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

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Answer 1 · 2018-06-06T11:31:39

$x = 21, y = 13$ $x=21, y=13$

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

$x = 2 y - 5$ $x=2y-5$

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

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

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

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

Divide by 5

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

$(y + 9) (y - 13) = 0$ $(y+9)(y-13)=0$

$y = - 9 or y = 13$ $y=-9 or y=13$

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

if $y = 13, x = 2 \times 13 - 5 = 21$ $y=13, x=2xx13-5=21$

Has to be the positive integers