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Q. 104

Expert-verified
Found in: Page 668

### Intermediate Algebra

Book edition OER 2017
Author(s) OPENSTAX
Pages 1346 pages
ISBN 9780998625720

# In the following exercises, (a) find the LCD for the given rational expressions (b) rewrite them as equivalent rational expressions with the lowest common denominator. $\frac{5}{{c}^{2}-4c+4},\frac{3c}{{c}^{2}-7c+10}$

(a) The LCD is $\left(c-2\right)\left(c-2\right)\left(c-5\right)$.

(b) The expressions with LCD are $\frac{5\left(c-5\right)}{\left(c-2\right)\left(c-2\right)\left(c-5\right)},\frac{3c}{\left(c-2\right)\left(c-2\right)\left(c-5\right)}$.

See the step by step solution

## Part (a) Step 1. Given Information

The given expressions are $\frac{5}{{c}^{2}-4c+4},\frac{3c}{{c}^{2}-7c+10}$.

## Part (a) Step 2. Find LCD

• Factor the denominators of both the given rational expressions.

${c}^{2}-4c+4=\left(c-2\right)\left(c-2\right)\phantom{\rule{0ex}{0ex}}{c}^{2}-7c+10=\left(c-2\right)\left(c-5\right)$

• So, the LCD is $\left(c-2\right)\left(c-2\right)\left(c-5\right)$.

## Part (b) Step 1. Given Information

The given expressions are $\frac{5}{{c}^{2}-4c+4},\frac{3c}{{c}^{2}-7c+10}$

## Part (b) Step 2. Express the expressions with LCD

• Multiply and divide the expressions by the missing factor to make the denominator equal to LCD.

$\frac{5}{{c}^{2}-4c+4}=\frac{5\left(c-5\right)}{\left(c-2\right)\left(c-2\right)\left(c-5\right)}\phantom{\rule{0ex}{0ex}}\frac{3c}{{c}^{2}-7c+10}=\frac{3c}{\left(c-2\right)\left(c-2\right)\left(c-5\right)}$

• So, the expressions with equal denominators are $\frac{5\left(c-5\right)}{\left(c-2\right)\left(c-2\right)\left(c-5\right)},\frac{3c}{\left(c-2\right)\left(c-2\right)\left(c-5\right)}$.