Question

Subtract:$\frac{z+3}{z+2}-\frac{z}{z+3}$

Short answer

The simplification of the expression is$\frac{3}{z+3}$.

Step-by-step solution

Given information

The given rational expression is$\frac{z+3}{z+2}-\frac{z}{z+3}$.

The simplification of the expression

$\frac{z+3}{z+2}-\frac{z}{z+3}$

The LCD of the rational expression is $\left(z+2\right)\left(z+3\right)$.

Change into equivalent fractions with the LCD $\left(z+2\right)\left(z+3\right)$.

role="math" localid="1648455874752" $$\begin{gathered} =\frac{\left(z+3\right)\left(z+2\right)}{\left(z+2\right)\left(z+3\right)}-\frac{z\left(z+2\right)}{\left(z+3\right)\left(z+2\right)} \\ =\frac{z^2+3z+2z+6}{\left(z+2\right)\left(z+3\right)}-\frac{z^2+2z}{\left(z+2\right)\left(z+3\right)} \\ Simplify \\ =\frac{z^2+5z+6-z^2-2z}{\left(z+2\right)\left(z+3\right)} \\ =\frac{3z+6}{\left(z+2\right)\left(z+3\right)} \\ =\frac{3\left(z+2\right)}{\left(z+2\right)\left(z+3\right)} \\ =\frac{3}{\left(z+3\right)} \end{gathered}$$