Answer:
Option 3:
elimination using multiplication (3,-1)
Step-by-step 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
[tex] \implies \frac{16y}{16}= \frac{16}{16} [/tex]
[tex]\implies y = - 1[/tex]
Putting the value of y into eqn(2)
[tex] \implies 2x + 2( -1 ) = 4[/tex]
[tex]\implies 2x - 2= 4[/tex]
Adding 2to both sides
[tex]\implies 2x - 2+ 2 = 4 + 2[/tex]
[tex]\implies 2x= 6[/tex]
Dividing through by 2
[tex]\implies \frac{2x}{2} = \frac{6}{2} [/tex]
[tex]\implies x = 3[/tex]
Hence, (x, y)=(3,-1)
