Solve for $v$ - simplify as much as possible Solve for $v$. Simplify the answer.
$$-3 = -\frac{8}{v-1}$$
Here is what I tried:
$$-3 = \frac{-8}{v-1} $$
$$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8) $$
$$24 = v-1$$ 
$$25 = v$$
 A: If I have a fraction, I always cross multiply to make things easiest.
$$ -3 = -\frac {8} {v-1} $$
$$ -3(v-1) = -8 $$
$$ v-1 = \frac {8} {3} $$
$$ v = \frac {11} {3} $$
A: I'd probably start by taking a reciprocal of both sides:
$$
\begin{align}
-3&=-\frac{8}{v-1}\\
-\frac{1}{3}&=-\frac{v-1}{8}\\
(-8)-\frac{1}{3}&=-\frac{v-1}{8}(-8)\\
\frac{8}{3}&=v-1\\
\frac{8}{3}+1&=v\\
\frac{11}{3}&=v
\end{align}
$$
A: Observe the solution, if any, must be $\ne 1$. I'd first change the signs and multiply both sides by $v-1$:
$$3v-3=8\iff v=\frac{11}3.$$
A: It may be instructive to, as an exercise, try a few problems similar to this one without skipping any steps (as follows). 
$$\begin{array}{lll}
-3&=&-\displaystyle\frac{8}{v-1}\\
-3&=&(-1)\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{-1}{1}\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{(-1)\cdot8}{1\cdot(v-1)}\\
-3&=&\displaystyle\frac{-8}{v-1}\\
\displaystyle\frac{-3}{1}&=&\displaystyle\frac{-8}{v-1}\\
\displaystyle\frac{-3}{1}\cdot\frac{v-1}{1}&=&\displaystyle\frac{-8}{v-1}\cdot\frac{v-1}{1}\\
\displaystyle\frac{-3(v-1)}{1\cdot 1}&=&\displaystyle\frac{-8(v-1)}{(v-1)\cdot 1}\\
\displaystyle\frac{-3(v-1)}{1}&=&\displaystyle\frac{-8(v-1)}{1\cdot(v-1)}\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\cdot\frac{v-1}{v-1}\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\cdot 1\\
-3(v-1)&=&\displaystyle\frac{-8}{1}\\
-3(v-1)&=&-8\\
\displaystyle\frac{-3(v-1)}{-3}&=&\displaystyle\frac{-8}{-3}\\
\displaystyle\frac{-3(v-1)}{-3\cdot 1}&=&\displaystyle\frac{(-1)\cdot8}{(-1)\cdot 3}\\
\displaystyle\frac{-3}{-3}\cdot \frac{v-1}{1}&=&\displaystyle\frac{(-1)}{(-1)}\cdot\frac{8}{3}\\
\displaystyle 1\cdot \frac{v-1}{1}&=&\displaystyle 1\cdot\frac{8}{3}\\
\displaystyle  \frac{v-1}{1}&=&\displaystyle \frac{8}{3}\\
\displaystyle v-1&=&\displaystyle \frac{8}{3}\\
\displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+1\\
\displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+\frac{3}{3}\\
\displaystyle (v+(-1))+1&=&\displaystyle \frac{8+3}{3}\\
\displaystyle v+((-1)+1)&=&\displaystyle \frac{11}{3}\\
\displaystyle v+0&=&\displaystyle \frac{11}{3}\\
\displaystyle v&=&\displaystyle \frac{11}{3}\\
\end{array}$$
