Problem of the Week

Updated at Dec 23, 2019 4:38 PM

This week we have another equation problem:

How would you solve \(\frac{{w}^{2}-3}{3}+5=\frac{28}{3}\)?

Let's start!



\[\frac{{w}^{2}-3}{3}+5=\frac{28}{3}\]

1
Simplify  \(\frac{{w}^{2}-3}{3}\)  to  \(-1+\frac{{w}^{2}}{3}\).
\[-1+\frac{{w}^{2}}{3}+5=\frac{28}{3}\]

2
Simplify  \(-1+\frac{{w}^{2}}{3}+5\)  to  \(\frac{{w}^{2}}{3}+4\).
\[\frac{{w}^{2}}{3}+4=\frac{28}{3}\]

3
Subtract \(4\) from both sides.
\[\frac{{w}^{2}}{3}=\frac{28}{3}-4\]

4
Simplify  \(\frac{28}{3}-4\)  to  \(\frac{16}{3}\).
\[\frac{{w}^{2}}{3}=\frac{16}{3}\]

5
Multiply both sides by \(3\).
\[{w}^{2}=\frac{16}{3}\times 3\]

6
Cancel \(3\).
\[{w}^{2}=16\]

7
Take the square root of both sides.
\[w=\pm \sqrt{16}\]

8
Since \(4\times 4=16\), the square root of \(16\) is \(4\).
\[w=\pm 4\]

Done
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