The answer is 6 and 8.

The numbers are consecutive even numbers. So, if one number is x then the other number, call it y, is x+2. So, y = x + 2.

The reciprocals of x and y are 1/x and 1/y and the sum is equal 7/24. So, 1/x + 1/y = 7/24.

Because you have two equations you can solve for the two unknown numbers.

We know

1/x + 1/y = 7/24

multiply both sides of the equation by x to get

1 + x/y = 7x/24

multiply both sides of the equation by y to get

y + x = 7xy/24

multiply both sides of the equation by 24 to get

24y + 24x = 7xy

we know y = x + 2, so change y to (x + 2) to get

24(x + 2) + 24x = 7x(x + 2)

use the distributive property of multiplication to get

24x + 48 +24x = 7x^2 + 14x

add the x terms together to simplify the equation to get

48x + 48 = 7x^2 + 14x

subtract 48x from both sides of the equation to get

48 = 7x^2 – 34x

subtract 48 from both sides of the equation to get

0 = 7x^2 – 34x - 48

this in the form of a quadratic equation which is ax^2 + bx + c = 0

the solution to a quadractic equation is

x = (-b + the square root of (b^2 – 4ac)) / 2a

or

x = (-b - the square root of (b^2 – 4ac)) / 2a

(which can be derived, but that will be the answer to another question)

in this case a = 7, b = -34, and c = -48

start with the first solution and substitute the values of a, b, and c to get

x = (-(-34) + the square root of (-34^2 - 4*7*(-48)) / 2*7

simplify the terms to get

x = (34 + the square root of (1156 + 1344)) / 14

or

x = (34 + the square root of (2500)) / 14

the square root of (2500) = 50 so

x = (34 + 50) / 14

x = 84/14

x = 6

and y = x + 2 = 6 + 2 = 8

you can try the second solution to see if it also works

x = (-(-34) - the square root of (-34^2 - 4*7*(-48)) / 2*7

x = (34 - 50) / 14

x = -16 / 14 = -8/7 which is not an even number, so this solution cannot satisfy the conditions of the stated problem

Check your work by substituting the values into the original equations

1/6 + 1/8 = 8/48 + 6/48 = 14/48 = 7/24

The numbers are consecutive even numbers. So, if one number is x then the other number, call it y, is x+2. So, y = x + 2.

The reciprocals of x and y are 1/x and 1/y and the sum is equal 7/24. So, 1/x + 1/y = 7/24.

Because you have two equations you can solve for the two unknown numbers.

We know

1/x + 1/y = 7/24

multiply both sides of the equation by x to get

1 + x/y = 7x/24

multiply both sides of the equation by y to get

y + x = 7xy/24

multiply both sides of the equation by 24 to get

24y + 24x = 7xy

we know y = x + 2, so change y to (x + 2) to get

24(x + 2) + 24x = 7x(x + 2)

use the distributive property of multiplication to get

24x + 48 +24x = 7x^2 + 14x

add the x terms together to simplify the equation to get

48x + 48 = 7x^2 + 14x

subtract 48x from both sides of the equation to get

48 = 7x^2 – 34x

subtract 48 from both sides of the equation to get

0 = 7x^2 – 34x - 48

this in the form of a quadratic equation which is ax^2 + bx + c = 0

the solution to a quadractic equation is

x = (-b + the square root of (b^2 – 4ac)) / 2a

or

x = (-b - the square root of (b^2 – 4ac)) / 2a

(which can be derived, but that will be the answer to another question)

in this case a = 7, b = -34, and c = -48

start with the first solution and substitute the values of a, b, and c to get

x = (-(-34) + the square root of (-34^2 - 4*7*(-48)) / 2*7

simplify the terms to get

x = (34 + the square root of (1156 + 1344)) / 14

or

x = (34 + the square root of (2500)) / 14

the square root of (2500) = 50 so

x = (34 + 50) / 14

x = 84/14

x = 6

and y = x + 2 = 6 + 2 = 8

you can try the second solution to see if it also works

x = (-(-34) - the square root of (-34^2 - 4*7*(-48)) / 2*7

x = (34 - 50) / 14

x = -16 / 14 = -8/7 which is not an even number, so this solution cannot satisfy the conditions of the stated problem

Check your work by substituting the values into the original equations

1/6 + 1/8 = 8/48 + 6/48 = 14/48 = 7/24