# How do you solve the following system of equations?: 10x – 30y = -1, 2x + y = 8 ?

$x = \frac{239}{70}$
$y = \frac{41}{35}$

#### Explanation:

From the given:

$10 x - 30 y = - 1 \text{ }$first equation
$2 x + y = 8 \text{ }$second equation

Solve by method of substitution:

$y = 8 - 2 x$
Use this in the first equation

$10 x - 30 y = - 1 \text{ }$first equation
#10x-30(8-2x)=-1

$10 x - 240 + 60 x = - 1$

$70 x = 239$

$x = \frac{239}{70}$

Now use this value to solve y:
Go back to $y = 8 - 2 x$

$y = 8 - 2 \left(\frac{239}{70}\right)$

$y = \frac{560 - 478}{70}$

$y = \frac{82}{70}$

$y = \frac{41}{35}$

checking
$10 x - 30 y = - 1 \text{ }$first equation
$10 \left(\frac{239}{70}\right) - 30 \left(\frac{41}{35}\right) = - 1$

$\frac{239}{7} - \frac{246}{7} = - 1$

$- \frac{7}{7} = - 1$
$- 1 = - 1$

checking:
$2 x + y = 8 \text{ }$second equation

$2 \left(\frac{239}{70}\right) + \frac{41}{35} = 8$

$\frac{239}{35} + \frac{41}{35} = 8$
$\frac{280}{35} = 8$

$8 = 8$

God bless....I hope the explanation is useful