## Solving Equations With fractions

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iceking
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### Solving Equations With fractions

Hey, I know how to solve equations with variables, but not when they involve fractions. I have a problem like this: R= $\frac{x(A+B)}{1}$ I need to "Solve for B". How would I do that? Please reply as soon as possible.

Thanks, iceking

stapel_eliz
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The steps for solving equations with fractions are the same as for solving equations without. Where are you getting stuck?

Note: Since this "fraction" is of the form "divided by one", you can ignore the denominator and simply solve "R = x(A + B)" in the usual manner.

iceking
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Joined: Sat Sep 26, 2009 12:46 am
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### Re: Solving Equations With fractions

The steps for solving equations with fractions are the same as for solving equations without. Where are you getting stuck?

Note: Since this "fraction" is of the form "divided by one", you can ignore the denominator and simply solve "R = x(A + B)" in the usual manner.
Thanks, so now I have the equation "R=xa+xb". Should I divide both sides by x? If so, then I would have the equation " $\frac{R}{X}$ = A+B" .
Then what? Would I then subtract "A" from both sides to get the equation $\frac{R}{X}$ -A=B ? The only answer choices I have are:
B=(R-A)/x and B=(A-Rx)/x and B=(R-Ax)/x . I need to figure out Which equation would get me to the value of "B"

stapel_eliz
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Work with the algebra, and see which of the answer-options matches your answer. The only difference is one of formatting: they did the addition/subtraction to isolate the B-term first, whereas you went the (usually harder) route of doing the multiplication/division first.

iceking
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### Re:

Work with the algebra, and see which of the answer-options matches your answer. The only difference is one of formatting: they did the addition/subtraction to isolate the B-term first, whereas you went the (usually harder) route of doing the multiplication/division first.

I tried the addition/subtraction this time, and I got the equation "xB=R-xA" Then I divided both sides by "x" " $\frac{xb}{x}$ = $\frac{R-xA}{x}$ " Then simplyfying that, I got: " B= $\frac{R-xA}{x}$ " Is this the correct equation?

stapel_eliz
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Looks good to me!

iceking
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