# How do you solve \frac { 4a + 5} { a + 8} = \frac { 4a + 6} { a + 9}?

Jun 20, 2018

$a = 1$

#### Explanation:

First of all, we must assume that $a \setminus \ne - 8$ and $a \setminus \ne - 9$, otherwise one of the two denominators would vanish.

With this assumption, we can multiply both sides by $\left(a + 8\right) \left(a + 9\right)$ to get

$\cancel{\left(a + 8\right)} \left(a + 9\right) \setminus \frac{4 a + 5}{\cancel{a + 8}} = \setminus \frac{4 a + 6}{\cancel{a + 9}} \left(a + 8\right) \cancel{\left(a + 9\right)}$

So, the equation is

$\left(a + 9\right) \left(4 a + 5\right) = \left(4 a + 6\right) \left(a + 8\right)$

Expand both sides to get

$\cancel{4 {a}^{2}} + 41 a + 45 = \cancel{4 {a}^{2}} + 38 a + 48$

Subtract $38 a$ from both sides:

$3 a + 45 = 48$

Subtract $45$ from both sides:

$3 a = 3$

Divide both sides by $3$:

$a = 1$

The solution respects the conditions $a \setminus \ne - 8$ and $a \setminus \ne - 9$, so we can accept it.