#[(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.