# How do you solve 3x+2y=5 and 2x+5y=-3?

Apr 19, 2016

x = 31/11
y = -19/11

#### Explanation:

Note that these are both line equations (no high powers like x^2, x^3). Also, there are two equations and two unknowns. If the two equations have different slopes then the lines intersect. The intersection point is the solution.

There are a couple approaches. One way is to isolate a variable in one equation and substitute it into the other. Another approach is to apply linear operations to the equations. Let's use method #2, and let's aim to make the coefficient of X the same.

Multiply eq1 through by 2:
6× + 4y = 10
Multiply eq2 through by 3:
6× + 15y = -9

The equations are different while the coefficient of X is the same so the slopes are different, and there is an answer.

We can subtract eq2 from eq1 term by term:

6x-6×+4y-15y=10+9
-11y=19, or y=-19/11

If we put y into each of the original equations we should get the same x.
Eq1:
3x = 5 - 2(-19/11)
3x = (55+38)/11
x = 93/33
x = 31/11

As a check, repeat with Eq2:
2x = -3 -5(-19/11)
2x = (-33+95)/11; 62/11
x = (62/11)/2; 31/11