EASY TEN POINTS PLZ ANSWER SIMPLIFY COMPLEX FRACTIONS!?

Question

1/ x^2 - x - 2

Minus

x/ x^2 - 5x +6

Can you please show your work so I can understand how to do this.

Thankyou, and ten points to person who gives the right answer and shows work.

Answer 1

Forgive me, but in this format I can't tell if your problem is supposed to be 1/(x^2-x-2)-x/(x^2-5x+6) or is it ((1/x^2)-x-2)-((x/x^2)-5x+6).  I could assume it's the 1st one since you say it's a complex fraction.

So remembering polynomials it would work out something like this:

I'll look at it for a few more minutes to see if there is anything else that will make it simpler.

Edit:

Ok, let's make both denominators of the right half the same thus:

[1/(x-2)]*[(1(x-3)/(x+1)(x-3))-(x(x+1)... Then we can add the numerators of that half

[1/(x-2)]*[(x-3)-(x^2+1x)/(x+1)(x-3)]

[1/(x-2)]*[(-x^2-3)/(x+1)(x-3)]

Hmmm, think I did something wrong since the numerator on the right won't factor out to help simplify further and I've run out of time.  Hope someone else can help you more.

Answer 2

i'm not sure if this is right but i think this is what you do: (he little stars mean multiplication by the way, i couldn't figure out how to do a dot.

simplify the first one:

since x squared is the same thing as x times x, you can write the bottom part like this: (x)(x)-x-2. then you can cancel out two of the x's (x-x=0x) so you have x-2. that can also be written as -2+x. since thats all below 1 you have: 1/-2+x.

simplify the second one:

the fraction can also be written like this: x/(x)(x)-(x*x*x*x*x) + 6. ok, you know how when you have the same number on the top and the bottom it means it equal to one? (3/3=1) well i think you can do the same thing with variables. so one of the x on top and one of the two positive x's on the bottom cancel out so you get: 1/x- (x*x*x*x*x)+6.

then i think you just subtract the x's to get: 1/-(4x)+6 or 1/6-(4x)

then you subtract 1/6-4x and 1/-2=x. an i don't know how to do that or even if those steps were exactly right.

hope i helped a little!  

Answer 3

1/( x^2 - x - 2)  -  x/ x^2 - 5x +6 =

Whenever you add (or subtract) two fractions with "unlike" denominators you have to find a common denominator ... or in this case create a common denominator.

Step 1: Factor each of your denominators

x^2 - x - 2 = (x-2)(x+1)

x^2 - 5x +6 = (x-2)(x-3)

The denominators share (x-2) in common but each of them also have another factor. The first fraction has (x-2)(x+1) as a denominator; multiply this fraction by (x-3)/(x-3) This is allowed because you are really just multiplying by 1. Now multiply the second fraction by (x+1)/(x+1). At this point you should have:

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

Find the product in your numerators

   1 (x-3) = x - 3

x (x+1) = x^2 + x

The denominator should also be expanded.

x^2 - x - 2 (x-3) = x^3 - 4x^2 + x + 6 just like x^2 - 5x + 6 (x+1) also equals x^3 - 4x^2 + x + 6. Now your equation looks like:

     (x - 3)/x^3 - 4x^2 + x + 6    -   (x^2 + x)/x^3 - 4x^2 + x + 6

Simplify the numerator: x - 3 - x^2 - x = - x^2 - 3

Put your common denominator back:

  - x^2 - 3/ x^3 - 4x^2 + x + 6    (solution)

I hope I helped!



    

Answer 4

1/(x^2 - x - 2) - x/(x^2 - 5x + 6)

Factor the denominators to derive the LCD

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

LCD = (x + 1)(x - 2)(x - 3)

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

[x - 3 - x^2 - x]/[(x + 1)(x - 2)(x - 3)]

-(x^2 + 3)/[(x + 1)(x - 2)(x - 3)]