# 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.1 Multiply the coefficient of the first term by the constant term.$7\times -9=-63$2 Ask: Which two numbers add up to $$-18$$ and multiply to $$-63$$?$$3$$ and $$-21$$3 Split $$-18p$$ as the sum of $$3p$$ and $$-21p$$.$7{p}^{2}+3p-21p-9$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$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 $$p$$.1 Ask: When will $$(7p+3)(p-3)$$ equal zero?When $$7p+3=0$$ or $$p-3=0$$2 Solve each of the 2 equations above.$p=-\frac{3}{7},3$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$p=-\frac{3}{7},3$Done Decimal Form: -0.428571, 3p=-3/7,3