The sum of the reciprocals of two consecutive even integers is 9/40. This can be represented by the equation shown. 1/x+1/x+2=9/40

use the rational equation to determine the integers. show all work

3 answers

First, we need to clear the denominators by multiplying both sides of the equation by the common denominator of x(x+2):

1/x + 1/(x+2) = 9/40

40x(x+2)(1/x) + 40x(x+2)(1/(x+2)) = 40x(x+2)(9/40)

40(x+2) + 40x = 9x(x+2)

Distribute the 40 on the left:

40x + 80 + 40x = 9x(x+2)

Combine like terms:

80x + 80 = 9x^2 + 18x

Move all terms to one side:

9x^2 - 62x - 80 = 0

Now we can factor this quadratic equation:

(9x + 8)(x - 10) = 0

Setting each factor equal to zero gives us two possible solutions:

9x + 8 = 0 or x - 10 = 0

Solving for x gives us:

x = -8/9 or x = 10

We can disregard the negative solution since we are looking for even integers. Thus, the two consecutive even integers are:

x = 10

x+2 = 12

Therefore, the two consecutive even integers are 10 and 12.
your wrong
I apologize for the error in my previous answer. Here is the correct solution:

The equation is:

1/x + 1/(x+2) = 9/40

To solve for x, we can cross multiply and simplify:

40(x+2) + 40x = 9x(x+2)

80x + 80 = 9x^2 + 18x

9x^2 - 62x - 80 = 0

We can factor the quadratic equation:

(9x + 4)(x - 10) = 0

Setting each factor equal to zero gives us two possible solutions:

9x + 4 = 0 or x - 10 = 0

Solving for x gives us:

x = -4/9 or x = 10

Since we are looking for consecutive even integers, x must be even. Thus, we can disregard the solution x = -4/9. The only valid solution is x = 10.

Therefore, the two consecutive even integers are:

x = 10

x+2 = 12

Hence, the two consecutive even integers that satisfy the given condition are 10 and 12.
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