How do you solve the system #y = x²# and #1 + y = 2x#?

1 Answer
Mar 9, 2016

Answer:

#x=1, y=1#

Explanation:

When trying to find values of x and y in a system, check first which value can be easily changed so that it becomes equal. In this case let's pick the y, but let us write first the system in its format.
#y=x^2#
#1+y=2x# ----------> make everything negative so it will be erased the y later

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

Now we have:
#-1=-2x+x^2# or #0= x^2-2x+1# which is equal to # 0= (x-1)^2#
Now we need to solve for x:
-we cancel the squared expression and use only #x-1# which #x=1#

Now we can solve for y by taking any of the two equations:

  • #y=x^2=1^2=1 #
    -#y=2x-1= 2*1-1=1#

You have now the solution for both values