Question 1101288
One positive number is 10 less than another positive number.
a = b- 10
 If the reciprocal of the smaller number is added to three times the reciprocal of the larger number, the sum is 1/4.
{{{1/a}}} + {{{3/b}}} = {{{1/4}}}
multiply by 4ab, cancel the denominators
4b + 12a = ab
replace a with (b-10)
4b + 12(b-10) = b(b-10)
4b + 12b - 120 = b^2 - 10b 
16b - 120 = b^2 - 10b
combine to form a quadratic equation on the right
0 = b^2- 10b - 16b + 120
b^2 - 26b + 120 = 0
factors to
(b-6)(b-20) = 0
two solutions
b = 6, not possible, a has to be positive
b = 20, this is our solution
then
a = 20 - 10
a = 10
 Find the two numbers: 10 & 20
:
:
:
see if that checks out
{{{1/10}}} + {{{3/20}}} = {{{1/4}}}
{{{2/20}}} + {{{3/20}}} = {{{1/4}}}
{{{5/20}}} = {{{1/4}}}