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 4m5(2+m2)=415\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}?

Let's begin!



4m5(2+m2)=415\frac{4m}{5(2+{m}^{2})}=\frac{4}{15}

1
Multiply both sides by 5(2+m2)5(2+{m}^{2}).
4m=415×5(2+m2)4m=\frac{4}{15}\times 5(2+{m}^{2})

2
Use this rule: ab×cd=acbd\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}.
4m=4×5(2+m2)154m=\frac{4\times 5(2+{m}^{2})}{15}

3
Simplify  4×5(2+m2)4\times 5(2+{m}^{2})  to  20(2+m2)20(2+{m}^{2}).
4m=20(2+m2)154m=\frac{20(2+{m}^{2})}{15}

4
Simplify  20(2+m2)15\frac{20(2+{m}^{2})}{15}  to  4(2+m2)3\frac{4(2+{m}^{2})}{3}.
4m=4(2+m2)34m=\frac{4(2+{m}^{2})}{3}

5
Multiply both sides by 33.
12m=4(2+m2)12m=4(2+{m}^{2})

6
Divide both sides by 44.
3m=2+m23m=2+{m}^{2}

7
Move all terms to one side.
3m−2−m2=03m-2-{m}^{2}=0

8
Multiply both sides by −1-1.
m2−3m+2=0{m}^{2}-3m+2=0

9
Factor m2−3m+2{m}^{2}-3m+2.
(m−2)(m−1)=0(m-2)(m-1)=0

10
Solve for mm.
m=2,1m=2,1

Done
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