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# what is the solution to this linear system?

This is the question: x+y=2
3x-2y=-9

multiply the first eqn by 2
2(x + y = 2) gives 2x + 2y = 4
add to 2nd eqn to eliminate the y
5x = -5 so x = -1
substitute in either original eqn
using the 1st eqn, y = 2-x = 2-(-1) = 3

Hi Tracey;
x+y=2
3x-2y=-9
The coefficient of x in the second equation is 3.
The coefficient of y in the second equation is -2.
Let's take the first equation and multiply both sides by either coefficient.  I randomly choose 3...
3(x+y)=(2)(3)
3x+3y=6

Let's subtract one equation from the other...
3x+3y=6
-(3x-2y=-9)
0+5y=15
Please remember that subtracting a negative number is the same as adding a positive number...
3y--2y=3y+2y=5y
6--9=6+9=15
5y=15
Let's divide both sides by 5...
(5y)/5=15/5
y=3

Let's input this result into either equation to establish the value of x.  I randomly choose the first.  I will use its original form...
x+y=2
x+3=2
Let's subtract 3 from both sides...
x+3-3=2-3
x=-1

Let's take the second equation and verify the results...
3x-2y=-9
[(3)(-1)]-[(2)(3)]=-9
-3-6=-9
-9=-9