How do you solve the system 3x2−y2=11, x2+2y=2?

1 Answer
Cem Sentin
Mar 7, 2018

Pairs of the solution are (−2√3,−5), (−2,−1), (2,−1) and (2√3,−5)

Explanation:

3⋅(x2+2y)−(3x2−y2)=2⋅3−11

3x2+6y−3x2+y2=−5

y2+6y=−5

y2+6y+5=0

(y+1)⋅(y+5)=0

y1=−5 and y2=−1

For y=−5, x2+2⋅(−5)=2, so x2=12

Hence x1=−2√3 and x2=2√3

For y=−1, x2+2⋅(−1)=2, so x2=4

Hence x3=−2 and x4=2

Thus, pairs of the solution are (−2√3,−5), (−2,−1), (2,−1) and (2√3,−5)

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