simplify complex fractions......

[Princezz3286]

the directions are to simplify completely.....
I am not too sure how to write these so you can see what the problem looks like exactally....... but I'll try to explain it......
1) 1+ 2/d -3
---------------
-2d/ d-3 - d

This problem reads 1 + 2 over (d -3), all over -2d over (d - 3). the -d is written after the fraction. I hope this kind of makes sense...... in my book it say the answer is -1/d but I keep getting something completely different....

to begin, on the top I multiply the 1/1 by (d - 3) to get the lcd so I can add them and then I can cancel the d - 3 from the top and the bottom and I am left with 2

for the bottom part..... I multiplied (-2d/d - 3) by d and I multiplied the (- d) by (d - 3) to get the lcd d(d - 3)
I got -d (2d) / d(d - 3) minus d (d - 3) so I end up with - d (2d) - d (d - 3)/ d (d - 3) and cancel the d (d - 3)
I get for the final simplification, - 2/ d (2d) where nothing can be cancelled or combined as they are not like terms.... but the answer in my book is - 1/d can someone please highlight what I did wrong here? I really struggle with these types of problems.......

 

[jwpaine]

\(\displaystyle \frac{\frac{1+2}{d -3} }{ \frac{-2d}{d-3}-d}\)

First, make the denominator into a single rational, recall: \(\displaystyle \frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{db}\)

\(\displaystyle \frac{-2d}{d-3}-d} = \frac{-2d - d(d-3)}{(d-3)} = \frac{-2d -d^2 + 3}{(d-3)}\) (don't factor yet, it'll be easier)

Now multiply the numerator of the original rational by the reciprocal of your new denominator. Recall: \(\displaystyle \frac{\frac{a}{b}}{c} = \frac{a}{b} x \frac{1}{c}\)

\(\displaystyle \frac{(1+2)}{(d -3)} \cdot \frac{(d-3)}{-2d -d^2 + 3}\)

Now you can cancel the (d-3) from both the numerator and denom.... then proceed to factor and cancel some more!

John

 

[Princezz3286]

in regards to this problem, does it matter that the top equation in my book had the 1 + in front of the fraction on the top? I noticed that you had it written correctly all except that...... Sorry for so many questions but I get really lost on these...... I am horrible at math in case you hadn't noticed!

Heather

 

[jwpaine]

Yes, it would matter.

Hopefully my above work can guide you in the right direction for the 1 being added to a rational.

John