# How do I solve these?

• Sep 2nd 2009, 01:52 AM
strigy
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. (Speechless)

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
• Sep 2nd 2009, 03:32 AM
HallsofIvy
Quote:

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)
$x^2- 6x+ 8= (x- 4)(x- 2)$ so this is
$\frac{x+ 17}{(x- 4)(x- 2)}+ \frac{x- 2}{x- 4}= \frac{x- 4}{x- 2}$

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

Quote:

32/t^2+3t+2 - 3 = t-3/t+1
Again, I am going to assume that you really mean $\frac{32}{t^2+ 3t+ 3}- 3= \frac{t- 3}{t+ 1}$. 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.

Quote:

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

Quote:

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. (Speechless)

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?!
• Sep 2nd 2009, 04:46 AM
strigy
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?