Question

Add$\frac{3z}{z-2}+\frac{1}{z+5}$.

Short answer

The sum is$\frac{3z^2+16z-2}{(z-2)(z+5)}$.

Step-by-step solution

Given Information 

The Given expression is$\frac{3z}{z-2}+\frac{1}{z+5}$.

Find the LCD of denominators. 

Since there are no common factors in denominators.

The LCD is$(z-2)(z+5)$.

Rewrite each fraction as an equivalent fraction with the LCD. 

$=\frac{3z(z+5)}{(z-2)(z+5)}+\frac{1(z-2)}{(z-2)(z+5)}$

Simplify numerator.

$=\frac{3z^2+15z}{(z-2)(z+5)}+\frac{z-2}{(z-2)(z+5)}$

Add and simplify the numerator. 

Take common denominator.

$=\frac{3z^2+15z+z-2}{(z-2)(z+5)}$

$=\frac{3z^2+16z-2}{(z-2)(z+5)}$