- Thread starter jp2003
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Adding and subtracting fractions is never a plesant task . . .

\(\displaystyle \L\frac{3}{x\,+\,3}\,+\,\frac{2}{x}\,-\,\frac{5}{2}\)

Answer in book is: \(\displaystyle \L\:\frac{12\,-\,5x\,-\,5x^2}{2x(x\,+\,3)}\)

To add and/or subtract fractions, the denominators must be the same.

Find the common denominator: \(\displaystyle \:2x(x\,+\,3)\)

Then "convert" the fractions so they each have the common denominator.

Multiply the first fraction by \(\displaystyle \frac{2x}{2x}:\;\;\;\L\frac{2x}{2x}\,\cdot\,\frac{3}{x\,+\,3} \;=\;\frac{6x}{2x(x\,+\,3)}\)

Multiply the second fraction by \(\displaystyle \frac{2(x+3)}{2(x+3)}:\;\;\;\L\frac{2(x+3)}{2(x+3)}\,\cdot\,\frac{2}{x} \;=\;\frac{4(x\,+\,3)}{2x(x\,+\,3)}\)

Multiply the third fraction by \(\displaystyle \frac{x(x+3)}{x(x+3)}:\;\;\;\L\frac{x(x+3)}{x(x+3)}\,\cdot\,\frac{5}{2}\;=\;\frac{5x(x\,+\,3)}{2x(x\,+\,3)}\)

The problem becomes: \(\displaystyle \L\:\frac{6x}{2x(x\,+\,3)}\,+\,\frac{4(x+3)}{2x(x\,+\,3)} \,-\,\frac{5x(x\,+\,3)}{2x(x\,+\,3)}\)

Then we have: \(\displaystyle \L\:\frac{6x\,+\,4(x\,+\,3)\,-\,5x(x\,+\,3)}{2x(x\,+\,3)} \:=\:\frac{6x\,+\,4x\,+\,12\,-\,5x^2\,-\,15x}{2x(x\,+\,3)}\)

And finally: \(\displaystyle \L\:\frac{12\,-\,5x\,-\,5x^2}{2x(x\,+\,3)}\)

if you go in stages; like:

2/x - 5/2 = (4 - 5x) / (2x)

Now bring in the other term:

3 / (x + 3) + (4 - 5x) / (2x) = you know what to do :wink:

Similarly, if you had 4 terms, you can:

do the 1st 2

do the next 2

do the above 2 :idea:

Much easier to "check back" if you end up with wrong answer;

take my word for it.