Rational addition?

Question

4/ x^2 -3x + 3/x^2-5x+6

x^2-3x makes x^2 - 5x +6 by (x-2)(x+3) right????

So how do I do the fancy thing where I add it to the top and bottom to get the LCD

Answer 1

you almost got it !!
(x^2 - 3x)= x(x-3)
and (x^2 - 5x + 6 ) = (x -3 )(x - 2)
therefore you have 4/[x(x-3)] + 3 / [(x -3 )(x - 2)]
now you take the LCM
that is, you multiply (x-2)to the nr & dr of the first term .
you multiply x to the nr and dr of the 2nd term
you will thus get ..
[4(x-2)]/[x(x-3)(x-2)] + 3x/[x(x-3)(x-2)]
=[4x+3x-2]/[x(x-3)(x-2)]
=[7x-2]/[x(x-3)(x-2)]

Answer 2

4/ x^2 -3x + 3/x^2-5x+6

Probably should be written as

4/(x^2 -3x) + 3/(x^2-5x+6)

to show the denominators properly

Factor the denominators in order to find the LCD

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

LCD is x(x - 2)(x - 3)

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

(4x - 8 + 3x)/[x(x - 2)(x - 3)]

(7x - 8)/[x(x - 2)(x - 3)]

Answer 3

4/(x^2-3x) + 3/(x^2 -5x +6)
4/[x(x-3)] + 3/[(x-2)(x-3)]
4/[x(x-3)(x-2)] + 3x/[x(x-3)(x-2)]
Now the denominators are the same and we can add, getting:
(4+3x)/[x(x-3)(x-2)]