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So the expression is:

$$\frac1{2b} + \frac b2$$

Apparently the answer is $\frac{1 + b^2}{ 2b}$.

I came up with $2b$ by multiplying the first fraction by $2$ and the second fraction by $2b.$

I got: $$\frac{2+ 2b^2}{2(2b)}$$ which simplifies to $\frac{2b^2}{2b}$ which is where I got $2b$ from. Can someone help me with this problem and the LCD?

user1176409
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SunGone
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    Khan academy is your friend-https://www.khanacademy.org/math/arithmetic/fraction-arithmetic/arith-review-common-denominators/v/finding-common-denominators – John_dydx Mar 20 '18 at 17:26
  • A lot of your notation is ambiguous, but I'm trying to put it into correct notation. Please check my edits. I'm particularly concerned about the expression after "I got:..," because your error is either in this step, or the very next step. – Thomas Andrews Mar 20 '18 at 17:36
  • How did you get ${2b^2\over2b}$ from ${2+2b^2\over2(2b)}$? It looks like you divided the denominator by $2$, but subtracted $2$ from the numerator. Or did you perhaps drop the $1$ term after dividing the numerator by $2$? – amd Mar 20 '18 at 18:47

1 Answers1

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If you have

$$ \frac{1}{2b}+\frac{b}{2} $$

the LCD of the denominators is $2b$; therefore, the first fraction can be left alone, and the second fraction multiplied by $b/b$ to yield

$$ \frac{1}{2b}+\frac{b^2}{2b} $$

which can then be combined to obtain

$$ \frac{1+b^2}{2b} $$

It seemed to me, incidentally, that you obtained

$$ \frac{2+2b^2}{2(2b)} $$

which wants only division by $2/2$ to get

$$ \frac{1+b^2}{2b} $$

as above. As a side note, I would say that writing out fractions "in-line" as opposed to "stacked" (as it were) makes it potentially confusing just exactly where the numerators and denominators begin and end. I don't know if you're doing that as you work out the problem, but if you are, it might lead to errors.

Brian Tung
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  • Oh my gosh it makes so much sense now! I can see it was just silly mistakes that led me astray. Thank you so much!! Your help is greatly appreciated. – SunGone Mar 20 '18 at 17:44