Question

Divid 15 into two parts such that the sum of their reciprocal is 3/10

Answer 1

From the information obtained from the question, two equations can be created:
 
Let x and z be the two numbers (parts)

$\frac{1}{z} + \frac{1}{x} = \frac{3}{10}$ .  .  .  . (1)

$z + x = 15$  .  .  .  . (2)

By transposing (2), make 'z' the subject of the equation
$z = 15 - x$  .  .  .  . (3)

By substituting (3) into equation (1) to find a value for x
$\frac{1}{(15 - x)} + \frac{1}{x} = \frac{3}{10}$

$\frac{15}{( 15 - x ) ( x )} = \frac{3}{10}$

$3 ( - x^{2} + 15 x ) = 150$

$3 x^{2} - 45x + 150 = 0$

⇒  $( x - 5 ) ( x - 10 ) = 0$

∴ either $( x - 5) = 0$        OR    $( x - 10 ) = 0$

Thus x = 5 or x = 10

By substituting the values of x into (2) to find z

     z + (5) = 15       OR      z + (10) = 15

    ⇒      z = 10      OR                 z = 5

So, the two numbers or two parts into which fifteen is divided to yield the desired results are 5 and 10.