Question

Which of the following is the simplified form of x plus 4 over 2 x plus 6 plus the fraction 3 over x squared minus 9 ?

Answer 1

\[\frac{ x - 4 }{ 2x + 6 } + \frac{ 3 }{ ?x^2 - 9}\]

Answer 2

\[\frac{x+4}{2x+6}+\frac{3}{x^2-9}\]?

Answer 3

Oops, there's a question mark in there, and it shouldn't be there

Answer 4

is x-4 or x+4?

Answer 5

in any case, start by factoring everything as much as possible, then make a common denominator for the two fractions so you can combine them

Answer 6

x-4

Answer 7

okay, because your problem statement says "x plus 4"

Answer 8

I'm sorry, it is x+4

Answer 9

I got this far:
\[\frac{ x+4 }{ 2(x+3) } + \frac{ 3 }{ (x-3)(x+3) }\]

Answer 10

okay, to make the common denominator you'll have to multiply the lh fraction by (x-3)/(x-3) and the rh fraction by 2/2

Answer 11

do you see why?

Answer 12

yes

Answer 13

good. have fun :-)

Answer 14

is it:
\[\frac{ 6+(x+4) }{ 2(x+3) }\]

Answer 15

\[x + 2 \over 2(x + 3)\]
\[ x - 2 \over 2(x - 3) \]
\[ x - 2 \over 2(x +3)\]
\[ x+ 2 \over 2(x- 3) \]

Answer 16

Hmm, looks like you went off the rails somewhere...

\[\frac{ x+4 }{ 2(x+3) } + \frac{ 3 }{ (x-3)(x+3) } = \frac{(x+4)} {2(x+3)}*\frac{(x-3)}{(x-3)} + \frac{3}{(x-3)(x+3)}*\frac{2}{2}\]\[=\frac{(x+4)(x-3) + 3*2}{2(x-3)(x+3) } = \frac{ x^2-3x+4x-12+6}{2(x-3)(x+3) }=\frac{x^2+x-6} {2(x-3)(x+3)}\]Now we factor the numerator, cancel common factors and we're done.

Answer 17

Its adding not multiplication

Answer 18

You have to multiply to get the common denominator in order to add

Answer 19

just like adding 1/4 and 1/7

Answer 20

\[\frac{1}{4} + \frac{1}{7} = \frac{1}{4}*\frac{7}{7} + \frac{1}{7}*\frac{4}{4} = \frac{7}{28} + \frac{4}{28} = \frac{11}{28}\]

Answer 21

But it has to be simplified so the (x-3)s would cancel each other, leaving: 6+(x+4)/2(x+3)

Answer 22

you're doing it wrong. what do you get when you factor \(x^2+x-6\)?