Question 465944
x-y=5
{{{(x+y)^2=169}}}
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solve this system
x=5+y
{{{((5+y)+y)^2=169}}}
{{{25+20y+4y^2=169}}}
{{{4y^2+20y-144=0}}} divide everything by 4
{{{y^2+5y-36=0}}}
(y+9)(y-4)=0
y=-9 y=4 Problem said 2 POSITIVE numbers, so y=-9 won't work.
Which means y=4 is one of the numbers, since we said x-y=5 then x-4=5 so you add 4 to each side and you get x=9
so the numbers are 9 and 4