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
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[4m=\frac{4\times 5(2+{m}^{2})}{15}\]

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

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

5
Multiply both sides by \(3\).
\[12m=4(2+{m}^{2})\]

6
Divide both sides by \(4\).
\[3m=2+{m}^{2}\]

7
Move all terms to one side.
\[3m-2-{m}^{2}=0\]

8
Multiply both sides by \(-1\).
\[{m}^{2}-3m+2=0\]

9
Factor \({m}^{2}-3m+2\).
\[(m-2)(m-1)=0\]

10
Solve for \(m\).
\[m=2,1\]

Done
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