Question

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.

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

Short answer

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

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

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.

$$\begin{gathered} c^2-4c+4=(c-2)(c-2) \\ {c}^2-7c+10=(c-2)(c-5) \end{gathered}$$

• So, the LCD is $(c-2)(c-2)(c-5)$.

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.

$$\begin{gathered} \frac{5}{c^2-4c+4}=\frac{5(c-5)}{(c-2)(c-2)(c-5)} \\ \frac{3c}{c^2-7c+10}=\frac{3c}{(c-2)(c-2)(c-5)} \end{gathered}$$

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