Question

Two numbers differ by 3. The sum of their reciprocals is seven tenths. How do you find the numbers?

Answer 1

Tthere are two solutions to a problem:
`(x_1, y_1)=(5,2)`
`(x_2, y_2)=(6/7,-15/7)`

Explanation:

This is a typical problem that can be solved using a system of two equations with two unknown variables.

Let the first unknown variable be `x` and the second `y`.

The difference between them is `3`, which results in the equation:
(1) `x-y=3`

Their reciprocals are `1/x` and `1/y`, the sum of which is `7/10`, which results in the equation:
(2) `1/x+1/y=7/10`
Incidentally, existence of reciprocals necessitates the restrictions:
`x!=0` and `y!=0`.

To solve this system, let's use the method of substitution.
From the first equation we can express `x` in terms of `y` and substitute into the second equation.

From equation (1) we can derive:
(3) `x = y+3`

Substitute it into equation (2):
(4) `1/(y+3) + 1/y = 7/10`
Incidentally this necessitates another restriction:
`y+3!=0`, that is `y!=-3`.

Using common denominator `10y(y+3)` and considering only numerators, we transform equation (4) into:
`10y+10(y+3)=7y(y+3)`

This is a quadratic equation that can be rewritten as:
`20y+30=7y^2+21y` or
`7y^2+y-30=0`

Two solutions to this equation are:
`y_(1,2)=(-1+-sqrt(1+840))/14`
or
`y_(1,2)=(-1+-29)/14`

So, we have two solutions for `y`:
`y_1=2` and `y_2=-30/14=-15/7`

Correspondingly, using `x=y+3`, we conclude that there are two solutions to a system:
`(x_1, y_1)=(5,2)`
`(x_2, y_2)=(6/7,-15/7)`

In both cases `x` is greater than `y` by `3`, so the first condition of a problem is satisfied.
Let's check the second condition:
(a) for a solution `(x_1, y_1)=(5,2)`:
`1/5+1/2=(2+5)/(5*2)=7/10` - checked
(b) for a solution `(x_2, y_2)=(6/7,-15/7)`:
`7/6-7/15=70/60-28/60=42/60=7/10` - checked

Both solutions are correct.