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

by Platinum
(119,140 points)
in Mathematics
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How would you solve \(\dfrac{p^{2}}{5(2+4 p)}=\dfrac{9}{70} ?\)

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Best answer
\[
\frac{p^{2}}{5(2+4 p)}=\frac{9}{70}
\]
1) Factor out the common term 2 .
\[
\frac{p^{2}}{5 \times 2(1+2 p)}=\frac{9}{70}
\]
2 Simplify \(5 \times 2(1+2 p)\) to \(10(1+2 p)\).
\[
\frac{p^{2}}{10(1+2 p)}=\frac{9}{70}
\]
(3) Multiply both sides by \(10(1+2 p)\).
\[
p^{2}=\frac{9}{70} \times 10(1+2 p)
\]
4\} Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{a c}{b d}\).
\[
p^{2}=\frac{9 \times 10(1+2 p)}{70}
\]

(5) Simplify \(9 \times 10(1+2 p)\) to \(90(1+2 p)\).
\[
p^{2}=\frac{90(1+2 p)}{70}
\]
6 Simplify \(\frac{90(1+2 p)}{70}\) to \(\frac{9(1+2 p)}{7}\).
\[
p^{2}=\frac{9(1+2 p)}{7}
\]
(7) Multiply both sides by 7 .
\[
7 p^{2}=9(1+2 p)
\]
8 Expand.
\[
7 p^{2}=9+18 p
\]
9 Move all terms to one side.
\[
7 p^{2}-9-18 p=0
\]

10 Split the second term in \(7 p^{2}-9-18 p\) into two terms.
\[
7 p^{2}+3 p-21 p-9=0
\]
(11) Factor out common terms in the first two terms, then in the last two terms.
\[
p(7 p+3)-3(7 p+3)=0
\]
12 Factor out the common term \(7 p+3\).
\[
(7 p+3)(p-3)=0
\]
13 Solve for \(p\).
\[
p=-\frac{3}{7}, 3
\]
by Platinum
(119,140 points)

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