Problem of the Week

Updated at Mar 28, 2022 4:33 PM

This week we have another equation problem:

How can we solve the equation 36−(3−z)2=12\frac{3}{6-{(3-z)}^{2}}=\frac{1}{2}?

Let's start!



36−(3−z)2=12\frac{3}{6-{(3-z)}^{2}}=\frac{1}{2}

1
Multiply both sides by 6−(3−z)26-{(3-z)}^{2}.
3=12(6−(3−z)2)3=\frac{1}{2}(6-{(3-z)}^{2})

2
Simplify  12(6−(3−z)2)\frac{1}{2}(6-{(3-z)}^{2})  to  6−(3−z)22\frac{6-{(3-z)}^{2}}{2}.
3=6−(3−z)223=\frac{6-{(3-z)}^{2}}{2}

3
Multiply both sides by 22.
3×2=6−(3−z)23\times 2=6-{(3-z)}^{2}

4
Simplify  3×23\times 2  to  66.
6=6−(3−z)26=6-{(3-z)}^{2}

5
Cancel 66 on both sides.
0=−(3−z)20=-{(3-z)}^{2}

6
Multiply both sides by −1-1.
0=(3−z)20={(3-z)}^{2}

7
Take the square root of both sides.
0=3−z0=3-z

8
Subtract 33 from both sides.
−3=−z-3=-z

9
Multiply both sides by −1-1.
3=z3=z

10
Switch sides.
z=3z=3

Done
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