Answer:
Option 3:
elimination using multiplication (3,-1)
Explanation:
The system of equations to be solved are
-5x+3y=-18.......eqn (1)
2x+2y=4. .......eqn (2)
We first multiply eqn(1) by 2 and eqn(2) by 5 to get eqn(3) and eqn(4) respectively.
This implies
-10x+6y=-36.......eqn (3)
10x+10y= 20.......eqn (4)
we then add eqn(3) and eqn(4) to obtain
16y=-16
We divide through by 16


Putting the value of y into eqn(2)


Adding 2to both sides


Dividing through by 2


Hence, (x, y)=(3,-1)