Rational Expressions

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Algebra II › Rational Expressions

Questions 1 - 10

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

Which value of makes the following expression undefined?

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

$\text{All real numbers}$

$\text{Cannot be determined}$

Explanation

A rational expression is undefined when the denominator is zero.

The denominator is zero when .

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.

$\frac{y+1}{8y+2} + \frac{3}{y} = \frac{5}{3}$

Solve for .

, $y=-\frac{9}{37}$

, $y=-\frac{7}{15}$

$y=-\frac{5}{2}$

Explanation

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

$\frac{y+1}{8y+2}\cdot \frac{y}{y}$ becomes $\frac{y^{2}+y}{(y)(8y+2)}$ .

$\frac{3}{y}\cdot \frac{8y+2}{8y+2}$ becomes $\frac{24y+6}{(y)(8y+2)}$ .

Now add the two fractions: $\frac{y^{2}+25y+6}{(y)(8y+2)}$

To solve, multiply both sides of the equation by , yielding

$y^{2}+25y+6 = \frac{(40y^{2}+10y)}{3}$ .

Multiply both sides by 3:

$3y^{2}+75y+18 = 40y^{2}+ 10y$

Move all terms to the same side:

$37y^{2}-65y-18 = 0$

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

Our solutions are therefore

and

$y=-\frac{9}{37}$ .

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 1 \frac{2}{5}$

$x = 4\textrm{ or }x = 7$

Explanation

Multiply both sides by :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

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

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

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

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

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

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

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

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

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.

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