Solving the system with \(\displaystyle{x}^{{2}}-{x}{y}-{1}={0}\) and

alparcero97oy

alparcero97oy

Answered question

2022-04-01

Solving the system with x2xy1=0 and 2xy4y2+3=0.

Answer & Explanation

strahujufl75

strahujufl75

Beginner2022-04-02Added 8 answers

xy=x21
2(x21)=4y23
2x2=4y21
4y2=2x2+1 (3)
From the first equation,
x21=xy
Let's square both sides and use (3),
(x2-1)2=x2y2=x2(12x2+14)
x42x2+1=12x4+14x2
12x494x2+1=0
2x49x2+4=0
(2x21)(x24)=0

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