subtracting rationals

• Sep 9th 2007, 04:00 PM
Jonboy
subtracting rationals
Hi Everyone! I would like you to check my work. :)

BTW my book is Precalculus, 3rd Edition by Bittinger, Beecher, Ellenbogen, and Penna

The directions are to add/subtract/divide and simplify if possible.

#38: $\displaystyle \frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{2b\,-\,3a}$

Get like denominators: $\displaystyle \;\frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

$\displaystyle \;=\frac{-3a}{-3a\,+\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

Simplify: $\displaystyle \;\frac{\,-\,5a}{-3a\,+\,2b}$

Would it be better to put my answer as: $\displaystyle \;\frac{5a}{3a\,-\,2b}$ ?

# 39: $\displaystyle \frac{9x\,+\,2}{3x^2\,-\,2x\,-\,8}\,+\,\frac{7}{3x^2\,+\,x\,-\,4}$

Factor the denominators: $\displaystyle \frac{9x\,+\,2}{(3x\,+\,4)(x\,-\,2)}\,+\,\frac{7}{(3x\,+\,4)(x\,-\,1)}$

Get a LCM of the denominators: $\displaystyle \frac{(9x\,+\,2)(x\,-\,1)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}\,+\,\frac{7x\,-\,14}{(3x\,+\,4)(x\,-\,1)(x\,+\,2)}$

$\displaystyle \;=\,\frac{9x^2\,-\,7x\,-\,2\,+\,7x\,-\,14}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Simplify: $\displaystyle \frac{(3x\,-\,4)(3x\,+\,4)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Which simplifies to:

$\displaystyle \frac{3x\,-\,4}{(x\,-\,2)(x\,-\,1)}$

My book agrees so I'm pretty sure.

#44: $\displaystyle \frac{6}{x\,+\,3}\,-\,\frac{x\,+\,4}{9\,-\,x^2}\,+\,\frac{2x\,-\,3}{9\,-\,6x\,+\,x^2}$

My answer is: $\displaystyle \frac{-8x^2\,+\,32x\,-\,41}{-(x\,-\,3)^2(x\,+\,3)}$

# 62: $\displaystyle \frac{1}{x^2}\,-\,\frac{1}{y^2}\div\frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2}$

Get the like denominators: $\displaystyle \frac{y^2\,-\,x^2}{x^2y^2}\,\div\,\frac{y^2\,-\,2x\,+\,x^2}{x^2y^2}$

$\displaystyle \;=\,\frac{y^2\,-\,x^2}{x^2y^2}\cdot\,\frac{x^2y^2}{y^2\,-\,2x\,+\,x^2}$

Which just simplifies to: $\displaystyle \frac{y^2\,-\,x^2}{y^2\,-\,2x\,+\,x^2}$

Correct?

Thank you all for your help!
• Sep 9th 2007, 04:05 PM
shilz222
$\displaystyle \frac{(3x-4)(3x+4)}{(3x+4)(x-2)(x-1)} = \frac{3x-4}{(x-1)(x-2)}$
• Sep 9th 2007, 04:08 PM
Jonboy
Quote:

Originally Posted by shilz222
$\displaystyle \frac{(3x-4)(3x+4)}{(3x+4)(x-2)(x-1)} = \frac{3x-4}{(x-1)(x-2)}$

yeah i just saw that; thanks :)
• Sep 9th 2007, 04:10 PM
Jhevon
Quote:

Originally Posted by Jonboy
Hi Everyone! I would like you to check my work. :)

BTW my book is Precalculus, 3rd Edition by Bittinger, Beecher, Ellenbogen, and Penna

The directions are to add/subtract/divide and simplify if possible.

#38: $\displaystyle \frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{2b\,-\,3a}$

Get like denominators: $\displaystyle \;\frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

$\displaystyle \;=\frac{-3a}{-3a\,+\,2b}\,-\,\frac{2a}{-3a\,+\,2b}$

Simplify: $\displaystyle \;\frac{\,-\,5a}{-3a\,+\,2b}$

Good!

Quote:

Would it be better to put my answer as: $\displaystyle \;\frac{5a}{3a\,-\,2b}$ ?
yes. it looks nicer, which is good

Quote:

# 39: $\displaystyle \frac{9x\,+\,2}{3x^2\,-\,2x\,-\,8}\,+\,\frac{7}{3x^2\,+\,x\,-\,4}$

Factor the denominators: $\displaystyle \frac{9x\,+\,2}{(3x\,+\,4)(x\,-\,2)}\,+\,\frac{7}{(3x\,+\,4)(x\,-\,1)}$

Get a LCM of the denominators: $\displaystyle \frac{(9x\,+\,2)(x\,-\,1)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}\,+\,\frac{7x\,-\,14}{(3x\,+\,4)(x\,-\,1)(x\,+\,2)}$

$\displaystyle \;=\,\frac{9x^2\,-\,7x\,-\,2\,+\,7x\,-\,14}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Simplify: $\displaystyle \frac{(3x\,-\,4)(3x\,+\,4)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}$

Which simplifies to:

$\displaystyle \frac{3x\,-\,4}{(x\,-\,2)(x\,-\,1)}$

My book agrees so I'm pretty sure.
yes, good. i was just about to tell you it was the difference of two squares, but you beat me to it. nice

By the way, you LaTex is way more complicated than it has to be. there is no need to put "\," after every term
• Sep 9th 2007, 04:29 PM
Jhevon
Quote:

Originally Posted by Jonboy
#44: $\displaystyle \frac{6}{x\,+\,3}\,-\,\frac{x\,+\,4}{9\,-\,x^2}\,+\,\frac{2x\,-\,3}{9\,-\,6x\,+\,x^2}$

My answer is: $\displaystyle \frac{-8x^2\,+\,32x\,-\,41}{-(x\,-\,3)^2(x\,+\,3)}$

my answer is $\displaystyle \frac {9x^2 - 32x + 33}{(x + 3)(x - 3)^2}$

Quote:

# 62: $\displaystyle \frac{1}{x^2}\,-\,\frac{1}{y^2}\div\frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2}$

Get the like denominators: $\displaystyle \frac{y^2\,-\,x^2}{x^2y^2}\,\div\,\frac{y^2\,-\,2x\,+\,x^2}{x^2y^2}$

$\displaystyle \;=\,\frac{y^2\,-\,x^2}{x^2y^2}\cdot\,\frac{x^2y^2}{y^2\,-\,2x\,+\,x^2}$

Which just simplifies to: $\displaystyle \frac{y^2\,-\,x^2}{y^2\,-\,2x\,+\,x^2}$

my answer is: $\displaystyle \frac {(x - y)^2 - x^4}{x^2 y^2} = \frac {\left( x - y + x^2 \right) \left( x - y - x^2 \right)}{x^2 y^2}$

remember the order of operations. we do division before addition and subtraction

the only way your answer would be valid is if the question originally said $\displaystyle \left( \frac{1}{x^2}\,-\,\frac{1}{y^2} \right) \div \left( \frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2} \right)$