If a ? b and 1/x + 1/a = 1/b, then x =

[CalledSomething]

Hello, everyone! I know that in the very back of my mind, I should and do know how to solve this problem, but I simply cannot remember how to. Please help me with this problem.

If a ? b and (1/x) + (1/a) = (1/b), then x =

I am taking an exam in a few hours, and was hoping that someone might help me with this as soon as they could.

Please help, and thank you for reading and/or helping me!

 

[galactus]

$\displaystyle \frac{1}{x}+\frac{1}{a}=\frac{1}{b}$

$\displaystyle \frac{1}{x}=\frac{1}{b}-\frac{1}{a}$

$\displaystyle \frac{1}{x}=\frac{a-b}{ab}$

Now, flip each one by taking reciprocals.

$\displaystyle x=\frac{ab}{a-b}$

 

[mawavoy]

I am not sure, but I think the solution goes like this.

1/x + 1/a = 1/b then 1/x = 1/a - 1/b then x (1/x) = (1/b - 1/a)x then 1= (1/b -1/a) x then

1/(1/b -1/a) = x then 1/(a-b)/ab = x . Note Galactus' solution is simpler.

 

[CalledSomething]

$\displaystyle \frac{1}{x}+\frac{1}{a}=\frac{1}{b}$

$\displaystyle \frac{1}{x}=\frac{1}{b}-\frac{1}{a}$

$\displaystyle \frac{1}{x}=\frac{a-b}{ab}$

Now, flip each one by taking reciprocals.

$\displaystyle x=\frac{ab}{a-b}$

I do not understand the third step. I'm sorry.

 

[CalledSomething]

Thank you both, though, for aiding me! <3

 

[galactus]

$\displaystyle \frac{1}{x}+\frac{1}{a}=\frac{1}{b}$

$\displaystyle \frac{1}{x}=\frac{1}{b}-\frac{1}{a}$

$\displaystyle \frac{1}{x}=\frac{a-b}{ab}$

Now, flip each one by taking reciprocals.

$\displaystyle x=\frac{ab}{a-b}$

I do not understand the third step. I'm sorry.

I just cross-multiplied the fraction. Remember that?. It's how you can add or subtract any fraction.

Say tou have $\displaystyle \frac{a}{b}-\frac{c}{d}$............any fraction.

Then, $\displaystyle \frac{ad-bc}{bd}$

That's all I done.

 

[CalledSomething]

Oh!! Aha, I do remember. Just hadn't used that in wuite a long while. Thank you so very much! You have truly helped me understand!

 

[lookagain]

1/(1/b -1/a) = x then 1/(a-b)/(ab) = x .

You need to have grouping symbols around the last denominator as in the altered quote box,
because of the order of operations.