Let f(x) = 1/(x+7)+15/(x-8). Find all a for which f(a)=f(a+15).

Question

Let f(x) = 1/(x+7)+15/(x-8). Find all a for which f(a)=f(a+15).

How do I solve this problem?

I understand this so far
f(a)=f(a+15)
1/(7+a) + 15/(a-8) = 1/(a+22) + 15/(a+7)

I am lost after that

This is a multiple choice question and I am not getting any of the choices

a.) -161/8
b.) 120/7
c.) 161/7
d.) -17

Answer 1

I did it in my head and I didn't get any of the answers either. I recommend you do as I told, recheck once more.

Answer 2

To solve the equation f(a) = f(a+15), you will need to simplify and find a common denominator for the fractions.

Starting with the equation you wrote:
1/(7+a) + 15/(a-8) = 1/(a+22) + 15/(a+7)

First, find the least common denominator to combine the fractions:

The least common denominator (LCD) is the product of the denominators of all the fractions, in this case:
(7+a) * (a-8) * (a+22) * (a+7)

Now, multiply both sides of the equation by the LCD:
[(7+a) * (a-8) * (a+22) * (a+7)] * [1/(7+a) + 15/(a-8)] = [(7+a) * (a-8) * (a+22) * (a+7)] * [1/(a+22) + 15/(a+7)]

Simplifying the fractions on both sides:

On the left side, the denominator (7+a) cancels out with the corresponding numerator 1/(7+a), and the denominator (a-8) cancels out with the corresponding numerator 15/(a-8). So the left side becomes:
(a-8) * 1 + 15 * (a+22)

On the right side, the denominator (a+22) cancels out with the corresponding numerator 1/(a+22), and the denominator (a+7) cancels out with the corresponding numerator 15/(a+7). So the right side becomes:
(a+22) * 1 + 15 * (a-8)

Simplifying further:

(a-8) + 15(a+22) = (a+22) + 15(a-8)

Expand and combine like terms:

a - 8 + 15a + 330 = a + 22 + 15a - 120

Combine like terms:

16a + 322 = 16a - 98

Now, subtract 16a from both sides to eliminate the variable:

322 = -98

This is a contradiction, which means there is no solution to the equation. Therefore, none of the answer choices (a, b, c, or d) are correct for this problem.

It seems there may be a mistake in either the original equation or the multiple-choice options provided. You may want to double-check the problem or seek clarification to ensure accuracy.