# How do you solve the following system of equations?: 4x+3y=8 , x - y= 10?

Feb 27, 2017

$x = 5 \frac{3}{7}$ and $y = - 4 \frac{4}{7}$

#### Explanation:

$4 x + 3 y = 8$
$x - y = 10$

To determine the values of $x$ and $y$, we must first determine from one equation a temporary value for either of the terms, and then substitute that temporary value in the other equation.

We determine a temporary value for $x$ from the second equation.

$x - y = 10$

Add $y$ to both sides.

$x = 10 + y$

In the first equation, substitute $x$ with $\textcolor{red}{\left(10 + y\right)}$.

$4 x + 3 y = 8$

$4 \textcolor{red}{\left(10 + y\right)} + 3 y = 8$

Open the brackets and simplify.

$40 + 4 y + 3 y = 8$

$40 + 7 y = 8$

Subtract $40$ from both sides.

$7 y = - 32$

Divide both sides by $7$.

$y = - \frac{32}{7} = - 4 \frac{4}{7}$

In the second equation, substitute $y$ with $\textcolor{b l u e}{- \frac{32}{7}}$.

$x - y = 10$

$x - \left(\textcolor{b l u e}{- \frac{32}{7}}\right) = 10$

Open the brackets and simplify. The product of two negatives is a positive.

$x + \frac{32}{7} = 10$

Multiply all terms by $7$.

$7 x + \left(7 \times \frac{32}{7}\right) = 70$

$7 x + \left(1 \cancel{7} \times \frac{32}{1 \cancel{7}}\right) = 70$

$7 x + 32 = 70$

subtract $32$ from each side.

$7 x = 38$

Divide both sides by $7$.

$x = \frac{38}{7} = 5 \frac{3}{7}$