How do you solve the system #x^2+y^2=5# and #y=3x+5#?

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Answer 1 · 2016-08-28T22:38:34

Solution set is #(-2,-1)# or #(-1,2)#

As #y=3x+5#, #x^2+y^2=5#

#hArrx^2+(3x+5)^2=5# or

#x^2×9x^2+2×3x×5+25=5# or

#10x^2+30x+20=0# or dividing by #10#, we get

#x^2+3x+2=0# or

#x^2+x+2x×2=0# or

#x(x+1)+2(x+1)=0# or

#(x+2)(x+1)=0# i.e.

#x=-2# or #x=-1#

Hence #y=3×(-2)+5=-1# or

#y=3×(-1)+5=2#

Hence solution set is #(-2,-1)# or #(-1,2)#