Solve the system of equations below.

5x+2y=9 2x-3y=15

A. (3,-3)

B. (-3,12)

C. (12,-3)

D. (-3,3)

A. (3, - 3)

Step-by-step explanation:

Let's use the addition/elimination for this to avoid fractions. You could solve one of the equations for y or x and then sub that value into the other equation, but you would have to deal with fractions along the way, so let's just not. Using the addition/elimination (they're the same thing; math books sometimes use one or the other), we line up our equations:

5x + 2y = 9

2x - 3y = 15

It looks like if we want to eliminate the y terms, we can multiply the top equation by 3 and the bottom equation by 2 to get a whole new set of equations:

15x + 6y = 27

4x - 6y = 30

Now we can see that when we add down the columns, the y's cancel each other out. 6y's minus 6y's leaves us with no y's. What we are left with is

19x = 57 so

x = 3.

Now we will sub that 3 back in to either of the original equations to solve for y:

2 (3) - 3y = 15 and

6 - 3y = 15 and

-3y = 9 so

y = - 3

The solution set then is (3, - 3)