# How do you solve x + y = 5 and 2x + Y = 8?

Jan 8, 2016

$\left(3 , 2\right)$

#### Explanation:

This could be solved with either elimination or substitution. I'll use elimination, since the category is "Linear Systems with Multiplication."

Multiply the first equation by $- 1$.

$- 1 \left(x + y = 5\right) \rightarrow - x - y = - 5$

We now have the two equations:

$\left\{\begin{matrix}- x - y = - 5 \\ 2 x + y = 8\end{matrix}\right.$

$- x + 2 x - y + y = - 5 + 8$

Simplify. Notice that the $y$ terms cancel, which was the point in multiplying the first equation by $- 1$ earlier.

$x = 3$

Now, substitute $x = 3$ into either of the original equations.

$3 + y = 5$
$y = 2$

This leaves us with $x = 3 , y = 2$, or the point $\left(3 , 2\right)$.