Question

the sum of the two numbers is 12 and the sum of its receprocal 3/8.find the numbers​

Answer 1

Answer:

If sum of two numbers is 12 and sum of its reciprocal is 3/8, then the numbers are 4 and 8

Step-by-step explanation:

Let x and y be two numbers

Given x+y = 12

$\frac{1}{x} +\frac{1}{y} =\frac{3}{8}$

$\frac{y+x}{xy} =\frac{3}{8}$

x+y=12

y=12-x

Substitute y as (12-x ) in the second equation.

$\frac{(12-x)+x}{x(12-x)} = \frac{3}{8} \\\\\frac{12}{12x-x^{2} } =\frac{3}{8} \\\\12*8=3(12x-x^{2} )\\\\4*8= 12x- x^{2} \\\\32=12x-x^{2} \\\\x^{2} -12x+32=0$

To solve the quadratic equation, we can split the equation .

$x^{2} -12x+32=0$

x(x-4)-8(x-4)=0

(x-4)(x-8)=0

So x-4=0

x=4

x-8 = 0

x =8

so x can be 4 or 8

When x is 4, y= 12-x

= 12-4 =8

When x= 8, y= 12-x

= 12-8= 4

Hence the two numbers are 8 and 4