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Ok So I am about to take the accuplacer college level math for a college but I do not understand this problem on the practice packet. Honestly I am blank on this problem and need help step by step on how to solve it.Please help me understand it. Thanks in advance

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

A. 1/b – 1/a

B. b – a

C. 1/ab

D. a – b/ab

E. ab/a – b

2 Answers 2

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It is asking you to solve for $x$, given that $$\frac{1}{x} +\frac{1}{a}=\frac{1}{b}.$$

To solve for $x$, first isolate $x$ by itself on one side; for example, move that $\frac{1}{a}$ to the right. That will give you an equation of the form $$\frac{1}{x} = \text{stuff}.$$

Do the operation on the right, and then take reciprocals (or cross-multiply) to get an expression for $x$ in terms of $a$ and $b$. Then figure out which of the five options given is that expression for $x$.

You can the work as an edit to your question and we can tell you if you are doing it right or not; that will help you learn better than me doing it for you.

(For extra points, figure out exactly on which step you need to assume $a\neq b$...)

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    Clearly explained and brief still! +1.2012-05-03
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    So the equal sign (=) with a slash (/) means if a and b are "not equal"2012-05-03
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    @Backtrack... Ehr... yes. If you were unsure about the symbols, you should have mentioned that explicitly.2012-05-03
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    Yes I will try this tomorrow morning and you guys can check my work. But anyway yes, the fractions are throwing me off. To isolate x should I divide 1/a by itself and also by 1/b? And then after that what? And also should I leave the letters how they are?2012-05-03
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    @Backtrack: To isolate $x$, you first *move* the $\frac{1}{a}$ to the right; This does **not** involve divisions of any kind, just additions/subtractions. Then you will have an expression on the right that has no $x$'s in it: just $1$s, $a$s, and $b$s. Do the operation, which will be a sum/subtraction of fraction, and write the answer as a single fraction. You should get something along the lines of $$\frac{1}{x} = \frac{\text{expression1 involving }a\text{ and }b}{\text{expression2 involving }a\text{ and }b}.$$ **Then** you can cross multiply to get $x$ equal to something.2012-05-03
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    You cannot invert $1/a$ by itself and $1/b$ by itself, because you can't just do stuff to one side of the equation (or to just *parts of it*) and expect the equality to be respected. For example, while $\frac{1}{x}+\frac{1}{3} = \frac{1}{6}$ (if $a=3$ and $b=6$, say) has the solution $x=2$, $\frac{1}{x}+3=6$ is *not* true with $x=2$.2012-05-03
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    OK is this correct so far 1/x + 1/a = 1/b Then: 1/x + 1/a - 1/a = 1/b - 1/a Then: 1/x = 1/b - 1/a Now what? 1/x = 2/ba?2012-05-03
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    @Backtrack: Last time I checked, we don't add fraction by doing $$\frac{1}{b}-\frac{1}{a} = \frac{1+1}{ab}.$$ Do we? Is $$\frac{1}{2}-\frac{1}{5} = \frac{2}{10}\ ?$$ If you want to put a common denominator on them, notice that $\frac{1}{b} = \frac{a}{ab}$ and $\frac{1}{a} = \frac{b}{ab}$.2012-05-03
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    OK so far I at 1/x = 1/a - 1/b What am I supposed to do after this? would it be 1/x= ab/ a^2b^2 Sorry for such novice questions but you are really helping me a lot2012-05-03
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    Do you know how to add/subtract fractions? $$\frac{a}{ab} - \frac{b}{ab}\neq\frac{ab}{a^2b^2}.$$How do you subtract fractions when they have the same denominator? How much is $\frac{5}{14}-\frac{3}{14}$? It's not $\frac{5\times 3}{14\times 14}$. We'are **subtracting**, not multiplying.2012-05-03
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    So it would just be 1/x= a-b/ab right2012-05-03
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    @Backtrack: Yes (modulo parentheses; technically, $a-b/ab$ means $$a - \frac{b}{ab}$$ and you mean $(a-b)/ab$, which means $$\frac{a-b}{ab}$$; and now you can cross-multiply (or take reciprocals on both sides) to solve for $x$. So which is the right answer, and why do you need $a\neq b$?2012-05-03
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    I'm drawing blanks right now. So if a=3 and b=6 for example. (3-6)/(3)(6) : 1/x= -3/18?2012-05-03
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    @Backtrack: Yes: note that $-\frac{3}{18} = -\frac{1}{6}$. And indeed, $-\frac{1}{6} + \frac{1}{3} = \frac{1}{6}$. But the point here is to express $x$ in terms of $a$ and $b$, once you have that $$\frac{1}{x} = \frac{a-b}{ab}.$$ Not to solve specific cases.2012-05-03
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    so the answer is D? But it wants x= not 1/x= so that can't be it. How do I get it to be x2012-05-03
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    @Backtrack: By my count, I've told you what the next step is... three times. Instead of guessing, read what I've written.2012-05-03
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    So I took the reciprocal and got x/1 or (x)= 18/-3 which would mean x = ab/(a-b) So the answer is E2012-05-03
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    @Backtrack: **Why** are you doing a case with values for $a$ and $b$? The problem is about **algebra**, not arithmetic. Yes, the answer is E. Don't think that you can plug in specific values and solve and then check which answer is "right" for those values. Different expressions may give the same value for specific $a$ and $b$.2012-05-03
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    So I should have just solved it with the letters without putting the a=3 and b=6 example right? So I get that I needed to solve for x. I just don't know why at the beginning it says if a and b are not equal?2012-05-03
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    @Backtrack: Yes, you should have just solved with algebra. If $a$ were equal to $b$, after the first step you would get $\frac{1}{x}=0$. Does that have *any* solution? Or, when you get to $\frac{1}{x}=\frac{a-b}{ab}$, if you try to cross-multiply that requires dividing by $a-b$. But if $a=b$, then $a-b=0$. And you know what happens if you divide by $0$? That's right: the universe explodes.2012-05-03
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    Yes that's true. Thank you very much for your help. This site really gives in depth explanation and expertise tutoring Sorry for bothering you so much. I will keep practicing2012-05-03
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    I noticed that in order to get common denominators on the right side of the equation (a/ab-b/ab) we needed to change the 1's on top to letters (a and b). Are we able to remove the number only when there is a 1 or can we do it with any number. And if not what are we supposed to do for example if the number on top is 22012-05-03
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    @Backtrack: You use *algebra*. $$\frac{a}{b}+\frac{c}{d} = \frac{ad}{bd} + \frac{cb}{db} = \frac{ad+bc}{bd}.$$ Doesn't matter if the numerators are "numbers" or "letters", it's algebra all the way down.2012-05-03
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You need to solve the equation for $x$, but I imagine the fractions are throwing you off. You can get rid of the fractions by multiplying both sides by $abx$ (why? because this is the least common multiple of $a,b$, and $x$), then it's simple algebra to rearrange and solve for $x$. Do you see why it's necessary for $a\neq b$?