solve the system of equation by comparison x+2y=4, x+y=2?

Answer 1

1st equation (solve x):
x + 2y = 4
x = 4 - 2y

2nd equation (solve y, substituting x with (4 - 2y):
(4 - 2y) + y = 2
4 - 2y + y = 2
- y = 2 - 4
- y = - 2
y = 2

Go back to the 1st equation substituting y with 2:
x = 4 - (2 * [2])
x = 4 - 4
x = 0

Proofs:
1st equation:
0 + (2 * 2) = 4
0 + 4 = 4
4 = 4

2nd equation:
0 + 2 = 2
2 = 2

Answer 2

X=0 Y=2

Answer 3

x + 2y = 4
x + y = 2

Subtracting the second from the first equation we get,

x + 2y - x - y = 4 - 2

y = 2

Substituting that value of y in the second equation,

x + 2 = 2

x = 2 - 2 = 0

So, x = 0 and y = 2 is the solution

Answer 4

x+2y=4
x=4-2y
and
x+y=2
x=2-y
now
Or 4-2y=2-y
or -2y+y=2-4
Or -y=-2
Or y=2
and solve for x
x=2-2
x=0

Answer 5

x + 2y = 4
x + y = 2
multiply the 2nd equation by 2
so,

2x + 2y = 4
so 2x + 2y = x + 2y
so x + 2y = 2y
so x = 0

x + 2y = 4
so 0 + 2y = 4
so y = 2

Answer 6

x+2y=4
Or x=4-2y

x+y=2
Or x=2-y

Or 4-2y=2-y
or -2y+y=2-4
Or -y=-2
Or y=2
Similarly solve for x

Answer 7

not sure wat u mean by "by comparision" but in x+y=2 you subtract x(or y) from each side so it would be y=2-x and then in the other equation you just put that in for y. so it would be x+2(2-x)=4. then distribute and solve

Answer 8

Rearrange terms
x = 4 - 2y; x = 2 - y
Compare
4 - 2y = 2 - y
and solve
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