rational expressions

robert mclaughlin

New member
Joined
Jan 12, 2013
Messages
16
1] [ x+3
____ - ______
x^2-9 x^2-2x-3

i get

x^2-2x-3 - x^2-27
_______________________
x^2+27

=

(x-1)
_____
(x+1)

i also get

1
_____
(x+3)

where am i mixing up?
 
1] [ x+3
____ - ______
x^2-9 x^2-2x-3

i get

x^2-2x-3 - x^2-27
_______________________
x^2+27

=

(x-1)
_____
(x+1)

i also get

1
_____
(x+3)

where am i mixing up?
Did you see that I answered your previous post and asked you not to write fractions this way? I told you how to write fractions so they can be understood. Use parentheses, brackets, and / for fractions if you do not know LaTeX. No one can understand the mess that you are writing.

If you do not know LaTeX, write 1 / (x + 3) for \(\displaystyle \dfrac{1}{x + 3}\)
 
rewrite rational expression

1 / (x^2-9) - (x+3) / (x^2-2x-3)
=
(x^2-2x-3) - (x^2-27) / (x^2+27)

=

(x-1) / (x+1)

im missing a step, somewhere, didnt see other post, ill look now
 
subtract and simplify

1/(x^2-9)- (x+3)/ (x^2-2x-3)


ok thats the question working on how to post the equations correctly
 
factor the equation

its too long to show all the steps, but the answer should be

(-x+x+10) / (x+3)(x+3)(x+1)



self study
class
hard chapter
 
its too long to show all the steps, but the answer should be

(-x+x+10) / (x+3)(x+3)(x+1)



self study
class
hard chapter
\(\displaystyle \dfrac{1}{x^2 - 9} - \dfrac{x + 3}{x^2 - 2x - 3} = \dfrac{1}{(x - 3)(x + 3)} - \dfrac{x + 3}{(x - 3)(x + 1)} =\)

\(\displaystyle \left(\dfrac{1}{(x - 3)(x + 3)} * \dfrac{x + 1}{x + 1}\right) - \left(\dfrac{x + 3}{(x - 3)(x + 1)} * \dfrac{x + 3}{x + 3}\right) = \dfrac{x + 1 - (x + 3)^2}{(x - 3)(x + 3)(x + 1)} = \dfrac{-(x^2 + 5x + 8)}{(x - 3)(x + 3)(x + 1)}\)

Learn to check your work. That habit will give you confidence and save your bacon on tests.

\(\displaystyle x = 4 \implies \dfrac{1}{4^2 - 9} - \dfrac{4 + 3}{4^2 - 2 * 4 - 3} = \dfrac{1}{7} - \dfrac{7}{5} = \dfrac{5 - 49}{35} = \dfrac{-44}{35} = \dfrac{-(16 + 20 + 8)}{7 * 1 * 5} = \dfrac{-(4^2 + 5 * 4 + 8)}{(4 + 3)(4 - 3)(4 + 1)}.\)

Let's try your answer

\(\displaystyle x = 4 \implies \dfrac{1}{4^2 - 9} - \dfrac{4 + 3}{4^2 - 2 * 4 - 3} = \dfrac{-44}{35} \ne \dfrac{- 4 + 4 + 10}{(4 + 3)(4 + 3)(4 + 1)} = \dfrac{2}{49}.\)
 
4/(4-a^2)-5/(6+3a)=

12/(12-a^2)-10/(12+3a)=


2/(12+3a)

not reduced

or 2/ (12+3a^2)


should be easy but ,,
I see where you are going off the rails.

\(\displaystyle \dfrac{4}{4 - a^2} - \dfrac{5}{6 + 3a}.\) The common denominator is NOT found by finding a common multiple of the constant terms in the denominators.

It has to be a common multiple of the denominators in their entirety.
 
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