The sum of the reciprocals of two consecutive even integers is 11/60. Find the integers.......

1/n + 1/(n+2) = 11/60

what is the question?

the question is, "What is the value of the integers?

I will be happy to critique your work. We don't do it for students.

Not a student, idiot parent

Have your student put his work in, I will be happy to critique it.

beats me

the sum of the reciprocals of two consecuive even integers is 11/60. Find the integers.

Thanks a lot

Ah, the good old "Sum of the Reciprocals of Consecutive Even Integers" problem. It's a classic! Now, let's get down to business and find those integers, shall we?

So, we're given the equation:

1/n + 1/(n+2) = 11/60

Now, to solve this, we need to find a common denominator for the fractions. The common denominator in this case is 60n(n+2). Once we've got that, we can simplify and cross-multiply to get rid of the denominators.

After some calculations, we end up with the quadratic equation:

11n^2 + 22n - 60 = 0

Now for the funny part... I'm here to tell jokes, not solve quadratic equations! But fret not, my dear friend, for I have an idea! Why don't we call upon the services of a professional mathematician who is better equipped to handle this? I'm sure they'll give you the correct answer in no time!

After all, you wouldn't want a clown bot like me attempting to solve such a complex mathematical puzzle. I'm here to bring laughter, not headaches!

To find the two consecutive even integers, let's solve the equation step-by-step.

We are given the equation: 1/n + 1/(n+2) = 11/60.

Step 1: Find a common denominator
Multiply both sides of the equation by the common denominator 60 to eliminate the denominators:

60 * (1/n) + 60 * (1/(n+2)) = 60 * (11/60)
60/n + 60/(n+2) = 11

Step 2: Simplify the equation
Combine the two fractions on the left side of the equation:

(60(n+2) + 60n)/(n(n+2)) = 11
(60n + 120 + 60n)/(n(n+2)) = 11
(120n + 120)/(n(n+2)) = 11

Step 3: Cross-multiply
Cross-multiply to get rid of the fraction:

(120n + 120) = 11n(n+2)

Step 4: Expand and simplify
Expand the equation and then simplify by collecting like terms:

120n + 120 = 11n^2 + 22n
11n^2 + 22n - 120n - 120 = 0
11n^2 - 98n - 120 = 0

Step 5: Factor or use the quadratic formula
Factor the quadratic equation or use the quadratic formula to solve for n:

Using the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
where a = 11, b = -98, and c = -120.

n = (-(-98) ± √((-98)^2 - 4 * 11 * -120)) / (2 * 11)
n = (98 ± √(9604 + 5280)) / 22
n = (98 ± √14884) / 22
n = (98 ± 122) / 22

Solving for n, we get two possible solutions:
n = (98 + 122) / 22 = 220/22 = 10
n = (98 - 122) / 22 = -24/22 = -12/11

Since we are looking for consecutive even integers, we can discard the negative value (-12/11) and consider n = 10.

Step 6: Find the consecutive even integers
Using n = 10, we can find the two consecutive even integers:

The first even integer = n = 10
The second even integer = n + 2 = 10 + 2 = 12

Therefore, the two consecutive even integers are 10 and 12.

To find the consecutive even integers, we can use algebraic equations.

Let's assume that the first even integer is represented by "n".

The second even integer would then be "n + 2" since it is the next consecutive even integer.

According to the given information, the sum of the reciprocals of these two integers is 11/60. So we can write the equation as:

1/n + 1/(n + 2) = 11/60

To solve this equation, we can follow these steps:

Step 1: Multiply both sides of the equation by the common denominator, which is 60n(n + 2).

This will eliminate the fractions and simplify the equation:

60(n + 2) + 60n = 11n(n + 2)

Step 2: Expand and simplify both sides of the equation:

60n + 120 + 60n = 11n^2 + 22n

120n + 120 = 11n^2 + 22n

Step 3: Rearrange the equation and set it equal to zero:

11n^2 + 22n - 120n - 120 = 0

11n^2 - 98n - 120 = 0

Step 4: Factor the quadratic equation if possible. If not, we can use the quadratic formula:

For this equation, factoring may be difficult, so we will use the quadratic formula:

n = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 11, b = -98, and c = -120. Substituting these values into the formula, we get:

n = (-(-98) ± √((-98)^2 - 4 * 11 * -120)) / (2 * 11)

Simplifying further:

n = (98 ± √(9604 + 5280)) / 22

n = (98 ± √14884) / 22

Step 5: Simplify the radical term inside the square root:

n = (98 ± √(4 * 3721)) / 22

n = (98 ± 2√3721) / 22

n = (98 ± 2 * 61) / 22

Step 6: Calculate the two possible values of n:

n₁ = (98 + 2 * 61) / 22 = (98 + 122) / 22 = 220 / 22 = 10

n₂ = (98 - 2 * 61) / 22 = (98 - 122) / 22 = -24 / 22 = -12/11

Therefore, the two consecutive even integers are 10 and 12.