Suddenly terrible at basic algebra

partyatmygaff

Ok back in the J.C. I was all As e.t.c. at algebra and related chapters. The problem is since this year has started its as if I've never done algebra in my life, I get simple algebraic fractions wrong and completely mess up when surds e.t.c get involved. The problem is is that I have my xmas exams tomorrow :eek: and I have no idea whats going wrong, trigonometry and coordinate geometry is going fine but algebra even to the most basic form has went completely kaput.
So yeah.... How the hell would I do
1blank spc3
---- +

---

x+1 (x+1)(2x-1)

The answer I am getting is
3

---

2x-1 whilst the correct answer according to the book is


2

---

2x-1


Anyone?

jumpguy

partyatmygaff wrote: »

Ok back in the J.C. I was all As e.t.c. at algebra and related chapters. The problem is since this year has started its as if I've never done algebra in my life, I get simple algebraic fractions wrong and completely mess up when surds e.t.c get involved. The problem is is that I have my xmas exams tomorrow :eek: and I have no idea whats going wrong, trigonometry and coordinate geometry is going fine but algebra even to the most basic form has went completely kaput.
So yeah.... How the hell would I do
1blank spc3
---- +

---

x+1 (x+1)(2x-1)

The answer I am getting is
3

---

2x-1 whilst the correct answer according to the book is


2

---

2x-1


Anyone?

I hate algebra too, and I dread surds. I got better by relentless practice.

Okay, first of all, your first question. How is the question asked? If you have to "solve" it, you need to find x. If you need to "express as a single fraction" or "express in simplest terms" you need to factorise it down.

1blank spc3
---- +

---

x+1 (x+1)(2x-1)

For this, you get the LCM, which is (x+1)(2x-1). Just get everything at the bottom line. There's no need to have two (x+1)s.
You then divide your LCM by the denominator (not numerator) and multiply your answer by the numerator. So, you divide (x+1) by (x+1)(2x-1) and you get (2x-1) because the x+1 and x+1 cancel. Multiply this by the numerator then and you get 1(2x-1). Do this for the other term then (x+1)(2x-1) divided by (x+1)(2x-1) is 1. Multiply by the answer by the numerator (3) and you get 1.

1(2x-1) + 3

---

(x+1)(2x-1)

2x+2

---

(x+1)(2x-1)


2(x+1)

---

(x+1)(2x-1)

The (x+1) above the line cancels with the (x+1) below the line and you're left with

2

---

2x-1



Any questions on what I did there, feel free to ask.

DancingQueen:)

My maths mock is on Wednesday!! :O
But em yeah i got

---

(2x-1)

step by step

1(2x-1) + 3 2x-1+3

---

=

---

(x+1) (2x-1) 2x(squared) -x+2x-1

= 2x+2

---

= 2(x+1)
(2x-1)(2x+1)

---

(2x-1)(x+1)

and cancel the x + 1

sorry if its hard to understand,
Good luck in your tests!

**Timbuk2**

It might be easier to see if [noparse][latex][/latex][/noparse] tags are used.

[latex]\displaystyle \frac{1}{x+1} + \frac{3}{(x+1)(2x-1)}[/latex]

[latex]\displaystyle \frac{(1)(2x-1)}{(x+1)(2x-1)} + \frac{3}{(x+1)(2x-1)}[/latex]

[latex]\displaystyle = \frac{(2x-1) + 3}{(x+1)(2x-1)}[/latex]

[latex]\displaystyle = \frac{2x+2}{(x+1)(2x-1)}[/latex]

[latex]\displaystyle = \frac{2(x+1)}{(x+1)(2x-1)}[/latex]

[latex]\displaystyle = \frac{2}{2x-1}[/latex]

clartharlear

Algebra is like running. No matter how fast you ran last year, if you haven't done any training, you will run slowly now. However, some diligent training will bring you back up to speed!

rebel girl 15

clartharlear wrote: »

Algebra is like running. No matter how fast you ran last year, if you haven't done any training, you will run slowly now. However, some diligent training will bring you back up to speed!

Completely agree - you need to practise it, and if it doesn't work, don't panic!

For maths, its not all about getting the right answer in the exam, the process of getting there is way more important! best of luck in the mock!

DancingQueen:)

rebel girl 15 wrote: »

best of luck in the mock!

Thank you! I actually like algebra its trigonometry that's the hardest for me but i've been practising like mad. Fingers crossed

Gael23

why dont you multiply the 3 by x+1?

MathsManiac

ryanf1 wrote: »

why dont you multiply the 3 by x+1?

eh...because there's no particular reason why you should.

(Is there some reason why you would like to do so?)

clartharlear

MathsManiac wrote: »

eh...because there's no particular reason why you should.

(Is there some reason why you would like to do so?)

