Problem of the Week

Updated at Apr 20, 2020 1:46 PM

For this week we've brought you this equation problem.

How would you solve the equation 10(3−n)2+n=203\frac{10(3-n)}{2+n}=\frac{20}{3}?

Here are the steps:



10(3−n)2+n=203\frac{10(3-n)}{2+n}=\frac{20}{3}

1
Multiply both sides by 2+n2+n.
10(3−n)=203(2+n)10(3-n)=\frac{20}{3}(2+n)

2
Simplify  203(2+n)\frac{20}{3}(2+n)  to  20(2+n)3\frac{20(2+n)}{3}.
10(3−n)=20(2+n)310(3-n)=\frac{20(2+n)}{3}

3
Multiply both sides by 33.
30(3−n)=20(2+n)30(3-n)=20(2+n)

4
Expand.
90−30n=40+20n90-30n=40+20n

5
Add 30n30n to both sides.
90=40+20n+30n90=40+20n+30n

6
Simplify  40+20n+30n40+20n+30n  to  40+50n40+50n.
90=40+50n90=40+50n

7
Subtract 4040 from both sides.
90−40=50n90-40=50n

8
Simplify  90−4090-40  to  5050.
50=50n50=50n

9
Divide both sides by 5050.
1=n1=n

10
Switch sides.
n=1n=1

Done
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