# How do you solve the following system?: 5x + 2y =1 , 8x-9y= 12

Jan 7, 2016

The solution set is: $S = \left\{\frac{33}{61} , - \frac{52}{61}\right\}$

#### Explanation:

Just isolate one constant($x \mathmr{and} y$) in one side on the equality and substitute in the other equality:
$5 x + 2 y = 1 \implies 5 x = 1 - 2 y \implies x = \frac{1 - 2 y}{5}$

Now, substitute:
$8 \cdot \frac{1 - 2 y}{5} - 9 y = 12$ In order to take the dividing $5$, multiply the hole equation by $5$:
$\cancel{5} \cdot \frac{8 - 16 y}{\cancel{5}} - 5 \cdot 9 y = 60$
$8 - 16 y - 45 y = 60$
$- 61 y = 52 \implies y = - \frac{52}{61}$

Now, just go back to the first equation and solve it:
$x = \frac{1 - 2 \cdot \left(- \frac{52}{61}\right)}{5} \implies x = \frac{1 + \frac{104}{61}}{5} \implies x = \frac{\frac{61}{61} + \frac{104}{61}}{5} \implies \frac{\frac{165}{61}}{5}$
$x = \frac{\frac{165}{61}}{5} = \frac{165}{61} \cdot \frac{1}{5} = \frac{33}{61}$

Then, the solution set is: $S = \left\{\frac{33}{61} , - \frac{52}{61}\right\}$