How do you solve the system #5x^2+3y^2=17# and #-x+y=-1#?

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Answer 1 · 2016-08-19T15:02:06

Solution set:
#{7/4, 3/4}, {-1, -2}#

#y = -1 + x#

#=>5x^2 + 3(-1 + x)^2 = 17#

#=>5x^2 + 3(1 - 2x + x^2) = 17#

#=>5x^2 + 3 - 6x + 3x^2 = 17#

#=>8x^2 - 6x - 14 = 0#

#=>2(4x^2 - 3x - 7) = 0#

#=>4x^2 +4x - 7x - 7 = 0#

#=>4x(x +1) - 7(x + 1) = 0#

#=>(4x - 7)(x + 1) = 0#

#=>x = 7/4 and -1#

Now, solving for y:

#y = -1 + 7/4" and " y = -1 - 1#

#y = 3/4" and "y = -2#

Hence, the solution set is #{7/4, 3/4} and {-1, -2}#.

Hopefully this helps!