the least common denominator. In this case, the least common denominator is . Let's multiply the fraction on the left by a value of one. Let's start with the left expression. We will multiply it by one in the following form:

$1=\frac{6}{6}$

$\frac{6}{6}\times\frac{6}{x}=\frac{36}{6x}$

Now convert the right fraction to a form with the same denominator. Again, multiply the fraction by a value of one in the following form:

$1=\frac{x}{x}$

$\frac{x}{x}\times\frac{\left(x+6\right)}{6}=\frac{x(x+6))}{6x}$

Now, we can add our two fractions by rewriting them as one whole fraction.

$\frac{36}{6x}+\frac{x(x+6)}{6x}=\frac{36+x(x+6))}{6x}$

Simplify the numerator.

$\frac{36+x(x+6)}{6x} = \frac{36+x^2+6x}{6x}$

$\frac{36+x^2+6x}{6x}= \frac{x^2+6x+36}{6x}$

This cannot be reduced or simplified any further; therefore, the answer is:

$\frac{x^2+6x+36}{6x}$

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Algebra II : Adding and Subtracting Rational Expressions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #90 : Rational Expressions

Example Question #1751 : Algebra Ii

Example Question #1752 : Algebra Ii

Example Question #1751 : Algebra Ii

Example Question #1754 : Algebra Ii

Example Question #1755 : Algebra Ii

Example Question #51 : Solving Rational Expressions

All Algebra II Resources

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