Question

Please solve the following system of equations
x+3y=8
2x-y=2

Answer 1

(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).