How do I simplify these polynomial fractions?

**quikwerk** · March 6th 2010, 06:54 PM

I think I might have gotten the answer to these but I don't know....

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

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

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

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

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

answer
(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

**Gusbob** · March 6th 2010, 07:10 PM

Originally Posted by quikwerk

I think I might have gotten the answer to these but I don't know....

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

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

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

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

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

answer
(x-y)+(x+y)

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

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

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

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

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

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

.

**quikwerk** · March 6th 2010, 07:41 PM

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".....
My bad, I'm really sorry.

**Gusbob** · March 6th 2010, 08:06 PM

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".....
My bad, I'm really sorry.

Same idea,

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

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

$12 = x^2 + 4x$

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

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

So solutions are x = -6 and x = 2

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