# How do solve the following linear system?:  x - 3y = -6 , 10x - 7y = -18 ?

Jan 27, 2016

$\left\{x , y\right\} = \left\{- \frac{12}{23} , \frac{42}{23}\right\}$

#### Explanation:

Solve by elimination

$x - 3 y = - 6$
$10 x - 7 y = - 18$

We can eliminate $10 x$ from the second equation by $x$ in the first equation if we multiply $x$ by $- 10$ to get $- 10 x$

$\rightarrow - 10 \left(x - 3 y = - 6\right)$

$\rightarrow - 10 x + 30 y = 60$

Now add both of the equations

$\rightarrow \left(- 10 x + 30 y = 60\right) + \left(10 x - 7 y = - 18\right)$

$\rightarrow 23 y = 42$

$y = \frac{42}{23}$

Now substitute the value of y to the first equation

$x - 3 \left(\frac{42}{23}\right) = - 6$

$x - \frac{126}{23} = - 6$

$x = - 6 + \frac{126}{23} = \frac{- 138 + 126}{23} = - \frac{12}{23}$

So,$\left\{x , y\right\} = \left\{- \frac{12}{23} , \frac{42}{23}\right\}$