# How do you solve 5/(y - 3) = (y + 7) /(2y - 6) + 1 and find any extraneous solutions?

Nov 29, 2017

$y = 3$ and it is extraneous. Basically, no real solutions.

#### Explanation:

First, multiply both sides by $\left(y - 3\right) \left(2 y - 6\right)$ to get rid of the fractions.

$\frac{5}{y - 3} = \frac{y + 7}{2 y - 6} + 1 \implies \left(y - 3\right) \left(2 y - 6\right) \times \frac{5}{y - 3} = \left(y - 3\right) \left(2 y - 6\right) \times \frac{y + 7}{2 y - 6} + 1 \left(y - 3\right) \left(2 y - 6\right)$

Which is equal to $\left(2 y - 6\right) 5 = \left(y - 3\right) \left(y + 7\right) + \left(y - 3\right) \left(2 y - 6\right)$

Distribute them out to get $10 y - 30 = {y}^{2} + 4 y - 21 + 2 {y}^{2} - 12 y + 18$

Combine like terms: $10 y - 30 = 3 {y}^{2} - 8 y - 3$

Use the inverse operation to "clear out" the left side of the equation.
$10 y - 30 = 3 {y}^{2} - 8 y - 3$ => $0 = 3 {y}^{2} - 18 y + 27$

When we factor this, we get $0 = 3 {\left(y - 3\right)}^{2}$ The only solution to this equation is when $0 = y - 3$ which means that $y = 3$.

To see whether this is extraneous, we plug the value in the equation.
$\frac{5}{y - 3} = \frac{y + 7}{2 y - 6} + 1$ becomes $\frac{5}{0} = \frac{y + 7}{0} + 1$ , indicating that the equation is undefined. Therefore, the solution is extraneous.