Difficult Systems Of Equations

Algebra Level 4

Let x x and y y be real numbers satisfying the system of equations below: x x 2 y 2 1 1 x = 4 , x 2 y x 2 y 2 1 + y = 2. \dfrac {x}{x^{2}y^{2} - 1} - \dfrac{1}{x} = 4, \qquad \dfrac {x^{2}y}{x^{2}y^{2} - 1} + y = 2. Find all possible values of x y , xy, and give your answer as their product.


The answer is -0.5.

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1 solution

The given system can be expressed as follows:

x x 2 y 2 1 1 x = 4 \frac {x}{x^{2}y^{2} -1} - \frac{1}{x} = 4

x 2 y x 2 y 2 1 + y = 2 \frac {x^{2}y}{x^{2}y^{2} -1} + y = 2

\Rightarrow

x 2 x 2 y 2 1 1 = 4 x \frac {x^{2}}{x^{2}y^{2} -1} - 1 = 4x

x 2 x 2 y 2 1 + 1 = 2 y \frac{x^2}{x^{2}y^{2}-1}+1 = \frac{2}{y}

We then have

4 x + 2 y = 2 x 2 x 2 y 2 1 2 x + 1 y = x 2 x 2 y 2 1 4x + \frac{2}{y} = \frac{2x^{2}}{x^{2}y^{2} -1} \Rightarrow 2x + \frac{1}{y} = \frac{x^2}{x^{2}y^{2} -1}

and

4 x 2 y = 2 2 x 1 y = 1 , 4x - \frac{2}{y} = -2 \Rightarrow 2x - \frac{1}{y} = -1,

which gives us

( 2 x + 1 y ) ( 2 x 1 y ) = x 2 x 2 y 2 1 (2x + \frac{1}{y})(2x - \frac{1}{y}) = \frac{-x^2}{x^{2}y^{2} - 1}

4 x 2 1 y 2 = x 2 x 2 y 2 1 4x^2 - \frac{1}{y^2} = \frac{-x^2}{x^{2}y^{2} -1}

4 x 2 y 2 1 y 2 = x 2 x 2 y 2 1 \frac{4x^{2}y^{2} - 1}{y^2} = \frac{-x^2}{x^{2}y^{2} -1}

( 4 x 2 y 2 1 ) ( x 2 y 2 1 ) = x 2 y 2 (4x^{2}y^{2}-1)(x^{2}y^{2}-1) = -x^{2}y^{2}

4 x 2 y 2 5 x 2 y 2 + 1 = x 2 y 2 4x^{2}y^{2} - 5x^{2}y^{2} + 1 = -x^{2}y^{2}

4 x 2 y 2 4 x 2 y 2 + 1 = 0 4x^{2}y^{2} - 4x^{2}y^{2} + 1 = 0

( 2 x 2 y 2 1 ) 2 = 0 x 2 y 2 = 1 2 x y = ± 1 2 (2x^{2}y^{2} - 1)^{2} = 0 \Rightarrow x^{2}y^{2} = \frac{1}{2} \Rightarrow xy = \pm\frac{1}{\sqrt{2}}

Therefore the product is 1 2 -\frac{1}{2} or 0.5 \boxed{-0.5}

Marvellous bashing

Ram Sita - 3 years, 6 months ago

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