# The sum of the squares of two positive numbers is 80 and the difference of the squares of the numbers is 48. How do you find the numbers?

Jul 11, 2018

I got $8 \mathmr{and} 4$

#### Explanation:

Let us call the numbers $x$ and $y$ we get:

${x}^{2} + {y}^{2} = 80$

${x}^{2} - {y}^{2} = 48$

let us add the two together:

$\left({x}^{2} + {x}^{2}\right) + \left({y}^{2} - {y}^{2}\right) = 80 + 48$

$2 {x}^{2} = 128$

${x}^{2} = \frac{128}{2} = 64$

$x = \pm \sqrt{64} = \pm 8$ Let us choose the positive one and substitute into the first equation for $x$:

${8}^{2} + {y}^{2} = 80$

${y}^{2} = 80 - 64 = 16$

$y = \pm \sqrt{16} = \pm 4$ let us choose again the positive one.