How do you find the exact solutions to the system #y-x=1# and #x^2+y^2=25#?

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Answer 1 · 2016-11-25T00:58:38

You substitute #y# as function of #x# in the equation of higher degree, using the equation of lower degree, then you solve the former.

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

Then
#x^2+y^2=25#

#x^2 +(1+x)^2 = 25#

#2x^2+2x+1=25#

#2x^2+2x-24 = 0#

#x= (-2 +-sqrt(4+192))/4=(-2+-14)/4#

#x_1=-4#
#x_2=3#

and

#y_1=-3#
#y_2=4#