How do you solve #y = x²# and #1 + y = 2x# using substitution?

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Answer 1 · 2016-05-03T01:10:57

We already know the value of y, so it's just a matter of substituting.

#y = x^2 -> 1 + x^2 = 2x#

#x^2 - 2x + 1 = 0#

#(x - 1)(x - 1) = 0#

#x = 1#

#y = x^2#

#y = (1)^2#

#y = 1#

The solution set is #{1, 1}#

Hopefully this helps!