Problem of the Week

Updated at Aug 16, 2021 1:38 PM

This week we have another equation problem:

How would you solve \(\frac{{p}^{2}}{5(2+4p)}=\frac{9}{70}\)?

Let's start!



\[\frac{{p}^{2}}{5(2+4p)}=\frac{9}{70}\]

1
Factor out the common term \(2\).
\[\frac{{p}^{2}}{5\times 2(1+2p)}=\frac{9}{70}\]

2
Simplify  \(5\times 2(1+2p)\)  to  \(10(1+2p)\).
\[\frac{{p}^{2}}{10(1+2p)}=\frac{9}{70}\]

3
Multiply both sides by \(10(1+2p)\).
\[{p}^{2}=\frac{9}{70}\times 10(1+2p)\]

4
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[{p}^{2}=\frac{9\times 10(1+2p)}{70}\]

5
Simplify  \(9\times 10(1+2p)\)  to  \(90(1+2p)\).
\[{p}^{2}=\frac{90(1+2p)}{70}\]

6
Simplify  \(\frac{90(1+2p)}{70}\)  to  \(\frac{9(1+2p)}{7}\).
\[{p}^{2}=\frac{9(1+2p)}{7}\]

7
Multiply both sides by \(7\).
\[7{p}^{2}=9(1+2p)\]

8
Expand.
\[7{p}^{2}=9+18p\]

9
Move all terms to one side.
\[7{p}^{2}-9-18p=0\]

10
Split the second term in \(7{p}^{2}-9-18p\) into two terms.
\[7{p}^{2}+3p-21p-9=0\]

11
Factor out common terms in the first two terms, then in the last two terms.
\[p(7p+3)-3(7p+3)=0\]

12
Factor out the common term \(7p+3\).
\[(7p+3)(p-3)=0\]

13
Solve for \(p\).
\[p=-\frac{3}{7},3\]

Done

Decimal Form: -0.428571, 3
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