Question

3x-2/2x^2-5x-3 (-) 1/x-3 subtract

Answer 1

\[\frac{3x-2}{2x^2-5x-3}-\frac{1}{x-3}\]?

Answer 2

yep

Answer 3

alright so first you want to factor the denominator of the first part, do you know how to do this?

Answer 4

No i don

Answer 5

alright so first you want to multiply your leading coefficient by -3

2*-3=-6

now you need to figure out all the factors of -6

1*-6
2*-3
3*-2
6*-1


now you need to figure out which add up to your middle coefficient hwich is -5

-6+1=-5

so your numbers are -6 and 1

replace the middle term with these two terms like this

\[2x^2-6x+1x-3\]

now you can group both of these

\[(2x^2-6x)+(x-3)\]

pul out 2x from th first equation you get

\[2x(x-3)+1(x-3)\]

your answer is going to be the function in parenthesis and the one outside soo

(2x+1)(x-3)

Answer 6

so now you have

\[\frac{3x-2}{(2x+1)(x-3)}-\frac{1}{x-3}\]

Answer 7

now when subtracting to fractions you must always create a common denominator...looking at this, your second denominator is missing (2x+1) so multiply it by 1 (2x+1/2x+1)

\[\frac{3x-2}{(2x+1)(x-3)}-\frac{1}{x-3}*\frac{2x+1}{2x+1}\]

multiplying it out you get

\[\frac{3x-2}{(2x+1)(x-3)}-\frac{2x+1}{(x-3)(2x+1)}\]

you can now combine like the fractions to get

\[\frac{3x-2-2x-1}{(2x+1)(x-3)}\]

add like terms you get

\[\frac{x-3}{(2x+1)(x-3)}\]

you can cancel out the x-3's so you're left with

\[\frac{1}{2x+1}\]