Problem of the Week

Updated at Nov 1, 2021 4:12 PM

How can we solve the equation 53(v+2)+4=323\frac{5}{3}(v+2)+4=\frac{32}{3}?

Below is the solution.



53(v+2)+4=323\frac{5}{3}(v+2)+4=\frac{32}{3}

1
Simplify  53(v+2)\frac{5}{3}(v+2)  to  5(v+2)3\frac{5(v+2)}{3}.
5(v+2)3+4=323\frac{5(v+2)}{3}+4=\frac{32}{3}

2
Subtract 44 from both sides.
5(v+2)3=323−4\frac{5(v+2)}{3}=\frac{32}{3}-4

3
Simplify  323−4\frac{32}{3}-4  to  203\frac{20}{3}.
5(v+2)3=203\frac{5(v+2)}{3}=\frac{20}{3}

4
Multiply both sides by 33.
5(v+2)=203×35(v+2)=\frac{20}{3}\times 3

5
Cancel 33.
5(v+2)=205(v+2)=20

6
Divide both sides by 55.
v+2=205v+2=\frac{20}{5}

7
Simplify  205\frac{20}{5}  to  44.
v+2=4v+2=4

8
Subtract 22 from both sides.
v=4−2v=4-2

9
Simplify  4−24-2  to  22.
v=2v=2

Done
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