Problem of the Week

Updated at Sep 17, 2018 5:18 PM

This week we have another equation problem:

How can we solve the equation (2+n)24(n−3)=9\frac{{(2+n)}^{2}}{4(n-3)}=9?

Let's start!



(2+n)24(n−3)=9\frac{{(2+n)}^{2}}{4(n-3)}=9

1
Multiply both sides by 4(n−3)4(n-3).
(2+n)2=9×4(n−3){(2+n)}^{2}=9\times 4(n-3)

2
Simplify  9×4(n−3)9\times 4(n-3)  to  36(n−3)36(n-3).
(2+n)2=36(n−3){(2+n)}^{2}=36(n-3)

3
Expand.
4+4n+n2=36n−1084+4n+{n}^{2}=36n-108

4
Move all terms to one side.
4+4n+n2−36n+108=04+4n+{n}^{2}-36n+108=0

5
Simplify  4+4n+n2−36n+1084+4n+{n}^{2}-36n+108  to  112−32n+n2112-32n+{n}^{2}.
112−32n+n2=0112-32n+{n}^{2}=0

6
Factor 112−32n+n2112-32n+{n}^{2}.
(n−28)(n−4)=0(n-28)(n-4)=0

7
Solve for nn.
n=28,4n=28,4

Done
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