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Maths C3 - Cancelling down Algebraic Fractions...Help with Q?

So I'm stuck on Exercise 1A Question q from the Edexcel C3 Modular textbook.

\frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}

(edited 7 years ago)

Reply 1

rayquaza17

Original post by Philip-flop

So I'm stuck on Exercise 1A Question q from the Edexcel C3 Modular textbook.

Could you please post the question? Most of us don't have access to the textbook. :smile:

Edit: I see you have done that now. Just to check, on the bottom is it

\frac{1}{2x}

(edited 7 years ago)

Reply 2

Philip-flop

Original post by rayquaza17

Could you please post the question? Most of us don't have access to the textbook. :smile:

Edit: I see you have done that now. Just to check, on the bottom is it

\frac{1}{2x}

I'm not quite sure how to type this properly. But the numbers in the denominator of the main fraction are fractions themselves e.g. 1x/2 is a half, 1/3 is a third...

\frac{3x^2-x-2}{1x/2+1/3}

Edit:

\frac{3x^2-x-2}{\frac{1x}{2}+\frac{1}{3}}

(edited 7 years ago)

Reply 3

NotNotBatman

\frac{3x^2 - x - 2}{\frac{1}{2}x +\frac{1}{3}}

First you want to get rid of the 2 and 3, so that it's a normal looking fraction, so multiply top and bottom by

2\times 3

, put the bottom level of the fraction over a common denominator first if that makes it easier.

Reply 4

Philip-flop

Ok so I managed to write the question out properly to Exercise 1A Question q from the C3 Modular textbook..

\frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}

(edited 7 years ago)

Reply 5

Philip-flop

Original post by NotNotBatman

\frac{3x^2 - x - 2}{\frac{1}{2}x +\frac{1}{3}}

First you want to get rid of the 2 and 3, so that it's a normal looking fraction, so multiply top and bottom by

2\times 3

, put the bottom level of the fraction over a common denominator first if that makes it easier.

Thank you. I will attempt to follow your steps now :smile:

Reply 6

Philip-flop

Original post by rayquaza17

Could you please post the question? Most of us don't have access to the textbook. :smile:

Edit: I see you have done that now. Just to check, on the bottom is it

\frac{1}{2x}

No I finally fixed my question. I meant it like this...

\frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}

Reply 7

B_9710

Original post by Philip-flop

No I finally fixed my question. I meant it like this...

\frac{3x^2-x-2}{\frac{1}{2}x+\frac{1}{3}}

Multiply top and bottom by 6 giving

\displaystyle \frac{6(3x^2-x-2)}{3x+2}

.
What can you do from here?

Reply 8

Philip-flop

I'm still stuck...

I ended up with...

\frac{(3x-2)(x+1)}{\frac{1}{6}(3x+2)}

Reply 9

Philip-flop

Original post by B_9710

Multiply top and bottom by 6 giving

\displaystyle \frac{6(3x^2-x-2)}{3x+2}

.
What can you do from here?

Oh yeah, I think I'm following now. How did you know whether to multiply top and bottom by 6? I'm terrible at LCM's :frown:

(edited 7 years ago)

Reply 10

B_9710

Original post by Philip-flop

Oh yeah, I think I'm following now. How did you know whether to multiply top and bottom by 6? I'm terrible at LCM's :frown:

Gets rid of the half and third in the denominator. Horrible with fractions in the denominator.

Reply 11

Philip-flop

I managed to get the answer now. Thank you everyone so much.

My next steps were..

\displaystyle \frac{6(3x^2-x-2)}{3x+2}

\frac{6(3x+2)(x-1)}{3x+2}

6(x-1)

Reply 12

Philip-flop

Yeah I realised that I factoried the numerator incorrectly that's when I managed to completely figure it out.

Thank you everyone that helped. I feel so silly for such a noob mistake!! :frown:

Reply 13

G.name

6X-6, I believe.