GCSE Math

ScaredofMath

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Good morning,

I have to sit a university entrance exam in three weeks. It consists of three parts: two written, one math.

It's been 12 years since I did any kind of mathematics from school and I never got any grades while there (English and Classics were my strong subjects).

I've got some of the past papers and started using BBC Bitesize GCSE revision online to prepare (as that's apparently the level I'm expected to be capable of) but I don't think I'll be able to learn enough to pass the math section with the time I have.

The university mentions needing an understanding of algebra and calculus. Going through the old exams I started to practice and attempt to learn some of the material.

Below is one of the questions and I am completely lost on how to even begin.

It states clearly that Full Algebraic working must be clearly shown and the problem below is worth 5 marks.

Screenshot_20200211-124748__01.jpg

Am I correct in assuming this is an expression not an equation?

Do I follow the Bodmas rule (Brackets, order, division, multiplication, addition, subtraction) to simplify this?

This was as far as I got trying to solve it on my own:

2x2-1 - (I don't know where to start or what to do next or if I got that part correct).

Any advice is welcomed, even if it's a link to a guide that'll help me learn how to solve such problems on my own (I'm working through the BBC Bitesize GCSE online learning but I really struggled with the algebra section when trying to apply it to a problem with two parts separated by a subtraction).

Thanks in advance.
 

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Subhotosh Khan

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Good morning,

I have to sit a university entrance exam in three weeks. It consists of three parts: two written, one math.

It's been 12 years since I did any kind of mathematics from school and I never got any grades while there (English and Classics were my strong subjects).

I've got some of the past papers and started using BBC Bitesize GCSE revision online to prepare (as that's apparently the level I'm expected to be capable of) but I don't think I'll be able to learn enough to pass the math section with the time I have.

The university mentions needing an understanding of algebra and calculus. Going through the old exams I started to practice and attempt to learn some of the material.

Below is one of the questions and I am completely lost on how to even begin.

It states clearly that Full Algebraic working must be clearly shown and the problem below is worth 5 marks.

View attachment 16585

Am I correct in assuming this is an expression not an equation?

Do I follow the Bodmas rule (Brackets, order, division, multiplication, addition, subtraction) to simplify this?

This was as far as I got trying to solve it on my own:

2x2-1 - (I don't know where to start or what to do next or if I got that part correct).

Any advice is welcomed, even if it's a link to a guide that'll help me learn how to solve such problems on my own (I'm working through the BBC Bitesize GCSE online learning but I really struggled with the algebra section when trying to apply it to a problem with two parts separated by a subtraction).

Thanks in advance.
Yes .... you have been asked to simplify an "expression"

Yes ... you'll follow BODMAS (in US that acronym is mostly referred to as PEMDAS) - where applicable.

Without using calculator, can you simplify:

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

Leave your final answer in fraction form and show each step.

This excercize is not wasted time - you will basically follow the same steps for your posted problem.
 

lev888

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Jan 16, 2018
Messages
1,417
Good morning,

I have to sit a university entrance exam in three weeks. It consists of three parts: two written, one math.

It's been 12 years since I did any kind of mathematics from school and I never got any grades while there (English and Classics were my strong subjects).

I've got some of the past papers and started using BBC Bitesize GCSE revision online to prepare (as that's apparently the level I'm expected to be capable of) but I don't think I'll be able to learn enough to pass the math section with the time I have.

The university mentions needing an understanding of algebra and calculus. Going through the old exams I started to practice and attempt to learn some of the material.

Below is one of the questions and I am completely lost on how to even begin.

It states clearly that Full Algebraic working must be clearly shown and the problem below is worth 5 marks.

View attachment 16585

Am I correct in assuming this is an expression not an equation?

Do I follow the Bodmas rule (Brackets, order, division, multiplication, addition, subtraction) to simplify this?

This was as far as I got trying to solve it on my own:

2x2-1 - (I don't know where to start or what to do next or if I got that part correct).

Any advice is welcomed, even if it's a link to a guide that'll help me learn how to solve such problems on my own (I'm working through the BBC Bitesize GCSE online learning but I really struggled with the algebra section when trying to apply it to a problem with two parts separated by a subtraction).

Thanks in advance.
This is not an equation - no equal sign.
Here's an example that looks similar:
 

ScaredofMath

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Feb 10, 2020
Messages
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Yes .... you have been asked to simplify an "expression"

Yes ... you'll follow BODMAS (in US that acronym is mostly referred to as PEMDAS) - where applicable.

