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$x, y$ are positive reals such that $y\sqrt{x^2 - y^2} = 48$ and $x + y + \sqrt{x^2 - y^2} = 24$. How to find $x$ and $y$? Squaring equations leads to complicated equations with polynomials of high degrees. Is there any way to omit it and solve it smarter?

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    $\begingroup$ Use $\sqrt{x^2-y^2}={48\over y}$ in the second equation and solve for $x$, then you can use this info in the first equation to have a single-variable equation in $y$. $\endgroup$ – abiessu Dec 8 '16 at 20:05
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Call $x^2-y^2=z^2 \quad (1)$ then:

$yz=48 \quad(2)$ and $x+y+z=24 \quad(3)$

From $(1)$ we have:

$$x^2=y^2+z^2=(y+z)^2-2yz$$

and using $(2)$ and $(3)$ we get:

$$x^2=(24-x)^2-2.48 \Rightarrow x=10$$

and then $y=8$ or $y=6$

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