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

Jun 17, 2015

Tthere are two solutions to a problem:
$\left({x}_{1} , {y}_{1}\right) = \left(5 , 2\right)$
$\left({x}_{2} , {y}_{2}\right) = \left(\frac{6}{7} , - \frac{15}{7}\right)$

#### 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 $\frac{1}{x}$ and $\frac{1}{y}$, the sum of which is $\frac{7}{10}$, which results in the equation:
(2) $\frac{1}{x} + \frac{1}{y} = \frac{7}{10}$
Incidentally, existence of reciprocals necessitates the restrictions:
$x \ne 0$ and $y \ne 0$.

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) $\frac{1}{y + 3} + \frac{1}{y} = \frac{7}{10}$
Incidentally this necessitates another restriction:
$y + 3 \ne 0$, that is $y \ne - 3$.

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

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

Two solutions to this equation are:
${y}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 840}}{14}$
or
${y}_{1 , 2} = \frac{- 1 \pm 29}{14}$

So, we have two solutions for $y$:
${y}_{1} = 2$ and ${y}_{2} = - \frac{30}{14} = - \frac{15}{7}$

Correspondingly, using $x = y + 3$, we conclude that there are two solutions to a system:
$\left({x}_{1} , {y}_{1}\right) = \left(5 , 2\right)$
$\left({x}_{2} , {y}_{2}\right) = \left(\frac{6}{7} , - \frac{15}{7}\right)$

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 $\left({x}_{1} , {y}_{1}\right) = \left(5 , 2\right)$:
$\frac{1}{5} + \frac{1}{2} = \frac{2 + 5}{5 \cdot 2} = \frac{7}{10}$ - checked
(b) for a solution $\left({x}_{2} , {y}_{2}\right) = \left(\frac{6}{7} , - \frac{15}{7}\right)$:
$\frac{7}{6} - \frac{7}{15} = \frac{70}{60} - \frac{28}{60} = \frac{42}{60} = \frac{7}{10}$ - checked

Both solutions are correct.