simplifying complex fractions

[adixon5]

I need help with this college algebra problem.

3/2+x - 4/2-x/1/x+2-3/x-2 ......it's a complex fraction but I couldn't write it out how it actually looks in the book

 

[Deleted member 4993]

I need help with this college algebra problem.

3/2+x - 4/2-x/1/x+2-3/x-2 ......it's a complex fraction but I couldn't write it out how it actually looks in the book

I cannot decipher your problem. Try to parenthesis - using PEMDAS to indicate hierarchy of operation. For example:

\(\displaystyle \dfrac{x}{x+1}\) can be written as x/(x+1) and \(\displaystyle \dfrac{x-1}{x+1}\) can be written as (x-1)/(x+1)

Following the above principle, please repost your problem.

 

[mmm4444bot]

3/2+x - 4/2-x/1/x+2-3/x-2


it's a complex fraction

It sure is, as typed.


You could show what appears in your book, were you to use grouping symbols around numerators and denominators.


For example, typing [3/(2+x) - 4/(2-x)] / [1/(x+2) - 3/(x-2)] would be \(\displaystyle \frac{\frac{3}{2+x} - \frac{4}{2-x}}{\frac{1}{x+2} - \frac{3}{x-2}}\)


Is this correct?

 

[soroban]

Hello, adixon5!

3/2+x - 4/2-x/1/x+2-3/x-2

Do you know what you have written?. \(\displaystyle \frac{3}{2} + x - \frac{4}{2} - \dfrac{\frac{x}{1}}{x} + 2 - \frac{3}{x} - 2\)

I believe mmm4444bot has the correct interpretation:.\(\displaystyle \dfrac{\dfrac{3}{2+x} - \dfrac{4}{2-x}}{\dfrac{1}{x+2} - \dfrac{3}{x-2}} \)

We have: .\(\displaystyle \dfrac{\dfrac{3}{x+2} + \dfrac{4}{x-2}}{\dfrac{1}{x+2} - \dfrac{3}{x-2}}\)

Multiply by \(\displaystyle \dfrac{(x-2)(x+2)}{(x-2)(x+2)}\!:\;\;\dfrac{(x-2)(x+2)\left(\dfrac{3}{x+2} + \dfrac{4}{x-2}\right)}{(x-2)(x+2)\left(\dfrac{1}{x+2} - \dfrac{3}{x-2}\right)} \)

Now, continue...

 

[Jason76]

I would solve each fraction problem (separately) on the top and bottom. Next, divide the answers. That would be the same as multiplying the top answer by the inverse (flipped version) of the bottom answer.