Without using calculator, can you simplify:

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

Leave your final answer in fraction form and show each step.

This excercize is not wasted time - you will basically follow the same steps for your posted problem.
Thanks for the reply; I needed to look up how to do this but applied the "rules" I found using pen and paper.

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

\(\displaystyle \frac{21*3} {21*7} - \frac{7*2}{7*21}\)

\(\displaystyle \frac{63} {147} - \frac{14}{147}\)

\(\displaystyle \frac{49} {147}\)

\(\displaystyle \frac{1} {3}\)

Would I use the same method to solve the expression I posted?
 
Last edited:

MarkFL

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The product of two denominators will be a common denominator, but if the two denominators share any common factors, it won't be the lowest common denominator (LCD). Getting the LCD makes your work easier, as you will work with simpler numbers/expressions. This is how I would simplify the arithmetic expression:

\(\displaystyle \frac{3}{7}-\frac{2}{21}=\frac{3}{7}-\frac{2}{3\cdot7}=\frac{3}{7}\cdot\frac{3}{3}-\frac{2}{3\cdot7}=\frac{9-2}{21}=\frac{7}{21}=\frac{1}{3}\)

The algebraic expression you posted can be simplified in similar fashion. :)
 

lev888

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Jan 16, 2018
Messages
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Thanks for the reply; I needed to look up how to do this but applied the "rules" I found using pen and paper.

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

\(\displaystyle \frac{21*3} {21*7} - \frac{7*2}{7*21}\)

\(\displaystyle \frac{63} {147} - \frac{14}{147}\)

\(\displaystyle \frac{49} {147}\)

\(\displaystyle \frac{1} {3}\)

Would I use the same method to solve the expression I posted?
Do you see a better way to get a common denominator? It would result in fewer calculations.
 

HallsofIvy

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Jan 27, 2012
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Yes, to add or subtract two fractions, whether "numerical" or "algebraic", you have to get a "common denominator". Since you change the denominator of a fraction, without changing the value, by multiplying both numerator and denominator by the same thing, the first problem is to determine what factors are in one denominator that are not in the other.

The denominator of the first fraction is x-1 and the denominator of the other fraction is \(\displaystyle x^2+ 2x- 3\). What we are hoping is that x- 1 is a factor of \(\displaystyle x^2+ 2x- 3\). If that is the case, then we must have \(\displaystyle (x- 1)(ax+ b)= ax^2+ (b- a)x- b= x^3+ 2x- 3\). Comparing coefficients, we must have a= 1 and b= 3. Is b- a= 3- 1 equal to 2? Yes, it is! So \(\displaystyle \frac{2x}{x-1}+ \frac{7x+1}{x^2+ 2x- 3}= \frac{2x}{x-1}+ \frac{7x+1}{(x-1)(x+ 3)}= \frac{2x(x+3)}{(x-1)(x+3)}- \frac{7x+ 1}{(x-1)(x+3)}= \frac{2x^2+ 6x- 7x- 1}{(x-1)(x+3)}= \frac{2x^2- x- 1}{x^2+ 2x- 3}\).

Notice that the numerator and denominator have the same degree so this is a "improper fraction". In fact, \(\displaystyle \frac{2x}{x-1}\) is an improper fraction and is equal to \(\displaystyle 2+ \frac{2}{x-1}\).
 

Subhotosh Khan

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Thanks for the reply; I needed to look up how to do this but applied the "rules" I found using pen and paper.

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

\(\displaystyle \frac{21*3} {21*7} - \frac{7*2}{7*21}\)

\(\displaystyle \frac{63} {147} - \frac{14}{147}\)

\(\displaystyle \frac{49} {147}\)

\(\displaystyle \frac{1} {3}\)

Would I use the same method to solve the expression I posted?
Excellent work!

Yes - you will use the same process (with slight modification)

For your OP, I would look at the denominators of those given fractions with "polynomials". We will need to decide - whether any of those can be factorized.

We see that we have (x -1) as the first numerator. This cannot be factorized any further.

We look at x2 + 2x - 3 ...... this could be factorized further. Factorize it - following procedure shown in response #7 (or any other suitable method for quadratic function).

Note the slight change below:

\(\displaystyle \frac{3}{7} - \frac{2}{21}\)

\(\displaystyle \frac{3}{7} - \frac{2}{7*3}\)

\(\displaystyle \frac{3 *3} {7*3} - \frac{2}{21}\)

\(\displaystyle \frac{9 - 2} {21}\)

\(\displaystyle \frac{7} {21}\)

\(\displaystyle \frac{1}{3}\)

In algebra, the step of factorization becomes a huge time-saver. If you do not factorize, you will get the same answer - but you will have more tedious work to do to get the same answer.
 

