# Problem of the Week

## Updated at Mar 28, 2022 4:33 PM

This week we have another equation problem:

How can we solve the equation $$\frac{3}{6-{(3-z)}^{2}}=\frac{1}{2}$$?

Let's start!

$\frac{3}{6-{(3-z)}^{2}}=\frac{1}{2}$

 1 Multiply both sides by $$6-{(3-z)}^{2}$$.$3=\frac{1}{2}(6-{(3-z)}^{2})$2 Simplify  $$\frac{1}{2}(6-{(3-z)}^{2})$$  to  $$\frac{6-{(3-z)}^{2}}{2}$$.$3=\frac{6-{(3-z)}^{2}}{2}$3 Multiply both sides by $$2$$.$3\times 2=6-{(3-z)}^{2}$4 Simplify  $$3\times 2$$  to  $$6$$.$6=6-{(3-z)}^{2}$5 Cancel $$6$$ on both sides.$0=-{(3-z)}^{2}$6 Multiply both sides by $$-1$$.$0={(3-z)}^{2}$7 Take the square root of both sides.$0=3-z$8 Subtract $$3$$ from both sides.$-3=-z$9 Multiply both sides by $$-1$$.$3=z$10 Switch sides.$z=3$Donez=3