# How do you solve the following linear system:  x + 3y= 8, 2x + y = 15 ?

Mar 9, 2016

$\textcolor{b l u e}{x = \frac{37}{5} \text{ ; } y = \frac{1}{5}}$

#### Explanation:

Given:
$x + 3 y = 8$............................(1)
$2 x + y = 15$..........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The most straight forward way of dealing with this one is to make the coefficients of y the same and then subtract so that you only have 1 variable.

Multiply equation (2) by 3 then subtract, giving:

$6 x + 3 y = 45. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$
$\underline{\textcolor{w h i t e}{.} x + 3 y = 8} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$
$5 x + 0 \textcolor{w h i t e}{.} = 37$

Divide both sides by 5 giving:

$\textcolor{b l u e}{x = \frac{37}{5}}$................................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for $x$ in equation (1)

$\frac{37}{5} + 3 y = 8$

Subtract $\frac{37}{5}$ from both sides

$3 y = 8 - \frac{37}{5} = \frac{3}{5}$

Divide both sides by 3

$\textcolor{b l u e}{y = \frac{1}{5}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
$\textcolor{b r o w n}{x + 3 y = 8} \textcolor{g r e e n}{\to \frac{37}{5} + 3 \left(\frac{1}{5}\right) = 8} \textcolor{red}{\text{ True}}$

$\textcolor{b r o w n}{2 x + y = 15} \textcolor{g r e e n}{\to 2 \left(\frac{37}{5}\right) + \frac{1}{5} = 15} \textcolor{red}{\text{ True}}$