Solve for $v$ - simplify as much as possible

Solve for $v$. Simplify the answer.

$$-3 = -\frac{8}{v-1}$$

Here is what I tried:

$$-3 = \frac{-8}{v-1}$$ $$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8)$$ $$24 = v-1$$ $$25 = v$$

• Have you tried anything? People on this site will be more willing to help if you show your work and explain where you're confused. – Michael Burr Aug 18 '15 at 18:58
• This is what I did: -3 = -8/v-1 -8 x -3 = -8/v-1 x -8 24 = v-1 25 = v But when I pluged it back into the first equation I didn't get -3 as my answer... – Angie Aug 18 '15 at 18:59
• I did one more edit to make it readable and understand your logic. Check it after it will be peer reviewed if I did everything correct. – Slowpoke Aug 18 '15 at 19:12
• $\frac{(-8)}{v-1}(-8)$ is not $v-1$. Rather, it is $\frac{64}{v-1}$. – Akiva Weinberger Aug 18 '15 at 19:58

$$-3 = -\frac {8} {v-1}$$ $$-3(v-1) = -8$$ $$v-1 = \frac {8} {3}$$ $$v = \frac {11} {3}$$
\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}
Observe the solution, if any, must be $\ne 1$. I'd first change the signs and multiply both sides by $v-1$: $$3v-3=8\iff v=\frac{11}3.$$
It may be instructive to, as an exercise, try a few problems similar to this one without skipping any steps (as follows). $$\begin{array}{lll} -3&=&-\displaystyle\frac{8}{v-1}\\ -3&=&(-1)\cdot\displaystyle\frac{8}{v-1}\\ -3&=&\displaystyle\frac{-1}{1}\cdot\displaystyle\frac{8}{v-1}\\ -3&=&\displaystyle\frac{(-1)\cdot8}{1\cdot(v-1)}\\ -3&=&\displaystyle\frac{-8}{v-1}\\ \displaystyle\frac{-3}{1}&=&\displaystyle\frac{-8}{v-1}\\ \displaystyle\frac{-3}{1}\cdot\frac{v-1}{1}&=&\displaystyle\frac{-8}{v-1}\cdot\frac{v-1}{1}\\ \displaystyle\frac{-3(v-1)}{1\cdot 1}&=&\displaystyle\frac{-8(v-1)}{(v-1)\cdot 1}\\ \displaystyle\frac{-3(v-1)}{1}&=&\displaystyle\frac{-8(v-1)}{1\cdot(v-1)}\\ -3(v-1)&=&\displaystyle\frac{-8}{1}\cdot\frac{v-1}{v-1}\\ -3(v-1)&=&\displaystyle\frac{-8}{1}\cdot 1\\ -3(v-1)&=&\displaystyle\frac{-8}{1}\\ -3(v-1)&=&-8\\ \displaystyle\frac{-3(v-1)}{-3}&=&\displaystyle\frac{-8}{-3}\\ \displaystyle\frac{-3(v-1)}{-3\cdot 1}&=&\displaystyle\frac{(-1)\cdot8}{(-1)\cdot 3}\\ \displaystyle\frac{-3}{-3}\cdot \frac{v-1}{1}&=&\displaystyle\frac{(-1)}{(-1)}\cdot\frac{8}{3}\\ \displaystyle 1\cdot \frac{v-1}{1}&=&\displaystyle 1\cdot\frac{8}{3}\\ \displaystyle \frac{v-1}{1}&=&\displaystyle \frac{8}{3}\\ \displaystyle v-1&=&\displaystyle \frac{8}{3}\\ \displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+1\\ \displaystyle (v-1)+1&=&\displaystyle \frac{8}{3}+\frac{3}{3}\\ \displaystyle (v+(-1))+1&=&\displaystyle \frac{8+3}{3}\\ \displaystyle v+((-1)+1)&=&\displaystyle \frac{11}{3}\\ \displaystyle v+0&=&\displaystyle \frac{11}{3}\\ \displaystyle v&=&\displaystyle \frac{11}{3}\\ \end{array}$$