# How do you solve the system x + y = 2 and 2x + y = -1?

Apr 5, 2018

$x = \frac{1}{3}$

$y = \frac{5}{3}$

#### Explanation:

from this equation

$x + y = 2$

$x = 2 - y$

put the value of $x$ in this equation

$2 x + y = - 1$

$2 \left(2 - y\right) + y = - 1$

$4 - 4 y + y = - 1$

$- 3 y = - 5$

$y = \frac{5}{3}$

$x + y = 2$

$x + \frac{5}{3} = 2$

$x = \frac{1}{3}$

Apr 5, 2018

$x = - 2 , y = 3$

#### Explanation:

This is a simultaneous equation question. In the interest of time, I will solve it using the elimination method:

$x + y = 1$

$2 x + y = - 1$

To solve using elimination, all we do is literally subtract the bottom line from the top line which gets rid of the y $\left(y - y = 0\right)$, giving us:

$- x = 2$

$x = - 2$

We then substitute ($x = - 2$) into either of the equations:

$x + y = 1$

$\left(- 2\right) + y = 1$

$y = 3$

And then check that this works, using substituting both values into the second equation.

$- 2 x + y = - 1$

$2 \left(- 2\right) + \left(3\right) = - 1$

$- 1 = - 1$, so we know the values are correct!