# How do you solve the following linear system:  6x+y =3, 2x+3y=5 ?

Mar 4, 2018

$x = \frac{1}{4} , y = \frac{3}{2}$

#### Explanation:

In this case, we can use substitution, but I find using elimination simpler. We can see that if we do a little work, subtracting the two equations will let us solve for $y$.

${E}_{1} : 6 x + y = 3$

${E}_{2} : 2 x + 3 y = 5$

${E}_{2} : 3 \left(2 x + 3 y\right) = 3 \cdot 5$

${E}_{2} : 6 x + 9 y = 15$

${E}_{1} - {E}_{2} : 6 x + y - \left(6 x + 9 y\right) = 3 - 15$

$6 x - 6 x + y - 9 y = - 12$

$- 8 y = - 12$

$y = \frac{- 12}{- 8} = \frac{3}{2}$

Now we plug in the solution to $y$ into ${E}_{1}$ to solve for $x$:

${E}_{1} : 6 x + \frac{3}{2} = 3$

$6 x = 3 - \frac{3}{2}$

$6 x = \frac{3}{2}$

$x = \frac{\frac{3}{2}}{6} = \frac{3}{12} = \frac{1}{4}$