Question 141093
ratio 3:2:5.
ind 2 numbers such that their sum, their difference, and their product have the ratio 3:2:5.
:
Let the two numbers be x & y:
{{{(x+y)/(x-y)}}} = {{{3/2}}}
Cross multiply:
2(x+y) = 3(x-y)
2x + 2y = 3x - 3y
3y + 2y = 3x - 2x
5y = x
:
and
{{{(x-y)/(xy)}}} = {{{2/5}}}
Cross multiply:
5(x-y) = 2xy
5x - 5y = 2xy
5x - 2xy = 5y
x(5-2y) = 5y
x = {{{5y/(5-2y)}}}
Substitute 5y for x
5y = {{{5y/(5-2y)}}}
Just looking at this we can assume
(5-2y) = 1
-2y = 1 - 5
-2y = -4
y = +2
:
Therefore:
x = 5(2)
x = 10
:
Testing the solution: x=10, y=2
{{{12/8}}} = {{{3/2}}}
and
{{{8/20}}} = {{{2/5}}}