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

1 Answer
Jan 18, 2017

Answer:

#(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.