How do I simplify these polynomial fractions?

• Mar 6th 2010, 06:54 PM
quikwerk
How do I simplify these polynomial fractions?
I think I might have gotten the answer to these but I don't know....

(9x^2-3x)/3x

3x-1
^^^^^^^^^^^

(x^2-5x+4/x-1

x-4
^^^^^^^^^^^^^^^^

(x-y)/(x+y)+((x+y)/(x-y)

(x-y)+(x+y)

^^^^^^^^^^^^^^^^

One last thing I'm unsure of is a question that asks me to solve an equation. I have the answer, I used logic to find it and now I can't figure out how to show working for it.

1/x-1/(4-x)=1/3

x=2
• Mar 6th 2010, 07:10 PM
Gusbob
Quote:

Originally Posted by quikwerk
I think I might have gotten the answer to these but I don't know....

(9x^2-3x)/3x

3x-1
^^^^^^^^^^^ Correct

(x^2-5x+4/x-1

x-4
^^^^^^^^^^^^^^^^ Correct

(x-y)/(x+y)+((x+y)/(x-y)

(x-y)+(x+y)

^^^^^^^^^^^^^^^^ No. It's $\displaystyle \frac{2(x^2 + y^2)}{x^2 - y^2}$

One last thing I'm unsure of is a question that asks me to solve an equation. I have the answer, I used logic to find it and now I can't figure out how to show working for it.

1/x-1/(4-x)=1/3

x=2

Try this:
$\displaystyle \frac{1}{x} \times \frac{4-x}{4-x} - \frac{1}{4-x} \times \frac{x}{x} = \frac{1}{3}$

$\displaystyle \frac{4-2x}{4x - x^2} = \frac{1}{3}$

$\displaystyle 12 - 6x = 4x - x^2$

$\displaystyle x^2 - 10x + 12 = 0$

Solve this using the quadratic equation to get $\displaystyle 5 \pm \sqrt{13}$

.
• Mar 6th 2010, 07:41 PM
quikwerk
Thanks for your help, but I think I typed out that last question badly it was supposed to read: 1/x-1/x+4 NOT "1/x-1/(4-x)=1/3".....
(Surprised)(Worried)
• Mar 6th 2010, 08:06 PM
Gusbob
Quote:

Originally Posted by quikwerk
Thanks for your help, but I think I typed out that last question badly it was supposed to read: 1/x-1/x+4 NOT "1/x-1/(4-x)=1/3".....
(Surprised)(Worried)

Same idea,

$\displaystyle \left( \frac{1}{x} \times \frac{x+4}{x+4} \right)+ \left( \frac{1}{x+4} \times \frac{x}{x} \right) = \frac{1}{3}$

$\displaystyle \frac{x+4 - x}{x(x+4)} = \frac{1}{3}$

$\displaystyle 12 = x^2 + 4x$

$\displaystyle x^2 + 4x - 12 = 0$

$\displaystyle (x+6)(x-2) = 0$

So solutions are x = -6 and x = 2