One positive integer is 5 less than twice another. The sum of their squares is 610. How do you find the integers?

Jun 6, 2018

$x = 21 , y = 13$

Explanation:

${x}^{2} + {y}^{2} = 610$

$x = 2 y - 5$

Substitute $x = 2 y - 5$ into ${x}^{2} + {y}^{2} = 610$

${\left(2 y - 5\right)}^{2} + {y}^{2} = 610$

$4 {y}^{2} - 20 y + 25 + {y}^{2} = 610$

$5 {y}^{2} - 20 y - 585 = 0$

Divide by 5

${y}^{2} - 4 y - 117 = 0$

$\left(y + 9\right) \left(y - 13\right) = 0$

$y = - 9 \mathmr{and} y = 13$

If $y = - 9 , x = 2 \times - 9 - 5 = - 23$

if $y = 13 , x = 2 \times 13 - 5 = 21$

Has to be the positive integers