If,

ab/(a+b) = 2

ac/(a+c)=5

bc/(b+c)=4

then find a+b+c.

I'd really appreciate a step by step solution.

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- Apr 18th 2013, 05:00 AM #1

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- Apr 18th 2013, 05:41 AM #2
## Re: This problem is ridiculous

Hi there,

We would solve for one variable and after that replace it in the original equation in order to have only one variable equation:

You can solve for a or for b, I will show you the example for both:

$\displaystyle \frac{ab}{a+b} = 2 \\ ab = 2a + 2b \\ a = \frac{2a}{b} + 2 \\ a-2= \frac{2a}{b} \\ \frac{2a}{(a-2)} = b \\ a*\frac{2a}{(a-2)} = 2a + 2 \frac{2a}{(a-2)}$

Now, solve for a, replace the value of a in the initial equation and solve for b, then try for the other examples,

dokrbb

hint - you will have $\displaystyle a =1$ and $\displaystyle b = -2$ , check them in the first equation

- Apr 18th 2013, 06:31 AM #3

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- Apr 18th 2013, 07:16 AM #4
## Re: This problem is ridiculous

This is not correct. The problem is that this equaton simply yields $\displaystyle 2a^2 = 2a^2 + 4a - 4a$, so it does not yield a solution for a.

Here's how to do it. Once you have $\displaystyle a = \frac {2b}{b-2}$, use the same process on the equation $\displaystyle \frac {bc}{b+c} = 4$ to get $\displaystyle c = \frac {4b} {b-4}$. Now you have equatons for a and c in terms of b; sub them into $\displaystyle \frac {ac}{a+c} = 5$ and solve for b. Then you can sub the solution back into the other two equation to get values for a and c.

Despite the suggestion in the previous post $\displaystyle a \ne 1$ and $\displaystyle b \ne -2$.

- Apr 18th 2013, 08:32 AM #5

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## Re: This problem is ridiculous

If you're having trouble following ebaines' post, here's the process to figuring out a, b, and c.

1. Isolate a in equation 1

2. sub it into equation 2

3. Isolate isolate b in equation 2

4. sub it into equation 3

5. equation 3 will only have the variable c. Isolating c will yield its value.

6. sub in your value for c into equation 2 to find the value of b

7. sub in your value for b into equation 1 to find the value of a