# How do you solve the rational equation (3a-2)/(2a+2)=3/(a-1)?

Jan 7, 2016

Multiply both sides by $\left(2 a + 2\right) \left(a - 1\right)$, simplify and solve to find:

$a = - \frac{1}{3}$ or $a = 4$

#### Explanation:

Multiply both sides by $\left(2 a + 2\right) \left(a - 1\right)$ to get:

$\left(3 a - 2\right) \left(a - 1\right) = 3 \left(2 a + 2\right)$

That is:

$3 {a}^{2} - 5 a + 2 = 6 a + 6$

Subtract $6 a + 6$ from both sides to get:

$3 {a}^{2} - 11 a - 4 = 0$

Factor this by noticing that $A C = 3 \cdot 4 = 12$ has factors $12$ and $1$ which differ by $B = 11$.

So:

$0 = 3 {a}^{2} - 11 a - 4$

$= \left(3 {a}^{2} - 12 a\right) + \left(a - 4\right)$

$= 3 a \left(a - 4\right) + 1 \left(a - 4\right)$

$= \left(3 a + 1\right) \left(a - 4\right)$

Hence $a = - \frac{1}{3}$ or $a = 4$

It remains to check that when we multiplied both sides by $\left(2 a + 2\right) \left(a - 1\right)$ we were not multiplying both sides by $0$.

That's OK since $\left(2 a + 2\right) = 0$ when $a = - 1$ and $\left(a - 1\right) = 0$ when $a = 1$, so neither of the solutions of the multiplied equation were introduced by the multiplication step.