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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

C3 Composite Functions

Hey, I'm struggling to simplify a composite function. This is for question 3 part a.
So far I have h(x)= (6/x+3) -2. How do I simplify this to give the answer -2x/x+3 (which is given to me)
Thanks :smile:

:smile:

Original post by mrmigme

Hey, I'm struggling to simplify a composite function. This is for question 3 part a.
So far I have h(x)= (6/x+3) -2. How do I simplify this to give the answer -2x/x+3 (which is given to me)
Thanks :smile:

:smile:

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

OP

Original post by EricPiphany

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

I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)

Original post by mrmigme

I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)

Clearly,

2 = \dfrac{2(x+3)}{x+3}, x \not= -3

.

Original post by mrmigme

I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)

You have to multiply top and bottom of fraction by the same number, which is essentially multiplying by 1.

Original post by mrmigme

I think I'm misunderstanding the basics of this topic lol. Why is it 2(x+3)/x+3 and not 2(6)/(x+3)

Do you remember how to add fractions? Like in GCSE? You'd need to make common denominators?

So to add

\displaystyle \frac{1}{2} + \frac{1}{4} = \frac{1 \times 2}{2 \times 2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{2+1}{4}= \frac{3}{4}

, right?

Here it's the same thing:

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

OP

Original post by Zacken

Do you remember how to add fractions? Like in GCSE? You'd need to make common denominators?

So to add

\displaystyle \frac{1}{2} + \frac{1}{4} = \frac{1 \times 2}{2 \times 2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{2+1}{4}= \frac{3}{4}

, right?

Here it's the same thing:

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

Bloody hell I can't believe I got caught up on how to do that. Thanks a lot!

Original post by mrmigme

Bloody hell I can't believe I got caught up on how to do that. Thanks a lot!

You're very welcome!

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