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Q. 105

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. $\frac{2}{3{d}^{2}+14d-5},\frac{5d}{3{d}^{2}-19d+6}$

(a) The LCD is $\left(3d-1\right)\left(d+5\right)\left(d-6\right)$.

(b) The expressions with LCD are $\frac{2\left(d-6\right)}{\left(3d-1\right)\left(d+5\right)\left(d-6\right)},\frac{5d\left(d+5\right)}{\left(3d-1\right)\left(d-6\right)\left(d+5\right)}$.

See the step by step solution

## Part (a) Step 1. Given Information

The given expressions are $\frac{2}{3{d}^{2}+14d-5},\frac{5d}{3{d}^{2}-19d+6}$.

## Part (a) Step 2. Find LCD

• Factor the denominators of both the given rational expressions.

$3{d}^{2}+14d-5=\left(3d-1\right)\left(d+5\right)\phantom{\rule{0ex}{0ex}}3{d}^{2}-19d+6=\left(3d-1\right)\left(d-6\right)$

• So, the LCD is $\left(3d-1\right)\left(d+5\right)\left(d-6\right)$.

## Part (b) Step 1. Given Information

The given expressions are $\frac{2}{3{d}^{2}+14d-5},\frac{5d}{3{d}^{2}-19d+6}$

## 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.

$\frac{2}{3{d}^{2}+14d-5}=\frac{2\left(d-6\right)}{\left(3d-1\right)\left(d+5\right)\left(d-6\right)}\phantom{\rule{0ex}{0ex}}\frac{5d}{3{d}^{2}-19d+6}=\frac{5d\left(d+5\right)}{\left(3d-1\right)\left(d-6\right)\left(d+5\right)}$

• So, the expressions with equal denominators are role="math" localid="1645521296602" $\frac{2\left(d-6\right)}{\left(3d-1\right)\left(d+5\right)\left(d-6\right)},\frac{5d\left(d+5\right)}{\left(3d-1\right)\left(d-6\right)\left(d+5\right)}$.