Just plug in the x and y values into both equations and see if they work. They must

work in BOTH equations to be a solution.

(3,2) means x = 3, y = 2 to

2x + 3y = 12 becomes

2(3) + 3(2) = 12

6 + 6 = 12

12 = 12 so it works for the first one

x - 4y = -5 becomes

3 - 4(2) = -5

3 - 8 = -5

-5 = -5

and it works in the 2nd one as well.

So (3,2) is the solution for the system of equations

The other way to do this is to find the solution to the system of equations, and see if it turns

out to be (3,2). You will eventually need to know how to do this anyhow, so here goes.

Take the 2nd equation

x - 4y = -5 and add 4y to both sides to make it

x = 4y - 5 Now you can go to the first equation and "substitute" 4y -5 in place of x

2x + 3y = 12 becomes

2(4y-5) + 3y = 12 or

8y - 10 + 3y = 12 or

11y - 10 = 12 add 10 to both sides to get

11y = 22 and divide both side by 11 to get

y = 2

since x = 4y - 5 you can plug in y =2 to find x

x = 4(2) - 5 = 8 - 5 = 3

so x = 3, y = 2 is the solution

This is just one of the ways to solve a system of equations, you'll end up learning another

methods call "elimination" as well.