Question

The sum of two numbers in 18. The sum of their
reciprocale is 1/4. Find the numbers.​

Answer 1

Let x and y be the two numbers.

The question states that :-

$x + y = 18 \\ \\ \\ \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{4}$

Now,

From first equation we have

$x + y = 18 \\ \\ \\y = 18 - x$

Putting down this in equation (ii)

Hence,

$\frac{1}{x} + \frac{1}{18 - x} = \frac{1}{4} \\ \\ \\ \frac{18 - x + x}{x(18 - x)} = \frac{1}{4} \\ \\ \\ \frac{18}{18x - {x}^{2} } = \frac{1}{4} \\ \\ \\18x - {x}^{2} = 4 \times 18 \\ \\ \\ {x}^{2} - 18x + 72 = 0 \\ \\ \\ {x}^{2} - 12x - 6x + 72 = 0 \\ \\ \\x(x - 12) - 6(x - 12) = 0 \\ \\ \\(x - 12)(x - 6) = 0 \\ \\ \\ \\ \\x - 12 = 0 \\ x = 12 \\ \\ \\ \\ \\x - 6 = 0 \\ x = 6$

Now,

If x=6

$y =18 - x \\ \\y = 18 - 6 = 12$

If x = 12

$y = 18 - x \\ \\ \\y = 18 - 12 = 6$

Thus,

The numbers are 6 and 12.

Answer 2

Step-by-step explanation:

Given,

sum of two numbers in 18.

Their reciprocal is 1/4

let us assume the numbers be x and y

x + y = 18 .......(i)

Now, its given sum of their reciprocal is 1/4

$\frac{1}{x} + \frac{1}{y} = \frac{1}{4} \\ \\ \frac{x + y}{xy} = \frac{1}{4} \\ \\ 4(x + y) = xy$

Now put the value of (x + y) from eqn(i)

4(18) = xy

xy = 72

$x = \frac{72}{y}$ .......(ii)

now put value of x in eqn.(i) [to get value of y ]

$\frac{72}{y} + y = 18 \\ \\ \frac{72 + {y}^{2} }{y} = 18 \\ \\ 72 + {y}^{2} = 18y \\ \\ {y}^{2} - 18y + 72 = 0$

here we get the quadratic equation in y

on solving ,

By middle-term splitting

${y}^{2} - 18y + 72 = 0 \\ \\ {y}^{2} - 6y - 12y + 72 = 0 \\ \\ y(y - 6) - 12(y - 6) = 0 \\ \\ (y - 6)(y - 12) = 0$

Now equate the factors to zero to get the roots of the equation.

$y - 12 = 0 \\ \\ y = 12 \\ \\ y - 6 = 0 \\ \\ y = 6$

here we get two values of y

y = 6,12

if y = 6 then x = 12

if y = 12 then x = 6

So, numbers are 6 and 12.