# What is The difference Between the squares of two numbers is 5? What is Three times the square of the first number increased by the square of the second number is 31? Find the numbers.

Jun 25, 2018

$x = \pm 3$, $y = \pm 2$

#### Explanation:

The way you wrote the problem is super confusing and I suggest you write questions with cleaner English as it will be beneficial for everyone.

Let $x$ be the first number and $y$ be the second number.

We know:
${x}^{2} - {y}^{2} = 5$ --- i
$3 {x}^{2} + {y}^{2} = 31$ --- ii

From ii,
$3 {x}^{2} + {y}^{2} = 31$
$3 {x}^{2} = 31 - {y}^{2}$
$3 {x}^{2} - 31 = - {y}^{2}$ --- iii

Substitute iii into i,
${x}^{2} - {y}^{2} = 5$
${x}^{2} + \left(- {y}^{2}\right) = 5$
${x}^{2} + \left(3 {x}^{2} - 31\right) = 5$
$4 {x}^{2} - 31 = 5$
$4 {x}^{2} = 36$
${x}^{2} = 9$
$x = \pm \sqrt{9}$
$x = \pm 3$ --- iv

Substitute iv into i,
${x}^{2} - {y}^{2} = 5$
${\left(\pm 3\right)}^{2} - {y}^{2} = 5$ [${\left(\pm a\right)}^{2} = {a}^{2}$]
$9 - {y}^{2} = 5$
$- {y}^{2} = - 4$
${y}^{2} = 4$
$y = \pm \sqrt{4}$
$y = \pm 2$
$\therefore \left(x , y\right) = \left(\pm 3 , \pm 2\right)$