# Problem of the Week

## Updated at Apr 22, 2024 2:49 PM

To get more practice in equation, we brought you this problem of the week:

How would you solve the equation $${(\frac{5}{3-4p})}^{2}=25$$?

Check out the solution below!

${(\frac{5}{3-4p})}^{2}=25$

 1 Use Division Distributive Property: $${(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}$$.$\frac{{5}^{2}}{{(3-4p)}^{2}}=25$2 Simplify  $${5}^{2}$$  to  $$25$$.$\frac{25}{{(3-4p)}^{2}}=25$3 Multiply both sides by $${(3-4p)}^{2}$$.$25=25{(3-4p)}^{2}$4 Divide both sides by $$25$$.$1={(3-4p)}^{2}$5 Take the square root of both sides.$\pm \sqrt{1}=3-4p$6 Simplify  $$\sqrt{1}$$  to  $$1$$.$\pm 1=3-4p$7 Switch sides.$3-4p=\pm 1$8 Break down the problem into these 2 equations.$3-4p=1$$3-4p=-1$9 Solve the 1st equation: $$3-4p=1$$.1 Subtract $$3$$ from both sides.$-4p=1-3$2 Simplify  $$1-3$$  to  $$-2$$.$-4p=-2$3 Divide both sides by $$-4$$.$p=\frac{-2}{-4}$4 Two negatives make a positive.$p=\frac{2}{4}$5 Simplify  $$\frac{2}{4}$$  to  $$\frac{1}{2}$$.$p=\frac{1}{2}$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$p=\frac{1}{2}$10 Solve the 2nd equation: $$3-4p=-1$$.1 Subtract $$3$$ from both sides.$-4p=-1-3$2 Simplify  $$-1-3$$  to  $$-4$$.$-4p=-4$3 Divide both sides by $$-4$$.$p=1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$p=1$11 Collect all solutions.$p=\frac{1}{2},1$DoneDecimal Form: 0.5, 1p=1/2,1