#1**0 **

I'll assume that b and c are non-zero for the purposes of solving, or the answer might be unobtainable. I also think I need to assume them to be finite.

(a+b)/b = a/b + b/b = 5

b/b = 1

a/b + b/b - b/b = 5 - 1

a/b = 4

First, I split the fraction into 2 parts. It isn't helpful immediately, but was helpful with something else, namely...

Second, I used to identity property of division. Namely, dividing a finite, non-zero quantity by itself will yield 1.

Then, I subtract the second equation from the first. This is a very finicky thing to do, but because I know that the values apply the same in both places, I'm doing it anyways.

I could have also written: \(\frac{a}{b} + \frac{b}{b} = \frac{a}{b} + 1 = 5\), then subtracted 1 from each side to get a/b = 4. Different method, same result.

(b+c)/c = b/c + c/c = 4

c/c = 1

b/c + c/c - c/c = 4 - 1

b/c = 3

The numbers and letters are different here, but the steps from above apply the same here.

(a+c)/c = n = a/c + c/c

c/c = 1

a/c + c/c - c/c = n - 1

a/c = n-1

a/c = a/b * b/c

n-1 = 4 * 3

n-1 = 12

n = 13

The important thing that happened here, is I created a "placeholder" variable I named *n*. I set it equal to the initial quantity that we were solving for, so as to reduce the number of times I write out (a+c)/c, and to also allow operations related to it easier.

The third part of this might look odd, but I chose not to change *n*, and leave (n-1) in place. This helps, as I note that a/c = n-1, and also equal to a/b * b/c. The *b* in the numerator and denominator cancel out, and the remaining values match up, thus n-1 = a/b * b/c. However, we don't have a or c, so we keep a/b and b/c, since we have those values and can then find a/c.

From there, we then find n, and with it, the value it represented, as seen below:

(a+c)/c = n = 13

helperid1839321 Aug 30, 2021