Problem of the Week

Updated at Jul 15, 2024 2:53 PM

This week we have another equation problem:

How would you solve \(\frac{4q+2}{{(q-3)}^{2}}=10\)?

Let's start!



\[\frac{4q+2}{{(q-3)}^{2}}=10\]

1
Factor out the common term \(2\).
\[\frac{2(2q+1)}{{(q-3)}^{2}}=10\]

2
Multiply both sides by \({(q-3)}^{2}\).
\[2(2q+1)=10{(q-3)}^{2}\]

3
Divide both sides by \(2\).
\[2q+1=5{(q-3)}^{2}\]

4
Expand.
\[2q+1=5{q}^{2}-30q+45\]

5
Move all terms to one side.
\[2q+1-5{q}^{2}+30q-45=0\]

6
Simplify  \(2q+1-5{q}^{2}+30q-45\)  to  \(32q-44-5{q}^{2}\).
\[32q-44-5{q}^{2}=0\]

7
Multiply both sides by \(-1\).
\[5{q}^{2}-32q+44=0\]

8
Split the second term in \(5{q}^{2}-32q+44\) into two terms.
\[5{q}^{2}-10q-22q+44=0\]

9
Factor out common terms in the first two terms, then in the last two terms.
\[5q(q-2)-22(q-2)=0\]

10
Factor out the common term \(q-2\).
\[(q-2)(5q-22)=0\]

11
Solve for \(q\).
\[q=2,\frac{22}{5}\]

Done

Decimal Form: 2, 4.4
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