Question

How do you solve `[(x^2+1, 5-y), (x+y,y-4)]=[(5,x), (5,3)]`?

Answer 1

`(x,y)=(-2,7)`

Explanation:

`[(x^2+1,color(white)("x"),5-y),(x+y,,y-4)]=[(5,color(white)("x"),x),(5,,3)]`

`rArr`
[1]`color(white)("XX")x^2+1=5`
[2]`color(white)("XX")5-y=x`
[3]`color(white)("XX")x+y=5`
[4]`color(white)("XX")y-4=3`

Examining the above,
we note that both [1] and [4] only involve a single variable
and of the two [4] would appear to be the easiest to solve:
`y-4=3color(white)("XX")rarrcolor(white)("XX")y=7`

We can now substitute `7` for `y` back into [2]
`5-7=xcolor(white)("XX")rarrcolor(white)("XX")x=-2`

We now have our solution `(x,y)=(-2,7)`
provided this is consistent with the other two equations:
[1]`color(white)("XX")x^2+1` with `x=-2color(white)("x")rarrcolor(white)("x")2^2+1=5` (correct)
[3]`color(white)("XX")x+y=5` with `x=-2` and `y=7color(white)("x")rarrcolor(white)("X")-2+7=5` (correct)

So `(x,y)=(-2,7)` is a consistent (valid) solution.