Problem of the Week

Updated at Dec 2, 2019 2:53 PM

This week we have another equation problem:

How would you solve 4(z−3)+25=2\frac{4(z-3)+2}{5}=2?

Let's start!



4(z−3)+25=2\frac{4(z-3)+2}{5}=2

1
Factor out the common term 22.
2(2(z−3)+1)5=2\frac{2(2(z-3)+1)}{5}=2

2
Multiply both sides by 55.
2(2(z−3)+1)=2×52(2(z-3)+1)=2\times 5

3
Simplify  2×52\times 5  to  1010.
2(2(z−3)+1)=102(2(z-3)+1)=10

4
Divide both sides by 22.
2(z−3)+1=1022(z-3)+1=\frac{10}{2}

5
Simplify  102\frac{10}{2}  to  55.
2(z−3)+1=52(z-3)+1=5

6
Subtract 11 from both sides.
2(z−3)=5−12(z-3)=5-1

7
Simplify  5−15-1  to  44.
2(z−3)=42(z-3)=4

8
Divide both sides by 22.
z−3=42z-3=\frac{4}{2}

9
Simplify  42\frac{4}{2}  to  22.
z−3=2z-3=2

10
Add 33 to both sides.
z=2+3z=2+3

11
Simplify  2+32+3  to  55.
z=5z=5

Done
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