# Thread: How do I solve these?

1. ## How do I solve these?

x+17/x^2-6x+8 + x-2/x-4 = x-4/x-2

32/t^2+3t+2 - 3 = t-3/t+1

r(3-r) + (2r+2)^2 = 0

Can somebody help me with these please our midterm is tomorrow and if questions like these turn up (which is very likely) then my goose is cooked.

EDIT: I grouped them together to make things clearer:
(x+17)/(x^2-6x+8) + (x-2)/(x-4) = (x-4)/(x-2)

32/(t^2+3t+2) - 3 = (t-3)/(t+1)

r(3-r) + (2r+2)^2 = 0

2. Originally Posted by strigy
x+17/x^2-6x+8 + x-2/x-4 = x-4/x-2
If I read this correctly (Parentheses would help. I assume you mean (x+17)/(x^2- 6x+ 8) rather than x+ (17/x^2)- 6x+ 8)
$\displaystyle x^2- 6x+ 8= (x- 4)(x- 2)$ so this is
$\displaystyle \frac{x+ 17}{(x- 4)(x- 2)}+ \frac{x- 2}{x- 4}= \frac{x- 4}{x- 2}$

Multiplying each term by (x- 4)(x- 2) gets rid of all the fractions.

32/t^2+3t+2 - 3 = t-3/t+1
Again, I am going to assume that you really mean $\displaystyle \frac{32}{t^2+ 3t+ 3}- 3= \frac{t- 3}{t+ 1}$. Once again, [tex]t^3+ 3t+ 2= (t+1)(t+ 2) so multiplying each term by (t+1)(t+2) gets rid of the fractions.

r(3-r) + (2r+2)^2 = 0
Just go ahead and multiply it out: $\displaystyle 3r- r^2+ 4r^2+ 8r+ 4= 0$ which is the same as 3r^2+ 11r+ 4= 0. Complete the square or use the quadratic formula.

Can somebody help me with these please our midterm is tomorrow and if questions like these turn up (which is very likely) then my goose is cooked.

EDIT: I grouped them together to make things clearer:
(x+17)/(x^2-6x+8) + (x-2)/(x-4) = (x-4)/(x-2)

32/(t^2+3t+2) - 3 = (t-3)/(t+1)

r(3-r) + (2r+2)^2 = 0
Why didn't you do that to begin with?!

3. Ok thanks HallsofIvy and sorry for the confusion. I got number 1.

I seem to be stuck at 2 and 3, though.

I'm stuck here 4t^2-11t+32=0 for number 2, and at 3r^2+11r+4=0 for number 3.

I tried quadratic but that didn't work. What do I do next?