Algebraic Fraction

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http://www.edexcel.com/migrationdocuments/QP%20Current%20GCSE/March%202011%20-%20MS/5MB2H_01_msc_20110413.pdf

Question 14

You need to understand the difference between

\dfrac{x-5}{x-7}

and x-5/x-7.

They both are not the same, writing it in LaTex makes it clear, but if you don't write in LaTex then you should write it as (x-5)/(x-7).

Reply 101

zed963

Original post by raheem94

You need to understand the difference between

\dfrac{x-5}{x-7}

and x-5/x-7.

They both are not the same, writing it in LaTex makes it clear, but if you don't write in LaTex then you should write it as (x-5)/(x-7).

Okay

Reply 102

TenOfThem

Original post by zed963

http://www.edexcel.com/migrationdocuments/QP%20Current%20GCSE/March%202011%20-%20MS/5MB2H_01_msc_20110413.pdf

Question 14

jeez

Fine do not learn brackets learn latex instead

\dfrac{x+5}{x-7} = (x+5)/(x-7) \not= x+5/x-7

x-5/x-7 = x-\dfrac{5}{x}-7

but "whatever"

Reply 103

steve2005

Original post by zed963

http://www.edexcel.com/migrationdocuments/QP Current GCSE/March 2011 - MS/5MB2H_01_msc_20110413.pdf

Question 14

Brackets are not needed because there is a LONG fraction line.

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Reply 104

Mr M

Study Forum Helper

Original post by TenOfThem

x-5/x-7 = x-\dfrac{5}{x}-7

Or possibly even

x-\frac{5}{x-7}

Certainly ambiguous.

Reply 105

zed963

Oh right Now I see what you guys all mean. It does have a huge impact on how you understand the question. From now on I will use the brackets. Lesson Learned.

Reply 106

raheem94

Original post by zed963

Okay

You still didn't answer my one question, you are in which year?

Reply 107

steve2005

Original post by Mr M

Or possibly even

x-\frac{5}{x-7}

Certainly ambiguous.

I would not say this was ambiguous. I would say it was wrong.

Reply 108

TenOfThem

Original post by raheem94

You still didn't answer my one question, you are in which year?

He is in Y10, taking GCSE, hoping to get an A*

Reply 109

zed963

Original post by TenOfThem

He is in Y10, taking GCSE, hoping to get an A*

Thank you for answering.

Reply 110

raheem94

Original post by zed963

Thank you for answering.

I think you need to start working hard on your basics. You tend to be trying to solve some complex questions, while not paying attention on the basics.

Reply 111

Instinct01

Hello

May I just ask... how do you simplify the algebraic fraction in the first post? I tried to cross multiply, but that failed miserably. Then, I realised you only include the (2x-3) once in the denominator because it's shown twice. Even then, how do you figure out what to include in the numerator?

Thanks.

Reply 112

raheem94

Original post by Instinct01

\displaystyle \frac{2x}{2x-3} - \frac7{x(2x-3)} = \frac{2x \times x}{x(2x-3)} - \frac7{x(2x-3)}

Reply 113

Notnek

Original post by Instinct01

To add or subtract fractions, they must first have a common denominator.

So the first step is to make the first fraction have a denominator of x(2x-3) which can be done by multiplying top and bottom of the fraction by x.

So you get:

\displaystyle \frac{2x^2}{x(2x-3)} - \frac{7}{x(2x-3)}

Can you finish it off?

(edited 11 years ago)

Reply 114

Instinct01

Thanks for the replies.

Oh dear. I spent a long time studying algebraic fractions and still can't do this. Ermm, don't you just multiply the denominator together?

(2x^2)-7/(2x^2-3)(2x^2-3) But if I expand that, it definitely doesn't look right.

Reply 115

Mr M

Study Forum Helper

Original post by Instinct01

You have made a mistake with the question unless the last (2x^2-3) is on the top of the fraction - ambiguous notation yet again.

Reply 116

TenOfThem

Original post by Instinct01

\dfrac{3}{5} + \dfrac{1}{10} = \dfrac{6}{10} + \dfrac{1}{10} = \dfrac{7}{10}

Reply 117

Notnek

Original post by Instinct01

Ermm, don't you just multiply the denominator together?

Why do you think that? You should only do something in algebra if you understand why you're doing it. Do you understand how to get to the fractions given in mine and Raheem's posts?

They have a common denominator and if you think back to elementary fractions e.g.

\displaystyle \frac{4}{7} + \frac{2}{7}

If fractions have a common denominator then you can add the numerators together and put the sum over the common denominator so you'd get:

\displaystyle \frac{4}{7} + \frac{2}{7} = \frac{6}{7}

You do exactly the same with algebraic fractions.

Reply 118

Instinct01

Oh my bad, that makes sense now. What a stupid mistake. Usually, I can do algebraic fractions fine if there are different denominators which you just multiply together. But here, as there was more than one term within the denominator the same, I got confused. I don't understand how you decide what goes on the numerator, though. Could someone explain please?

Edit: Actually, I think I understand now. Can I have an example to do instead please?

(edited 11 years ago)

Reply 119

Notnek

Original post by Instinct01

Oh my bad, that makes sense now. What a stupid mistake. Usually, I can do algebraic fractions fine if there are different denominators which you just multiply together.

Could you give an example of a question where you could follow this method i.e. multiplying the denominators?

I'd just like to check your understanding.