Question

solving rational equations:
x+2 x - 1 4x+2
---- = ---- - -----
x x x^2-3x
I know how to make them all have a common denominator x(x-3)
but when it comes to solving I get
x^2+5x+6=x^2-4x-3x-3x-2
which then comes up to be x^2+3x-11
This isn't right. What did I do wrong?

Answer 1

maybe it would be easier to multiply both sides by
\[x(x-3)\]

Answer 2

i did

Answer 3

but i can't figure out what i did wrong because i get the wrong answer... when i multiply x+2 and x-3 i get x^2+5x+5
when i multiply x-1 and x-3 i get x^2-4x-3

Answer 4

you get
\[(x+2)(x-3)=(x-1)(x-3)-4x+2\] i htink

Answer 5

hold on

\[(x+2)(x-3)=x^2-x-6\] ,maybe that is the mistake

Answer 6

also
\[(x-1)(x-3)=x^2-4x+3\]

Answer 7

oh!!!!!! i put (x+2) and postive (x+3)!!

Answer 8

that was a waste of time i was looking just at how i solved it and not how i multiplied it. Thanks

Answer 9

\[(x+2)(x-3)=(x-1)(x-3)-4x+2\]
\[x^2-x-6=x^2-4x+3-4x+2\]
\[x^2-x-6=x^2-8x+5\] etc

Answer 10

yw

Answer 11

when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative?
for examples
x-1 4x+2
---- - -----
x(x-3) x(x-3)
would you make it into a plus sign, and then make 4x and 2 negative?
-4x-2

Answer 12

bump