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: \(1-m=3\).
\[m=-2\]

16
Solve the 2nd equation: \(1-m=-3\).
\[m=4\]

17
Collect all solutions.
\[m=-2,4\]

Done
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