Answer:
(2,2)
Explanation:
Given equations are
x+3y=8 eq(1)
2x-y=2 eq(2)
we'll solve the system of equation by substitution.
Adding -3y to both sides of eq(1),we get
x+3y-3y=8-3y
x+0=8-3y
x=8-3y eq(3)
Put above value of x in eq(2),we get
2(8-3y)-y=2
16-6y-y=2
16-7y=2
Adding -16 to both sides of above equation,we get
-16+16-7y=-16+2
0-7y=-14
-7y=-14
Dividing by -7 to both sides of above equation,we get
-7y/-7=-14/-7
y=2
Putting y=2 in eq(3),we get
x=8-3(2)
x=8-6
x=2
hence, the solution for this system of equations is (2,2).