# Problem of the Week

## Updated at Oct 12, 2020 10:33 AM

This week we have another equation problem:

How can we solve the equation $$6{(\frac{4}{4v})}^{2}=\frac{3}{2}$$?

Let's start!

$6{(\frac{4}{4v})}^{2}=\frac{3}{2}$

 1 Cancel $$4$$.$6{(\frac{1}{v})}^{2}=\frac{3}{2}$2 Use Division Distributive Property: $${(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}$$.$6\times \frac{1}{{v}^{2}}=\frac{3}{2}$3 Simplify  $$6\times \frac{1}{{v}^{2}}$$  to  $$\frac{6}{{v}^{2}}$$.$\frac{6}{{v}^{2}}=\frac{3}{2}$4 Multiply both sides by $${v}^{2}$$.$6=\frac{3}{2}{v}^{2}$5 Simplify  $$\frac{3}{2}{v}^{2}$$  to  $$\frac{3{v}^{2}}{2}$$.$6=\frac{3{v}^{2}}{2}$6 Multiply both sides by $$2$$.$6\times 2=3{v}^{2}$7 Simplify  $$6\times 2$$  to  $$12$$.$12=3{v}^{2}$8 Divide both sides by $$3$$.$\frac{12}{3}={v}^{2}$9 Simplify  $$\frac{12}{3}$$  to  $$4$$.$4={v}^{2}$10 Take the square root of both sides.$\pm \sqrt{4}=v$11 Since $$2\times 2=4$$, the square root of $$4$$ is $$2$$.$\pm 2=v$12 Switch sides.$v=\pm 2$Done v=2,-2