Problem of the Week

Updated at Feb 11, 2019 4:37 PM

How would you solve the equation (4+4m)×3−m5=125(4+4m)\times \frac{3-m}{5}=\frac{12}{5}?

Below is the solution.



(4+4m)×3−m5=125(4+4m)\times \frac{3-m}{5}=\frac{12}{5}

1
Use this rule: a×bc=abca \times \frac{b}{c}=\frac{ab}{c}.
(4+4m)(3−m)5=125\frac{(4+4m)(3-m)}{5}=\frac{12}{5}

2
Factor out the common term 44.
4(1+m)(3−m)5=125\frac{4(1+m)(3-m)}{5}=\frac{12}{5}

3
Multiply both sides by 55.
4(1+m)(3−m)=124(1+m)(3-m)=12

4
Expand.
12−4m+12m−4m2=1212-4m+12m-4{m}^{2}=12

5
Simplify  12−4m+12m−4m212-4m+12m-4{m}^{2}  to  12+8m−4m212+8m-4{m}^{2}.
12+8m−4m2=1212+8m-4{m}^{2}=12

6
Cancel 1212 on both sides.
8m−4m2=08m-4{m}^{2}=0

7
Factor out the common term 4m4m.
4m(2−m)=04m(2-m)=0

8
Solve for mm.
m=0,2m=0,2

Done
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