Answer :
let's call the smaller number x:
y-x=5
xy=104
y-x=5 can be rewritten as y=x+5
Substituting this into the second equation gives:
x(x+5)=104
x^2+5x=104
x^2+5x-104=0
Then simply solve as a quadratic:
x^2+13x-8x-104=0
x(x+13) -8(x+13)=0
(x-8)(x+13) = 0
x=8,-13
the question said the numbers are positive, so x=8
if x=8,
y=x+5
y=13
So the two numbers are 8 and 13.
y-x=5
xy=104
y-x=5 can be rewritten as y=x+5
Substituting this into the second equation gives:
x(x+5)=104
x^2+5x=104
x^2+5x-104=0
Then simply solve as a quadratic:
x^2+13x-8x-104=0
x(x+13) -8(x+13)=0
(x-8)(x+13) = 0
x=8,-13
the question said the numbers are positive, so x=8
if x=8,
y=x+5
y=13
So the two numbers are 8 and 13.
The best way to do this is to find the factors of 104. For this I got 1, 2, 4, 8, 13, 26, 52 and 104. From this, you are then able to look at the two that have a difference of five, which in this case, is 8 and 13. You should then double check that when these are multipled, they are equal to 104, which they are. I'm not 100% sure if this is the answer that you are looking for, but the two numbers are 8 and 13
