# Problem of the Week

## Updated at Dec 23, 2019 4:38 PM

This week we have another equation problem:

How would you solve $$\frac{{w}^{2}-3}{3}+5=\frac{28}{3}$$?

Let's start!

$\frac{{w}^{2}-3}{3}+5=\frac{28}{3}$

 1 Simplify  $$\frac{{w}^{2}-3}{3}$$  to  $$-1+\frac{{w}^{2}}{3}$$.$-1+\frac{{w}^{2}}{3}+5=\frac{28}{3}$2 Simplify  $$-1+\frac{{w}^{2}}{3}+5$$  to  $$\frac{{w}^{2}}{3}+4$$.$\frac{{w}^{2}}{3}+4=\frac{28}{3}$3 Subtract $$4$$ from both sides.$\frac{{w}^{2}}{3}=\frac{28}{3}-4$4 Simplify  $$\frac{28}{3}-4$$  to  $$\frac{16}{3}$$.$\frac{{w}^{2}}{3}=\frac{16}{3}$5 Multiply both sides by $$3$$.${w}^{2}=\frac{16}{3}\times 3$6 Cancel $$3$$.${w}^{2}=16$7 Take the square root of both sides.$w=\pm \sqrt{16}$8 Since $$4\times 4=16$$, the square root of $$16$$ is $$4$$.$w=\pm 4$Donew=4,-4