Help me!! :(

[firemouse]

1. Solve: 6/(x(3-2x)) - 4/(3 - 2x) = 1

2. Solve: s = vt + (at^2)/2 for t




Had alot of trouble with these, could use some help. Explaining how you got the answer would help alot to. Thanks guys.

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3. Graph the parabola y = 3x^2 8x - 11 between x = -4 x = 2 use the graphs to find the approximate roots.

for number 3, I dont have a graphing calculator. Is there any free online graphing calcs i can download. I think I could graph if i had a calc... but I really have no idea without a trusty graphic calc (and if i tried it without I'd probably mess up and this is for marks)


if you can help me with any of these questions it would be greatly apperciated. thanks a ton guys.

 

[Denis]

1. Solve: 6/(x(3-2x)) - 4/(3 - 2x) = 1

That's odd; equation is different here!
Let a = 3-2x
6/ax - 4/a - 1 = 0
6 - 4x - ax = 0 : multiply by ax
6 - 4x - x(3-2x) = 0 : substitute back
6 - 4x - 3x + 2x^2 = 0
2x^2 - 7x + 6 = 0
(2x - 3)(x - 2) = 0
x = 3/2 or x = 2

2. Solve: s = vt + (at^2)/2 for t

2s = 2vt + at^2 : multiply by 2
at^2 + 2vt - 2s = 0
t = [-2v +- sqrt(4v^2 - 100a)] / (2a)
t = [-2v +- 2sqrt(v^2 - 25a)] / (2a)
t = [-v +- sqrt(v^2 - 25a)] / a : 2's cancel out