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

Jan 9, 2016

Solving using substitution

#### Explanation:

2x + y = -5
y = -5 - 2x
3x - 5(-5 - 2x) = 2
3x + 25 + 10x = 2
13x = 2 - 25
13x = -23
x = $- \frac{23}{13}$

Substitute $- \frac{23}{13}$ for x.

2($- \frac{23}{13}$) + y = -5
$- \frac{46}{13}$ + y = -5
y = $- \frac{19}{13}$

The solution set is ($- \frac{23}{13}$, $- \frac{19}{13}$).

Below are a few exercises to practice yourself. At the end, I put a linear-quadratic system a for a little bit of a challenge should you want it... Be warned of more than one solution set in the linear-quadratic system. Feel free to ask for help with the linear-quadratic system.

a) 2x + 3y = 6
x - 5y = -11

b) 4x - 7y = -12
2x - 14y = 3

c) ${x}^{2}$ + 12x = -y - 13
-2x - y = 4

Enjoy!