Problem of the Week

Updated at May 12, 2025 3:20 PM

This week's problem comes from the equation category.

How would you solve \(3-4\times \frac{5}{{q}^{2}}=\frac{7}{4}\)?

Let's begin!



\[3-4\times \frac{5}{{q}^{2}}=\frac{7}{4}\]

1
Simplify  \(4\times \frac{5}{{q}^{2}}\)  to  \(\frac{20}{{q}^{2}}\).
\[3-\frac{20}{{q}^{2}}=\frac{7}{4}\]

2
Subtract \(3\) from both sides.
\[-\frac{20}{{q}^{2}}=\frac{7}{4}-3\]

3
Simplify  \(\frac{7}{4}-3\)  to  \(-\frac{5}{4}\).
\[-\frac{20}{{q}^{2}}=-\frac{5}{4}\]

4
Multiply both sides by \({q}^{2}\).
\[-20=-\frac{5}{4}{q}^{2}\]

5
Simplify  \(\frac{5}{4}{q}^{2}\)  to  \(\frac{5{q}^{2}}{4}\).
\[-20=-\frac{5{q}^{2}}{4}\]

6
Multiply both sides by \(4\).
\[-20\times 4=-5{q}^{2}\]

7
Simplify  \(-20\times 4\)  to  \(-80\).
\[-80=-5{q}^{2}\]

8
Divide both sides by \(-5\).
\[\frac{-80}{-5}={q}^{2}\]

9
Two negatives make a positive.
\[\frac{80}{5}={q}^{2}\]

10
Simplify  \(\frac{80}{5}\)  to  \(16\).
\[16={q}^{2}\]

11
Take the square root of both sides.
\[\pm \sqrt{16}=q\]

12
Since \(4\times 4=16\), the square root of \(16\) is \(4\).
\[\pm 4=q\]

13
Switch sides.
\[q=\pm 4\]

Done
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