ScaredofMath

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I think I'm slowly beginning to understand bits of this.

I need to simplify the expression but x-1, 2x, and 7x-1 can't be factorised any further?

\(\displaystyle x^2+ 2x- 3\) can be factorised by turning/expanding it into a multiplication? This looks like (x - 1) (x + 3) because x*x = x2 -1 * 3 = -3 and -1 + 3 = 2.

To get the lowest common denominator we need to use the factorised denominator from one side and multiply the other denominator and the numerator as well.

Once we have (x - 1) (x + 3) do we only use (x + 3) to find the lowest common denominator? What about the (x - 1)? Does that cancel out so we have we \(\displaystyle \frac{2x(x+3)}{(x-1)(x+3)}\) instead of \(\displaystyle \frac{2x(x+3)}{(x-1)(x-1)(x+3)}\)

From there the equation can be compiled? Which is how you end up with

\(\displaystyle \frac{2x^2+ 6x- 7x- 1}{(x-1)(x+3)}\)

Can you leave the expression above as it is our does it not count as fully simplified until it's

\(\displaystyle \frac{2x^2- x- 1}{x^2+ 2x- 3}\).

I just wanted to say thanks to all of you for the help and links to videos. They've been really helpful.

Would it be possible if one of you could please set a similar problem to this one so that I can attempt to solve it?

Thanks again.
 

Otis

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… \(\displaystyle \frac{2x^2+ 6x- 7x- 1}{(x-1)(x+3)}\)

Can you leave the expression above as it is …
Hello SoM. No, you ought to combine the like-terms in the numerator (6x and -7x).

Unless you've been instructed otherwise, multiplying out the factors in the denominator is somewhat optional. Some teachers care; others don't.

… Would it be possible if one of you could please set a similar problem to this one so that I can attempt to solve it?
\[\frac{5x + 9}{x^{2} - 2x + 1} - \frac{x + 15}{x - 1}\]

Also, if you google keywords practice rational expressions, you'll find practice worksheets like this one (scroll to the bottom of the page). Or, click the 'Images' tab on google's search-results page, and you'll be taken to images of various worksheets.

Cheers 😎
 

ScaredofMath

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\[\frac{5x + 9}{x^{2} - 2x + 1} - \frac{x + 15}{x - 1}\]

Also, if you google keywords practice rational expressions, you'll find practice worksheets like this one (scroll to the bottom of the page). Or, click the 'Images' tab on google's search-results page, and you'll be taken to images of various worksheets.

Cheers 😎
Thanks for the problem and the link to the practice expressions!

Here's what I've got so far.

\[\frac{5x + 9}{x^{2} - 2x + 1} - \frac{x + 15}{x - 1}\]

\[\frac{5x + 9}{(x - 1) (x - 1)} - \frac{x + 15}{x - 1}\]

\[\frac{5x + 9 (x - 1)}{(x - 1) (x - 1)} - \frac{x + 15 (x - 1)}{(x - 1) (x - 1)}\]

\[\frac{5x + 9 (x - 1)}{(x - 1)^2} - \frac{x + 15 (x - 1)}{(x - 1)^2}\]

\[\frac{5x^2 + 9x - 5x - 9 - x^2 + 15x - x - 15}{(x - 1)^2}\]

\[\frac{4x^2 + 18x - 24}{(x - 1)^2}\]

Am I close? I think that's the answer but I'm not sure if I've to stop there or if I've to do something like:

\[\frac{2 (2x^2 + 9x - 12)}{(x - 1)^2}\]
 

Jomo

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The first and 2nd line are fine.
In the 3rd line you tried to multiply the left numerator by (x-1). Why did you do that?
Also 5x+9(x-1) = 5x +9x-9 = 14x -9 NOT 5x^2 + 9x - 5x - 9!!! Yes, you need to multiply the whole numerator by (x-1) (PROVIDED that you multiply the denominator by (x-1)) but you need to denote it correctly. You should have written (5x+9)(x-1). Same comment for x+15(x-1)

It seems to me that the only mistake you made was in the left term by multiply the numerator by (x-1) and NOT multiplying the denominator by (x-1).

You can always check your own work by pick some values for x and see if you get the same result for the original expression and your simplified expression.

