# How do you solve the following linear system:  3y+8x=7 , 8x=y+1 ?

Apr 15, 2018

$x = \left(\frac{5}{16}\right)$ and $y = \left(\frac{3}{2}\right)$

#### Explanation:

Firstly, re-arrange the equations so that the $x$ and $y$ and their respective coefficients are on the lhs of the equal sign, and the constants on the rhs. Number the equations accordingly:

$8 x + 3 y = 7$ $\left(1\right)$
$8 x - y = 1$ $\left(2\right)$

subtract $\left(2\right)$ from $\left(1\right)$:

$8 x - 8 x = 0$ immediately allows us to solve for $y$:

$3 y - \left(- y\right) = 4 y$ and $7 - 1 = 6$

$4 y = 6$
$y = \left(\frac{6}{4}\right) = \left(\frac{3}{2}\right)$
Next, substitute $y = \left(\frac{3}{2}\right)$ back into equation $\left(2\right)$, giving:
$8 x - \left(\frac{3}{2}\right) = 1$
$8 x = 1 + \left(\frac{3}{2}\right) = \left(\frac{5}{2}\right)$
$x = \frac{\frac{5}{2}}{8}$
$x = \left(\frac{5}{16}\right)$
The solution is: $x = \left(\frac{5}{16}\right)$ and $y = \left(\frac{3}{2}\right)$

CHECK SOLUTION (IMPORTANT): Substitute these values back into equations $\left(1\right)$ and $\left(2\right)$ to make sure you arrive at $7$ for equation $\left(1\right)$ and $1$ for equation $\left(2\right)$.