# How do you solve the system of equations 10x + 8y = 2 and - 2x - 4y = 6?

Jun 22, 2018

$x = \frac{7}{3} \mathmr{and} y = - \frac{8}{3}$

#### Explanation:

From first equation: $y = \frac{2 - 10 x}{8} = \frac{\cancel{2} \left(1 - 5 x\right)}{\cancel{8}} = \frac{1 - 5 x}{4}$

Substitute to second equation: $- 2 x - \cancel{4} \frac{1 - 5 x}{\cancel{4}} = 6$

So: $- 2 x - 1 + 5 x = 6 \to x = \frac{7}{3}$

Then: $y = \frac{2 - 10 \left(\frac{7}{3}\right)}{8} = \left(2 - \frac{70}{3}\right) \cdot \frac{1}{8} = - \frac{\cancel{64}}{3} \cdot \frac{1}{\cancel{8}} = - \frac{8}{3}$

Jun 22, 2018

$x = \frac{7}{3} , y = - \frac{8}{3}$

#### Explanation:

Multiplying the second equation by  5 and adding to the first we get

$- 6 y = 16$
so

$y = - \frac{8}{3}$
so we get

$- x + \frac{16}{3} = 3$

$x = \frac{7}{3}$