Pick x=0: Then the original expression is 9 - (-15) = 24. The simplified expression yields -24/1=-24. Not the same so something is wrong.
 

HallsofIvy

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Thanks for the problem and the link to the practice expressions!

Here's what I've got so far.

\[\frac{5x + 9}{x^{2} - 2x + 1} - \frac{x + 15}{x - 1}\]

\[\frac{5x + 9}{(x - 1) (x - 1)} - \frac{x + 15}{x - 1}\]

\[\frac{5x + 9 (x - 1)}{(x - 1) (x - 1)} - \frac{x + 15 (x - 1)}{(x - 1) (x - 1)}\]

\[\frac{5x + 9 (x - 1)}{(x - 1)^2} - \frac{x + 15 (x - 1)}{(x - 1)^2}\]
Here's your error. You multiplied the numerator and denominator of the right side by x- 1 is the same as multiplying by 1 so does not change the value. But you also multiplied only the numerator on the r
right side by x- 1 which does change the value.
You multiply numerator and denominator of the right side in order to have "x- 1" in the numerator. There was already an "x- 1" in the denominator on the left so you don't need to multiply there.
This should be
\[\frac{5x + 9 }{(x - 1) (x - 1)} - \frac{(x + 15) (x - 1)}{(x - 1) (x - 1)}\]
The first and 2nd line are fine.
In the 3rd line you tried to multiply the left numerator by (x-1). Why did you do that?
Also 5x+9(x-1) = 5x +9x-9 = 14x -9 NOT 5x^2 + 9x - 5x - 9!!! Yes, you need to multiply the whole numerator by (x-1) (PROVIDED that you multiply the denominator by (x-1)) but you need to denote it correctly. You should have written (5x+9)(x-1). Same comment for x+15(x-1)

It seems to me that the only mistake you made was in the left term by multiply the numerator by (x-1) and NOT multiplying the denominator by (x-1).

You can always check your own work by pick some values for x and see if you get the same result for the original expression and your simplified expression.

Pick x=0: Then the original expression is 9 - (-15) = 24. The simplified expression yields -24/1=-24. Not the same so something is wrong.

You also should have parentheses around the "x+ 15" as I show.

\[\frac{5x^2 + 9x - 5x - 9 - x^2 + 15x - x - 15}{(x - 1)^2}\]

\[\frac{4x^2 + 18x - 24}{(x - 1)^2}\]

Am I close? I think that's the answer but I'm not sure if I've to stop there or if I've to do something like:

\[\frac{2 (2x^2 + 9x - 12)}{(x - 1)^2}\]
 

ScaredofMath

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This should be
\[\frac{5x + 9 }{(x - 1) (x - 1)} - \frac{(x + 15) (x - 1)}{(x - 1) (x - 1)}\]

You also should have parentheses around the "x+ 15" as I show.
Just to clear this part up. Since (x - 1) (x - 1) is already the denominator of (5x + 9) we don't need to do

\[\frac{5x + 9 }{(x - 1) (x - 1) (x - 1)}\]

Should there be an (x - 1) beside (5x + 9) on the left expression?

\[\frac{(5x + 9) (x - 1)}{(x - 1) (x - 1)}\]

I'm asking because I thought we would need to multiply numerator and denominator by (x - 1) and (5x + 9) (x - 1) would be 5x2 - 9? If this doesn't work can someone explain to me why? Is that what was mentioned earlier that multiplying by (x - 1) here is like multiplying by 1 on its own?

Should the numerator in that case end up as

\[\frac{5x - 9 }{(x - 1)^3} - \frac{x^2 - x + 15x - 15}{(x - 1)^3}\]

Or am I way off?

I thought that to find the LCD I need to multiply numerator and denominator of the left expression by whatever it is missing from the factorised denominator of the right expression? And then the same for the opposite side? Multiply the numerator and denominator of the right expression by whatever it is missing from the denominator of the left expression?

So should I end up with

\[\frac{(5x + 9) (x - 1) }{(x - 1) (x - 1)} - \frac{(x + 15) (x - 1)}{(x - 1) (x - 1)}\]

\[\frac{5x^2 - 5x + 9x - 9 }{(x - 1)^2} - \frac{x^2 - x + 15x - 15}{(x - 1)^2}\]

Or should I end up with

\[\frac{(5x + 9) (x - 1) }{(x - 1) (x - 1) (x - 1)} -

\frac{(x + 15) (x - 1)}{(x - 1) (x - 1) (x - 1)}\]

\[\frac{5x^2 - 5x + 9x - 9 }{(x - 1)^3} - \frac{x^2 - x + 15x - 15}{(x - 1)^3}\]

Unless I'm still mistaken about the numerators, in which case I'll redo those again.

