Problem of the Week

Updated at Nov 11, 2019 2:40 PM

This week's problem comes from the equation category.

How would you solve the equation 6+(3(3−z))2=426+{(3(3-z))}^{2}=42?

Let's begin!



6+(3(3−z))2=426+{(3(3-z))}^{2}=42

1
Use Multiplication Distributive Property: (xy)a=xaya{(xy)}^{a}={x}^{a}{y}^{a}.
6+32(3−z)2=426+{3}^{2}{(3-z)}^{2}=42

2
Simplify  32{3}^{2}  to  99.
6+9(3−z)2=426+9{(3-z)}^{2}=42

3
Subtract 66 from both sides.
9(3−z)2=42−69{(3-z)}^{2}=42-6

4
Simplify  42−642-6  to  3636.
9(3−z)2=369{(3-z)}^{2}=36

5
Divide both sides by 99.
(3−z)2=369{(3-z)}^{2}=\frac{36}{9}

6
Simplify  369\frac{36}{9}  to  44.
(3−z)2=4{(3-z)}^{2}=4

7
Take the square root of both sides.
3−z=±43-z=\pm \sqrt{4}

8
Since 2×2=42\times 2=4, the square root of 44 is 22.
3−z=±23-z=\pm 2

9
Break down the problem into these 2 equations.
3−z=23-z=2
3−z=−23-z=-2

10
Solve the 1st equation: 3−z=23-z=2.
z=1z=1

11
Solve the 2nd equation: 3−z=−23-z=-2.
z=5z=5

12
Collect all solutions.
z=1,5z=1,5

Done
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