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

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

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