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How would you solve the equation $\dfrac{(v+2)^{2}-6}{3}=1$ ?
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1 Multiply both sides by 3 .
$(v+2)^{2}-6=1 \times 3$
2 Simplify $1 \times 3$ to 3 .
$(v+2)^{2}-6=3$
(3) Add 6 to both sides.
$(v+2)^{2}=3+6$
4) Simplify $3+6$ to 9 .
$(v+2)^{2}=9$
5 Take the square root of both sides.
$v+2=\pm \sqrt{9}$
6 Since $3 \times 3=9$, the square root of 9 is 3 .
$v+2=\pm 3$

7. Break down the problem into these 2 equations.
\begin{aligned} &v+2=3 \\ &v+2=-3 \end{aligned}
$\mathbf{8}$ Solve the 1 st equation: $v+2=3$.
$v=1$
9 Solve the 2 nd equation: $v+2=-3$.
$v=-5$
10 Collect all solutions.
$v=1,-5$
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