Question

How do you solve algebraically `x+y=2` and `x-y=4`?

Answer 1

If you have 2 equations you can add the left sides of the equations together and the right sides of the equations together to get a new valid equation

`((,x+y,=,2),(+,x-y,=,4),(,2x+-0y,=,6))`

`2x = 6`
`x=3`

Substituting `x=3` back into either of our original allows us to solve for `y`
For example using `x+y=2`
we get
`(3)+y = 2`

`y=-1`

Answer 2

• A very simple way to solve this is by ADDING the two equations.

• The two equations are
  `x+y=2` ----(1)
  `x-y=4` ----(2)

Adding the Left Hand Sides will give us:

`(x+y)+(x-y)`
`= x+cancel(y)+x-cancel(y)`
`=2x`

Adding the Right Hand Sides will give us:
`2 + 4 = 6`

That gives us
`2x = 6`
Dividing both sides by 2, we get
`(cancel(2)x)/cancel(2) = 6/2`
`x = 3`

Substituting `x=3` in (1) we get
`3+y = 2`
Subtracting 3 from both sides gives us
`cancel(3) + y - cancel(3) = 2 - 3`
`y = -1`

The Solution for these two equations is `x = 3 and y = -1`

• Once we arrive at a solution, it is a good idea to VERIFY our answer

Substituting `x=3 , y=-1` in (1) we get
Left Hand Side: `x+y = 3+(-1) = 2`(Right Hand Side)

Substituting `x=3 , y=-1` in (2) we get
Left Hand Side: `x-y = 3-(-1) = 3+1 = 4`(Right Hand Side)

We have verified our answer, and we can be sure that it's correct.