How do you solve the system #x+2y^2=-6# and #x+8y=0#?

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Answer 1 · 2016-07-17T08:59:21

Solution set is #(-8,1)# or #(-24,3)#.

The set of equations can be solved using substitution method. The second equation gives us #x=-8y# and putting this in first we get

#(-8y)+2y^2=-6# or

#2y^2-8y+6=0# and diving by #2# we get

#y^2-4y+3=0# or

#y^2-3y-y+3=0# or

#y(y-3)-1(y-3)=0# or

#(y-1)(y-3)=0# i.e. either

#y-1=0# or #y=1# and #x=-8# or

#y-3=0# or #y=3# and #x=-24#

Hence solution set is #(-8,1)# or #(-24,3)#.