How do you find the exact solutions to the system #y=2x+1# and #2x^2+y^2=11#?

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Answer 1 · 2016-11-02T02:58:44

We can directly substitute equation #1# into equation #2#.

#y = 2x + 1 -> 2x^2 + (2x + 1)^2 = 11#

#2x^2 + 4x^2 + 4x + 1 = 11#

#6x^2 + 4x - 10 = 0#

#2(3x^2 + 2x - 5) = 0#

#3x^2 + 2x - 5 = 0#

#3x^2 -3x + 5x - 5 = 0#

#3x(x - 1) + 5(x - 1) =0#

#(3x + 5)(x - 1) = 0#

#x = -5/3 and 1#

Case 1: #x = -5/3#

#y = 2x + 1#

#y = 2(-5/3) + 1#

#y= -10/3 + 1#

#y = -7/3#

Case 2: #x = 1#

#y = 2x + 1#

#y = 2(1) + 1#

#y = 3#

Hence, the solution sets are #{1, 3}# and #{-5/3, -7/3}#.

Hopefully this helps!