This one stumps me.

michaelcaba said:
1/x + 1/a = 1/b. Solve for x.

Can someone show me the steps?

Thanks!
Click to expand...
I would start by getting 1/x = 1/b - 1/a. Then get a common denominator for the rhs. Then take reciprocals of both sides since the reciprocal of 1/x is x.
 
OK. The first step is 1/x = 1/b - 1/a. I got that. Then is it x(1/x) = x(1/b - 1/a)? This seems to be equal to x = x/b-a. But then what?
 
Woops! After 1/x = 1/b - 1/a, it should be 1/x = (a-b)/(ab), then x= (ab)/(a-b), correct?
 
You were FINE on the first step and then on your next step in post 4 although I don't think it is the easiest way to go. And then you multiplied x * (1/x) and got x. That is wrong, but you caught it in post 5. Then you went off the rails because you forgot all about common denominators, which you probably learned in third grade. Continuing from

[MATH]1= x * \left ( \dfrac{1}{b} - \dfrac{1}{a} \right ) =[/MATH]
[MATH]x * \left ( \dfrac{a}{ab} - \dfrac{b}{ab} \right ) = \dfrac{x(a - b)}{ab}.[/MATH]
[MATH]1 = \dfrac{x(a - b)}{ab} \implies ab = x(a - b).[/MATH]
Now (a - b) cannot be zero.

[MATH]\therefore x = \dfrac{ab}{a - b}.[/MATH]
EDIT: OK you got it right while I was writing. Way to go. Want to see an easier way?
 
JeffM said:
You were FINE on the first step and then on your next step in post 4 although I don't think it is the easiest way to go. And then you multiplied x * (1/x) and got x. That is wrong, but you caught it in post 5. Then you went off the rails because you forgot all about common denominators, which you probably learned in third grade. Continuing from

[MATH]1= x * \left ( \dfrac{1}{b} - \dfrac{1}{a} \right ) =[/MATH]
[MATH]x * \left ( \dfrac{a}{ab} - \dfrac{b}{ab} \right ) = \dfrac{x(a - b)}{ab}.[/MATH]
[MATH]1 = \dfrac{x(a - b)}{ab} \implies ab = x(a - b).[/MATH]
Now (a - b) cannot be zero.

[MATH]\therefore x = \dfrac{ab}{a - b}.[/MATH]
EDIT: OK you got it right while I was writing. Way to go. Want to see an easier way?
Click to expand...
 
I start with a theorem

[MATH]\text {IF } p,\ q,\ r, \ \text { and } s \in \mathbb R, \ pqrs \ne 0, \text { and } \dfrac{p}{q} = \dfrac{r}{s},[/MATH]
[MATH]\text {THEN } \dfrac{q}{p} = \dfrac{s}{r}.[/MATH]
You can prove that quickly.

[MATH](qs) * \dfrac{p}{q} = (qs) * \dfrac{r}{s} \implies ps = qr \implies \dfrac{1}{pr} * ps = \dfrac{1}{pr} * qr \implies \dfrac{s}{r} = \dfrac{q}{p}.[/MATH]
In English, if two expressions are equal and not zero, their reciprocals are also equal.

Now lets apply that rather obvious theorem to your problem.

[MATH]\text {ISOLATE THE UNKNOWN } \dfrac{1}{x} + \dfrac{1}{a} = \dfrac{1}{b} \implies = \dfrac{1}{x} = \dfrac{1}{b} - \dfrac{1}{a}.[/MATH]
[MATH]\text {USE COMMON DENOMINATORS } \dfrac{1}{x} = \dfrac{1}{b} - \dfrac{1}{a} = \dfrac{a}{ab} - \dfrac{b}{ab} = \dfrac{a - b}{ab}.[/MATH]
[MATH]\text {USE NON-ZERO RECIPROCALS THEOREM } \dfrac{1}{x} = \dfrac{a - b}{ab} \implies x = \dfrac{ab}{a - b}.[/MATH]
Basically three easy steps. However, there is a condition on the reciprocals theorem. How do we know that (a - b) is not zero?
 
One of the givens was a is not equal to b. Thus (a-b) cannot equal 0, I believe.

Thank you.
 
Well, that was not given in your first post. Thus if it was given in your text, I did not deduce it from that undisclosed information.

[MATH]a - b = 0 \text { and } \dfrac{1}{x} = \dfrac{a - b}{ab} \implies \dfrac{1}{x} = 0 \implies 1 = x * 0 = 0.[/MATH]
But 1 is not equal to zero.
 
michaelcaba said:
Woops! After 1/x = 1/b - 1/a, it should be 1/x = (a-b)/(ab), then x= (ab)/(a-b), correct?
Click to expand...
Yes it is--great job. I still want you to learn more from this! You can check your work somewhat by picking different values for a and b. Suppose a=1 and b=2. Then 1/x=1/2-1/1=-1/2 = (a-b)/(ab) = (1-2)/(1*2) =-1/2 =-1/2 ! (of course x=-2)

Also note that x(1/x) = 1 NOT x. If you multiply two (non zero) numbers and the product is one of the two numbers, THEN that other number BETTER be 1! Since x is not 1 and 1/x is not 1, then their product can't be x (or 1/x)
 
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