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Consistency of Linear System of Equations

Solve the fol...

Question

Solve the following equation.
$x^{2} y^{2} - 2 x + y^{2} = 0, 2 x^{2} - 4 x + 3 + y^{3} = 0$ .

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Solution

The given system of equations is equivalent to $y^{2} = 2 x / (1 + x^{2}), 2 (x - 1)^{2} + 1 + y^{3} = 0$ .
Since $2 x / (1 + x^{2}) \leq 1$ for all real x, it follows from the first of these equations that $y^{2} \leq 1$ or $- 1 \leq y \leq 1$ .
Again from the second equation of the system, we get
$y^{3} = - 1 - 2 (x - 1)^{2} \leq - 1$
Now since $y \leq - 1$ and $y \geq - 1$ , we must have $y = - 1$
Then $x = 1$ . This the solution is $x = 1$ , $y = - 1$ .

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Consistency of Linear System of Equations

Standard XII Mathematics

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