Sorry for the obvious stupidity here; this kind of math is something I've never encountered before. You've all probably just explained it above and I'm not following properly.
 

MarkFL

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I'll walk you through combining the terms. We begin with:

\(\displaystyle \frac{5x+9}{x^2-2x+1}-\frac{x+15}{x-1}\)

Factor the first denominator:

\(\displaystyle \frac{5x+9}{(x-1)^2}-\frac{x+15}{x-1}\)

Now, we see that the second term needs to be multiplied by 1 in the form of \(\displaystyle \frac{x-1}{x-1}\) so that both terms have the same denominator:

\(\displaystyle \frac{5x+9}{(x-1)^2}-\frac{(x+15)(x-1)}{(x-1)^2}\)

And now we can combine terms as follows:

\(\displaystyle \frac{(5x+9)-(x+15)(x-1)}{(x-1)^2}\)

Expand the numerator:

\(\displaystyle \frac{5x+9-(x^2+14x-15)}{(x-1)^2}\)

Distribute the negative sign:

\(\displaystyle \frac{5x+9-x^2-14x+15}{(x-1)^2}\)

Combine like terms:

\(\displaystyle \frac{-x^2-9x+24}{(x-1)^2}\)

I'd factor a negative sign from the numerator and put it out front:

\(\displaystyle -\frac{x^2+9x-24}{(x-1)^2}\)
 

Marcus Clayson

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Dear Scared of Math

I am new to this forum, and don't really understand the protocols, so forgive me if I do something wrong here.

I am both concerned and bemused by your thread and the responses from the experts, all of which are great, and I really admire the fact that there are so many mathematicians out there who are willing to put their time and energy into helping others. However, I think that they are expecting too much too soon. With all due respect to the other respondents, you are never going to understand how to solve algebraic fractions unless you have first mastered adding and subtracting numerical fractions. The respondents seem to think that you can make the leap without this first step.

I don't know if you are English or American - you write "Scared of Math", which is an Americanism. I presume that you are what we in England call a "mature student" - you are going to university after some time out of formal education. I have spent years teaching maths to people who want to go to university, who have, like I was when I went to university as a mature student, been away from maths for years and who, like me, hated maths at school. (I studied Economics at undergraduate and post graduate level).

The problem that you were asked to solve is a high level one. I do not know which subject you wish to study, but this is what you would find on the old English AS Level courses. What I say to my students, ALL of whom struggle with the addition and subtraction of fractions, is first of all, type "Corbettmaths" into your search engine; go to "Videos and worksheets" section and look at the videos about adding fractions; then do the questions on the worksheets; then check the WORKED solutions and make sure that you understand the processes. See also "Hegartymaths" GCSE fractions videos on "You Tube". What I used as the basis for my courses was the material from MASH - https://www.sheffield.ac.uk/mash:
download the materials about fractions - there is plenty of explanatory material and worksheets; look also - LATER - at the material about algebraic fractions - WATCH THE VIDEOS. Although they are old and old - fashioned, I think that they are wonderfully simple and clear.

In my experience, the universities that I have dealt with - Wolverhampton and Birmingham, BCU and Aston - have dedicated maths departments for students who were not "good" at maths at school, or who have been away from maths for a long time. Make sure that you find any similar departments at your chosen university and use them/it.

You might also benefit from doing some work on basic algebra before trying to tackle algebraic fractions. "Corbettmaths", "Hegartymaths" and MASH have lots of excellent material to introduce you to the basics and then move you on. You said that Classics was one of your strong points. Remember that algebra is a foreign language like ancient Greek, and you have to learn the grammar, the syntax before the lexis. This might take some time. Some of my students never master finding the HCF and LCM of numerical fractions - it is difficult and it takes time, patience and perseverance. Many of my students are amazed by the end of the course about what they can do compared to what they could do before they started; some, like me, come to understand that mathematics is the key to the universe and everything in it. Just as getting through life is difficult, getting to grips with maths is difficult and there are no short cuts, but if you stick with it, you can and shall see the infinite horizon that is mathematics. What turned me on to it? Reading a paper by a Professor of Fluid Mechanics about how he invented those little sachets that you get vinegar and sauce in at service stations: maths solves the seemingly trivial in ways we can only begin to imagine, but which should inspire in us a sense of wonder. Good luck.
 
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