Help me solve the simultaneous equation No. 2

Billy OMBIMA

New member
Consider
\( x^2+y^2=5 ......(1) \)
\[ xy=2 .........(2)\]
By substitution, \( y=2/x \) putting which in (1):

\[ x^2 +(2/x)^2=5 \]
Simplify this further you get,
\[ x^2+4/x^2=5 \]
Multiply through by \( x^2 \):
\[ x^4-5x^2+4=0 \]
Factorising this, we get;
\[x^4-4x^2-x^2+4=0\]
\[x^2(x^2-4)-(x^2-4)=0\]
\[(x^2-1)(x^2-4)=0\]
\[(x-1)(x+1)(x-2)(x+2)=0\]
\[x=\pm1 , \pm2\]

Therefore, since \(y=2/x\),
\[y= \pm1, \pm2\]
 
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