Question

How do you solve the system of equations `2x+2y=4` and `12-3x=3y`?

Answer 1

No solutions

Explanation:

`2x + 2y = 4`
`-3x + 12=3y`

We need to solve `2x + 2y = 4` for `x`

Let's start by adding `color(red)(-2y)` to both sides

`2x + 2y color(red)(-2y) = 4 color(red)(-2y)`

`2x = -2y + 4`

`x = (-2y+4)/2`

`x= -y + 2`

Then, substitute `-y+2` for `x` in `-3x + 12=3y`

`-3x + 12 = 3y`

`-3(-y + 2)+12 = 3y`

`3y - 6 +12 = 3y`

`3y + 6 = 3y`

Then add `-3y` to both sides

`3y - 3y + 6 = 3y - 3y`

`6= 0`

Finally, add `-6` to both sides

`6 - 6 = 0 - 6`

`0=-6`

Thus,

There is no solutions.

Answer 2

No solution

Explanation:

Rewriting the second equation gives `3x+3y=12`. Dividing each term by `3` then yields `x+y=4`.

For the first equation `2x+2y=4`, we can divide each term by `2`, yielding `x+y=2`.

Then, we have that `x+y=2` and `x+y=4`. But how could this ever be the case? `x+y` should always give the same value for a fixed `x` and fixed `y`. Thus there are no solutions.