# How do solve the following linear system?:  x-4y=2 , 2x+6y=5 ?

Feb 17, 2016

$\left(\frac{16}{7} , \frac{1}{14}\right)$

#### Explanation:

In order to solve systems of linear equations, you can either use the

1. Elimination Method
2. Substitution Method
3. Graphing

In this case, I would use the elimination method but note that you may use any method that you are comfortable with and still arrive at the same answer.

[Solution]
$x - 4 y = 2$
$2 x + 6 y = 5$

Multiply 2 to both sides of the first equation...

$2 x - 8 y = 4$

Doing this makes the coefficient of x in the two equations the same. This enables us to subtract one equation from the other to eliminate the $x$ variable

[Subtracting one equation from the other]
$2 x + 6 y = 5$
$2 x - 8 y = 4$

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

Now that we know the value of $y$, we can now evaluate any of the two equations with the value of $y$ to get the value of $x$

$x - 4 \left(\frac{1}{14}\right) = 2$
$x - \frac{2}{7} = 2$
$x = 2 + \frac{2}{7}$
$x = \frac{16}{7}$

[Checking]
Substituting to equation 1 -> $x - 4 y = 2$
$\frac{16}{7} - 4 \left(\frac{1}{14}\right) = 2$
$\frac{16}{7} - \frac{2}{7} = 2$
$\frac{14}{7} = 2$
$2 = 2$

Substituting to equation 2 -> $2 x + 6 y = 5$
$2 \left(\frac{16}{7}\right) + 6 \left(\frac{1}{14}\right) = 5$
$\frac{32}{7} + \frac{3}{7} = 5$
$\frac{35}{7} = 5$
$5 = 5$

$\left(\frac{16}{7} , \frac{1}{14}\right)$