# How do solve the following linear system?:  6x-4y=-1 , 3y = 15 -x ?

Jul 10, 2018

$x = \frac{57}{22} , y = \frac{91}{22}$

#### Explanation:

Writing the second equation in the form

$x = 15 - 3 y$
plugging this in the first equation

$6 \left(15 - 3 y\right) - 4 y = - 1$
expanding we get

$90 - 18 y - 4 y = - 1$
substracting $90$ and combining like Terms

$- 22 y = - 91$
so
$y = \frac{91}{22}$

so
$x = 15 - 3 \cdot \frac{91}{22}$

$x = \frac{15 \cdot 22 - 3 \cdot 91}{22} = \frac{57}{22}$

$x = \frac{57}{22}$, $y = \frac{91}{22}$

#### Explanation:

The given linear equations

$6 x - 4 y = - 1 \setminus \ldots \ldots \ldots . \left(1\right)$

$3 y = 15 - x$

or

$x + 3 y = 15 \setminus \ldots \ldots \ldots . \left(2\right)$

Multiplying (2) by $6$ & subtracting from (1) as follows

$6 x - 4 y - 6 \left(x + 3 y\right) = - 1 - 6 \setminus \times 15$

$- 22 y = - 91$

$y = \frac{91}{22}$

setting $y = \frac{91}{22}$ in (2), we get

$x = 15 - 3 y$

$= 15 - 3 \left(\frac{91}{22}\right)$

$= \frac{57}{22}$