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Algebraic Fraction Help.

Hi, would somebody be able to demonstrate how these type of questions are done. I don't mean to be needy, but could you annotate so I can get a proper understanding.

Thanks!
:smile:
Simplify 6(x+5)22(x+5) \frac{6(x+5)^2}{2 (x+5)}

Simplify x29x2+3x \frac{x^2 - 9}{x^2 + 3x}
(edited 11 years ago)

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Reply 1
You look for common factors

With ordinary numbers you know that

1242=6×26×7=27\dfrac{12}{42} = \dfrac{6\times2}{6\times7} = \dfrac{2}{7}


You know that you can cancel the 6 x because it is in the numerator and the denominator



In the same way

(x+2)(x5)(x+2)(2x+1)\dfrac{(x+2)(x-5)}{(x+2)(2x+1)}

has (x+2) multiplying in both the numerator and denominator so it can be cancelled

(x+2)(x5)(x+2)(2x+1)=(x+2)×(x5)(x+2)×(2x+1)=x52x+1\dfrac{(x+2)(x-5)}{(x+2)(2x+1)} = \dfrac{(x+2)\times(x-5)}{(x+2)\times(2x+1)} = \dfrac{x-5}{2x+1}
First one, cancel out common factor of x + 5

Second one factorise the numerator(difference of two squares) and the denominator and then cancel out the common factor of x + 3
Reply 3
Original post by TenOfThem
You look for common factors

With ordinary numbers you know that

1242=6×26×7=27\dfrac{12}{42} = \dfrac{6\times2}{6\times7} = \dfrac{2}{7}


You know that you can cancel the 6 x because it is in the numerator and the denominator



In the same way

(x+2)(x5)(x+2)(2x+1)\dfrac{(x+2)(x-5)}{(x+2)(2x+1)}

has (x+2) multiplying in both the numerator and denominator so it can be cancelled

(x+2)(x5)(x+2)(2x+1)=(x+2)×(x5)(x+2)×(2x+1)=x52x+1\dfrac{(x+2)(x-5)}{(x+2)(2x+1)} = \dfrac{(x+2)\times(x-5)}{(x+2)\times(2x+1)} = \dfrac{x-5}{2x+1}


Original post by Mr M
First one, cancel out common factor of x + 5

Second one factorise the numerator(difference of two squares) and the denominator and then cancel out the common factor of x + 3


For the first one, does that become 6(x+5)2 \frac{6(x+5)}{2}
Reply 4
Original post by TheMysteryMan
For the first one, does that become 6(x+5)2 \frac{6(x+5)}{2}


Apart from the fact that 2 goes into 6, yes
Original post by TheMysteryMan
For the first one, does that become 6(x+5)2 \frac{6(x+5)}{2}

Yes, but it can be made even simpler. Expand the brackets and try to simplify further.
Reply 6
Original post by CharlieBoardman
Yes, but it can be made even simpler. Expand the brackets and try to simplify further.

6(x+5)2=6x+302 \frac{6(x+5)}{2} = \frac{6x+30}{2}
3x+15 ?
Reply 7
Remember to exclude values of x that make the denominator zero - from the domain:
first one: x =/= -5
second one: x =/= 0, -3 in your final answer!
Original post by TheMysteryMan
6(x+5)2=6x+302 \frac{6(x+5)}{2} = \frac{6x+30}{2}
3x+15 ?

Correct :smile:
Original post by TheMysteryMan
6(x+5)2=6x+302 \frac{6(x+5)}{2} = \frac{6x+30}{2}
3x+15 ?


Yes but I would suggest dividing the 6 by 2 would be a better move.
Original post by tomctutor
Remember to exclude values of x that make the denominator zero - from the domain:
first one: x =/= -5
second one: x =/= 0, -3 in your final answer!


At this level, that is not necessary.
Original post by tomctutor
Remember to exclude values of x that make the denominator zero - from the domain:
first one: x =/= -5
second one: x =/= 0, -3 in your final answer!


You lost me :s-smilie:
For the 2nd one, am I factorising top and bottom first?
Original post by TheMysteryMan
You lost me :s-smilie:


don't worry ... he was answering a more complex question
Original post by TheMysteryMan
You lost me :s-smilie:

Don't worry, you do not need to know that yet.
Original post by TenOfThem
don't worry ... he was answering a more complex question



Original post by CharlieBoardman
Don't worry, you do not need to know that yet.

Ok, would you be able to explain the 2nd one a little more to me, not sure where to start. Am I factorising out the top and bottom?
Original post by TheMysteryMan
Ok, would you be able to explain the 2nd one a little more to me, not sure where to start. Am I factorising out the top and bottom?

Yes, but try to look for the 'difference of two squares' somewhere. Once the numerator and denominator have been factorised correctly, they should have a term in common which you will be able to cancel.
(edited 11 years ago)
I don't think i've been taught that, only the quadratic formula, factorising and graphs.
Original post by TheMysteryMan
I don't think i've been taught that, only the quadratic formula, factorising and graphs.


You've not been taught the difference of two squares?

x^2-4=x^2-2^2=(x+2)(x-2)

Hence difference of squares?
Is the answer -3?

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