# Problem of the Week

## Updated at Dec 18, 2023 9:20 AM

This week's problem comes from the equation category.

How would you solve $$\frac{6}{{(4-4m)}^{2}}=\frac{1}{24}$$?

Let's begin!

$\frac{6}{{(4-4m)}^{2}}=\frac{1}{24}$

 1 Factor out the common term $$4$$.$\frac{6}{{(4(1-m))}^{2}}=\frac{1}{24}$2 Use Multiplication Distributive Property: $${(xy)}^{a}={x}^{a}{y}^{a}$$.$\frac{6}{{4}^{2}{(1-m)}^{2}}=\frac{1}{24}$3 Simplify  $${4}^{2}$$  to  $$16$$.$\frac{6}{16{(1-m)}^{2}}=\frac{1}{24}$4 Simplify  $$\frac{6}{16{(1-m)}^{2}}$$  to  $$\frac{3}{8{(1-m)}^{2}}$$.$\frac{3}{8{(1-m)}^{2}}=\frac{1}{24}$5 Multiply both sides by $$8{(1-m)}^{2}$$.$3=\frac{1}{24}\times 8{(1-m)}^{2}$6 Use this rule: $$\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}$$.$3=\frac{1\times 8{(1-m)}^{2}}{24}$7 Simplify  $$1\times 8{(1-m)}^{2}$$  to  $$8{(1-m)}^{2}$$.$3=\frac{8{(1-m)}^{2}}{24}$8 Simplify  $$\frac{8{(1-m)}^{2}}{24}$$  to  $$\frac{{(1-m)}^{2}}{3}$$.$3=\frac{{(1-m)}^{2}}{3}$9 Multiply both sides by $$3$$.$3\times 3={(1-m)}^{2}$10 Simplify  $$3\times 3$$  to  $$9$$.$9={(1-m)}^{2}$11 Take the square root of both sides.$\pm \sqrt{9}=1-m$12 Since $$3\times 3=9$$, the square root of $$9$$ is $$3$$.$\pm 3=1-m$13 Switch sides.$1-m=\pm 3$14 Break down the problem into these 2 equations.$1-m=3$$1-m=-3$15 Solve the 1st equation: $$1-m=3$$.1 Subtract $$1$$ from both sides.$-m=3-1$2 Simplify  $$3-1$$  to  $$2$$.$-m=2$3 Multiply both sides by $$-1$$.$m=-2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$m=-2$16 Solve the 2nd equation: $$1-m=-3$$.1 Subtract $$1$$ from both sides.$-m=-3-1$2 Simplify  $$-3-1$$  to  $$-4$$.$-m=-4$3 Multiply both sides by $$-1$$.$m=4$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$m=4$17 Collect all solutions.$m=-2,4$Donem=-2,4