How do you solve the system of equations #x^2+y^2=11# and #x+y=15# by substitution?

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Answer 1 · 2017-05-28T02:45:36

See below

We set #x=15-y# (from the second equation).

Then,

#x^2+y^2=11#

#(15-y)^2+y^2=11#

#225-30y+2y^2=11#

#2y^2-30y+214=0#

#y^2-15y+107=0# (this is very ugly :/)

We know that #x# is the opposite term in the quadratic formula, as the two equations are symmetrical.

#y=frac(15pmsqrt(225-4(107)))(2)=frac(15pmsqrt(-203))(2)#

So, #y=frac(15+isqrt(203))(2)# and #x=frac(15-isqrt(203))(2)#