#### jp2003

##### New member

3....... 2....... 5
___ + __ _ ___
x+3.. x... ..... 2

answer in book is 12 _ 5x _ 5x^2
.................... ..._______________
........... .............. ......2x(x+3)

thank you

#### soroban

##### Elite Member
Hello, jp2003!

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

#### Denis

##### Senior Member
Just a suggestion; I find it easier and faster and a bit less confusing
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.