# How do you solve the following system?: -1/2x+5y=-20 , 3x-2y=1

$x = - \frac{105}{42}$ and $y = - \frac{119}{28}$

#### Explanation:

The given system
$- \frac{1}{2} x + 5 y = - 20$ and $3 x - 2 y = 1$

Let us use the method of substitution

From the second equation $3 x - 2 y = 1$

$x = \frac{1 + 2 y}{3}$

We use the second in the first equation

$- \frac{1}{2} x + 5 y = - 20$

$- \frac{1}{2} \left(\frac{1 + 2 y}{3}\right) + 5 y = - 20$

$- \left(\frac{1 + 2 y}{3}\right) + 10 y = - 40$

$\left(\frac{1 + 2 y}{3}\right) - 10 y = 40$

$1 + 2 y - 30 y = 120$

$- 28 y = 119$

$y = - \frac{119}{28}$
~~~~~~~~~~~~~~~~~~~~~~~

Solve for $x$

$x = \frac{1 + 2 \left(- \frac{119}{28}\right)}{3}$

$x = \frac{28 - 238}{84}$

$x = - \frac{210}{84}$

$x = - \frac{105}{42}$
~~~~~~~~~~~~~~~~~~~~~~

Checking:at $x = - \frac{105}{42}$ and $y = - \frac{119}{28}$

$- \frac{1}{2} x + 5 y = - 20$
$- \frac{1}{2} \left(- \frac{105}{42}\right) + 5 \left(- \frac{119}{28}\right) = - 20$

$\frac{105}{84} - \frac{595}{28} = - 20$

$\frac{105 - 1785}{84} = - 20$

$\frac{- 1680}{84} = - 20$

#-20=-20

Also

$3 x - 2 y = 1$

$3 \left(- \frac{105}{42}\right) - 2 \left(- \frac{119}{28}\right) = 1$

$- \frac{105}{14} + \frac{119}{14} = 1$

$\frac{14}{14} = 1$

$1 = 1$

Correct ! at $x = - \frac{105}{42}$ and $y = - \frac{119}{28}$

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