Solve the system by addition.?

Question

Solve the system by addition. 2x + y = 4 x + y = 3

Answer 1

2x + y = 4
x + y = 3

Subtract the 2 equations (so the y's cancel out), so you are left with
2x-x =x
y-y=0
4-3=1

x=1

Sub this back into the 2nd equation
x + y = 3
(1) + y = 3
y=2

Therefore, x=1, y=2

Answer 2

y = 3 x = 0

Answer 3

The two equations:
2x + y = 4
x + y = 3

1) Take the x's, y's and numbers away from each other:

2x + y = 4
-
x + y = 3

take the bottom x off the top 2x's (which leaves one x at the top and zero at the bottom),
take one y from the bottom off the top y (which leaves zero y's)
and take the bottom 3 off the 4 at the top (which leaves 1 at the top).

So now we're left with x = 1

Now you know what x is!




Put x in to ANY of the two equations.
(I'll use x + y = 3)

so:
1 + y = 3

rearrange it to get y = 3 - 1
so y = 2



Then check if x=1, y=2 works for both equations!

Answer 4

Is there a comma between 4 and x? So that
2x + y = 4--------------[1]
x + y = 3----------------[2]

subtract [2] from [1]
x = 1
substitute to [2]
1 + y = 3
y = 2

Answer 5

2x + y = 4 ........(1)
x + y = 3 ........(2)

multiply -1 on both side. It makes:
-x - y = -3 ......(3)

(1) + (3)
(2x +y) - x - y = 4 -3
x=1

substitue x with 1 in (2)
y=2

Answer 6

2x + y = 4
- x - y = - 3-----------ADD
x = 1

2 + y = 4
y = 2

x = 1 , y = 2