# How do you solve (3n)/(n-1)+(6n-9)/(n-1)=6?

Jun 23, 2017

empty set

#### Explanation:

Add the two fractions as they have the same denominator.

$\frac{3 n}{n - 1} + \frac{6 n - 9}{n - 1} = \frac{9 n - 9}{n - 1} = 9 \ne 6$

Now multiply both sides by (n-1)

$\frac{\left(n - 1\right) \times \left(9 n - 9\right)}{n - 1} = 6 \times \left(n - 1\right)$ This gives

$\left(9 n - 9\right) = 6 \left(n - 1\right)$ the 9 can be factored out giving

$9 \left(n - 1\right) = 6 \left(n - 1\right)$ now subtract 6(n-1) from both sides

$9 \left(n - 1\right) - 6 \left(n - 1\right) = 6 \left(n - 1\right) - 6 \left(n - 1\right)$

$3 \left(n - 1\right) = 0$ distribute the 3 across the parenthesis

$3 n - 3 = 0$ add 3 to both sides.

$3 n - 3 + 3 = 0 + 3$

$3 n = 3$ divide both sides by 3

$\frac{3 n}{3} = \frac{3}{3}$ which gives

$n = 1$ But this value for $n$ is forbidden, else you divide by zero