 # Problem of the Week Updated at Apr 22, 2019 5:15 PM

This week's problem comes from the equation category.

How can we solve the equation $$\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}$$?

Let's begin!

$\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}$

 1 Multiply both sides by $$5(2+{m}^{2})$$.$4m=\frac{4}{15}\times 5(2+{m}^{2})$2 Simplify $$\frac{4}{15}\times 5(2+{m}^{2})$$ to $$\frac{4(2+{m}^{2})}{3}$$.$4m=\frac{4(2+{m}^{2})}{3}$3 Multiply both sides by $$3$$.$12m=4(2+{m}^{2})$4 Divide both sides by $$4$$.$3m=2+{m}^{2}$5 Move all terms to one side.$3m-2-{m}^{2}=0$6 Multiply both sides by $$-1$$.${m}^{2}-3m+2=0$7 How?Factor $${m}^{2}-3m+2$$.1 Ask: Which two numbers add up to $$-3$$ and multiply to $$2$$?$$-2$$ and $$-1$$2 Rewrite the expression using the above.$(m-2)(m-1)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$(m-2)(m-1)=0$8 How?Solve for $$m$$.1 Ask: When will $$(m-2)(m-1)$$ equal zero?When $$m-2=0$$ or $$m-1=0$$2 Solve each of the 2 equations above.$m=2,1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$m=2,1$Done m=2,1