Problem of the Week

Updated at Sep 7, 2020 12:31 PM

For this week we've brought you this equation problem.

How would you solve the equation \(\frac{3(3-z)}{{(\frac{5}{z})}^{2}}=-6\)?

Here are the steps:



\[\frac{3(3-z)}{{(\frac{5}{z})}^{2}}=-6\]

1
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{3(3-z)}{\frac{{5}^{2}}{{z}^{2}}}=-6\]

2
Simplify  \({5}^{2}\)  to  \(25\).
\[\frac{3(3-z)}{\frac{25}{{z}^{2}}}=-6\]

3
Invert and multiply.
\[3(3-z)\times \frac{{z}^{2}}{25}=-6\]

4
Simplify  \(3(3-z)\times \frac{{z}^{2}}{25}\)  to  \(\frac{3(3-z){z}^{2}}{25}\).
\[\frac{3(3-z){z}^{2}}{25}=-6\]

5
Regroup terms.
\[\frac{3{z}^{2}(3-z)}{25}=-6\]

6
Multiply both sides by \(25\).
\[3{z}^{2}(3-z)=-150\]

7
Expand.
\[9{z}^{2}-3{z}^{3}=-150\]

8
Move all terms to one side.
\[9{z}^{2}-3{z}^{3}+150=0\]

9
Factor out the common term \(3\).
\[3(3{z}^{2}-{z}^{3}+50)=0\]

10
Factor \(3{z}^{2}-{z}^{3}+50\) using Polynomial Division.
\[3(-{z}^{2}-2z-10)(z-5)=0\]

11
Solve for \(z\).
\[z=5\]

12
Use the Quadratic Formula.
\[z=\frac{2+6\imath }{-2},\frac{2-6\imath }{-2}\]

13
Collect all solutions from the previous steps.
\[z=5,\frac{2+6\imath }{-2},\frac{2-6\imath }{-2}\]

14
Simplify solutions.
\[z=5,-1-3\imath ,-1+3\imath \]

Done
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