Algebra II - Adding and Subtracting Rational Expressions

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

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

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

$\frac{1}{12}x$

$\frac{1}{12x}$

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

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 .

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

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

$\frac{49}{12x}$

$\frac{49}{12}x$

$\frac{49}{24x}$

$\frac{14}{3x}$

$\frac{14}{3}x$

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: $\frac{49}{12x}$

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

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

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

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

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

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

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

Simplify: $\frac{5}{x}-\frac{2x}{x+7}$

$-\frac{2x^2-5x-35}{x^2+7x}$

$-\frac{2x^2+5x-35}{x^2+7x}$

$-\frac{2x^2-5x-35}{x+7}$

$\frac{2x^2-5x-35}{x+7}$

$\frac{2x-5}{x-7}$

Explanation

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

Multiply the denominators together.

Convert both fractions by multiplying both the top and bottom by what was multiplied to get the denominator. Rewrite the fractions and combine as one single fraction.

$\frac{5(x+7)}{x(x+7)}-\frac{2x(x)}{(x)(x+7)} = \frac{5x+35-2x^2}{x^2+7x}$

Re-order the terms.

$\frac{-2x^2+5x+35}{x^2+7x}$

Pull out a common factor of negative one on the numerator.

$\frac{-1(2x^2-5x-35)}{x^2+7x} = -\frac{2x^2-5x-35}{x^2+7x}$

The answer is: $-\frac{2x^2-5x-35}{x^2+7x}$

$\frac{2x-1}{xy}+\frac{4y+3}{y^2}$

$\frac{6xy+3x-y}{xy^2}$

$\frac{6xy+3x-y}{xy}$

$\frac{6xy+3x-y}{y^2}$

$\frac{3xy+3x-y}{xy^2}$

$\frac{3}{xy^2}$

Explanation

First, find the common denominator, which is . Then, make sure to offset each numerator. Multiply $\frac{2x-1}{xy}$ by y to get $\frac{(2x-1)y}{xy^2}$ . Multiply $\frac{4y+3}{y^2}$ by x to get $\frac{(4y+3)x}{xy^2}$ . Then, combine numerators to get . Then, put the numerator over the denominator to get your answer: $\frac{6xy+3x-y}{xy^2}$ .

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

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

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

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

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

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

Explanation

To combine these rational expressions, first find the common denominator. In this case, it is . Then, offset the second equation so that you get the correct denominator: $\frac{4}{x}=\frac{4x}{x^2}$ . Then, combine the numerators: . Put your numerator over the denominator for your answer: $\frac{3x+2}{x^2}$ .

Subtract:

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

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

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

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

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

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

Explanation

First let us find a common denominator as follows:

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

Now we can subtract the numerators which gives us : $2x^{2}+2x-3$

So the final answer is $\frac{2x^{2}+2x-3}{\left ( x+1 \right )\left ( x-1 \right )}$

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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation

Notice that can be factorized by the difference of squares. This means that the second fraction will need to multiply the quantity of 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: $-\frac{x^2-3x-2}{x^2-1}$

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

$\frac{1}{30x}$

$\frac{1}{30}x$

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

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

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

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: $\frac{1}{30x}$

Adding and Subtracting Rational Expressions

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