Algebra II - Solving Rational Expressions

Solve for , given the equation below.

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

No solutions

Explanation

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

Begin by cross-multiplying.

Distribute the on the left side and expand the polynomial on the right.

Combine like terms and rearrange to set the equation equal to zero.

Now we can isolate and solve for by adding to both sides.

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

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

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

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

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

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

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

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

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

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

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

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

Explanation

Rewrite the left fraction using common factors.

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

Cancel out common terms.

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

Factorize the bottom term.

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

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

Simplify: $\frac{x}{x-9}\div\frac{x-9}{x-18}$

$\frac{x^2-18}{x^2-18+81}$

$\frac{x+9}{x^2-18+81}$

$\frac{x}{x-18}$

$\frac{x-18}{x}$

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

Explanation

In order to simplify the rational expression, we will need to rewrite the expression as a multiplication sign and take the reciprocal of the second term.

$\frac{x}{x-9}\div\frac{x-9}{x-18}= \frac{x}{x-9} \times \frac{x-18}{x-9}$

Simplify the numerator.

Simplify the denominator by FOIL method.

Divide the numerator with the denominator.

The answer is: $\frac{x^2-18}{x^2-18+81}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation

Solve for .

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

Explanation

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

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

Next, divide by four on both sides.

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

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

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

Solving Rational Expressions

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