Question

In the following exercises, add or subtract.

$\frac{z^2+6z}{z^2-25}-\frac{3z+20}{25-z^2}$

Short answer

The required result is$\frac{z+4}{z-5}$.

Step-by-step solution

Step 1. Given Information

The given expression is$\frac{z^2+6z}{z^2-25}-\frac{3z+20}{25-z^2}$.

Step 2. Multiply the numerator and denominator of second expression by -1.

• Multiply the numerator and denominator of the second term by$-1$.

$\frac{z^2+6z}{z^2-25}-\frac{3z+20}{25-z^2}=\frac{z^2+6z}{z^2-25}-\frac{-3z-20}{z^2-25}$

Step 3. Subtract the numerators

• As the denominators of the rational expressions is equal, subtract the numerators and place them over a common denominator.

$\begin{array}{rcl}\frac{z^2+6z}{z^2-25}-\frac{-3z-20}{z^2-25}& =& \frac{z^2+6z-\left(-3x-20\right)}{z^2-25}\\ & =& \frac{z^2+6z+3z+20}{z^2-25}\\ & =& \frac{z^2+9z+20}{z^2-25}\\ & & \end{array}$

Step 4. Factor the numerator and denominator

• Factor the expression in the numerator and denominator and then cancel out the common terms in numerator and denominator.

$\begin{array}{rcl}\frac{z^2+9z+20}{z^2-25}& =& \frac{\left(z+4\right)\cancel{\left(z+5\right)}}{\cancel{\left(z+5\right)}\left(z-5\right)}\\ & =& \frac{z+4}{z-5}\end{array}$