I presume he'd like to do so because 'cross multiplying' is taught as an automatic 'go-to' method when these kind of fractions come up. It would work, but it'd mean unneeded effort, which wouldn't be needed with a slightly more thoughtful look at the initial problem.

If your post was meant as sarcasm, it might be a little mean, MM, if he meant it as a genuine question...

Gael23

You cross multiply the 1 by 2x-1 so why not do the same with the 3?
clartharlear thats exactly what I would do.

jumpguy

I'd be weary of "cross multiplying" (unless it's an equation). It'd be awkward if there was a minus instead of a plus.

rebel girl 15

jumpguy wrote: »

I'd be weary of "cross multiplying" (unless it's an equation). It'd be awkward if there was a minus instead of a plus.

Can't cross multiply because it is not an equation. remember doing this in college - problem because you don't know if x is positive or negative, which can cause two very different answers. Common Denominator is the way to go as above

you are in essence multiplying the first fraction by 1, to get the same denominator as the second fraction, which is already over the common denominator!

Making It Bad

If you "cross-multiply" something which is not an equation you kill maths. You might as well just start dividing by zero when you're at it.

clartharlear

If you post a smart comment without quite getting the point, God kills a kitten.

The technique of getting a common denominator of two unrelated fractions can be called cross-multiplying.

Even in the smart way that jumpguy did it, to get the LCM, you multiply top and bottom of the first fraction by (2x-1). There's no guarantee that (2x-1)>0

Making It Bad

clartharlear wrote: »

If you post a smart comment without quite getting the point, God kills a kitten.

The technique of getting a common denominator of two unrelated fractions can be called cross-multiplying.

Even in the smart way that jumpguy did it, to get the LCM, you multiply top and bottom of the first fraction by (2x-1). There's no guarantee that (2x-1)>0

Fortunately for me then that I'm an Atheist.

I don't know how you were thought but for me and I'm assuming most people "cross-multiplying" means literally cross-multiplying so if you had this

[latex]\displaystyle \frac{1}{x+1} = \frac{3}{(x+1)(2x-1)}[/latex]

you would multiply the [latex]\frac{1}{x+1}[/latex] by [latex]{(x+1)(2x-1)}[/latex] and you would multiply the [latex]\frac{3}{(x+1)(2x-1)}[/latex] by [latex]{x+1}[/latex] to "kill" the fraction.

The way which 'ryanf1' suggests that you multiply each fraction by a number but then neglect to mutliply denominator by that same number thus changing the fraction.

Getting a common denominator is not cross multiplying and calling it so makes no sense.

KnifeWRENCH

ryanf1 wrote: »

You cross multiply the 1 by 2x-1 so why not do the same with the 3?
clartharlear thats exactly what I would do.

You only cross multiply if you have an equation. Here, you're not trying to find the value of x, just trying to add two fractions and writing the solution in it's simplest form.

You mutiply the first term above and below by (2x-1) because you want both terms to have the same denominator, so that you can then add them without needing to do anything extra to the second term.

jumpguy

You're right, you only cross-multiply if you have an equation. I've heard ways of doing factorization questions and my classmates (mistakenly) call is cross-multiplication. I assume it's just an more rote-learned method of the LCM. I prefer the LCM way.

I wonder how partyatmygaff got on in his exam in the end...

MathsManiac

clartharlear wrote: »

If your post was meant as sarcasm, it might be a little mean, MM, if he meant it as a genuine question...

Not intended to be sarcastic at all. Since the denominator of the second fraction was left entirely unchanged, I couldn't see why there would be any reason to multiply the numerator by anything.

Subsequent posts make it clear that there was some confusion caused by this "cross-multiplication" stuff, which some people apparently use for solving equations.

Apologies to ryanf1 if I caused offence.

Gael23

Making It Bad wrote: »

Fortunately for me then that I'm an Atheist.

I don't know how you were thought but for me and I'm assuming most people "cross-multiplying" means literally cross-multiplying so if you had this

[latex]\displaystyle \frac{1}{x+1} = \frac{3}{(x+1)(2x-1)}[/latex]

you would multiply the [latex]\frac{1}{x+1}[/latex] by [latex]{(x+1)(2x-1)}[/latex] and you would multiply the [latex]\frac{3}{(x+1)(2x-1)}[/latex] by [latex]{x+1}[/latex] to "kill" the fraction.

The way which 'ryanf1' suggests that you multiply each fraction by a number but then neglect to mutliply denominator by that same number thus changing the fraction.

Getting a common denominator is not cross multiplying and calling it so makes no sense.

I should have explained better, I meant to multiply the bottom by the top to cancel the fraction.

clartharlear

MathsManiac wrote: »

Not intended to be sarcastic at all. Since the denominator of the second fraction was left entirely unchanged, I couldn't see why there would be any reason to multiply the numerator by anything.

Subsequent posts make it clear that there was some confusion caused by this "cross-multiplication" stuff, which some people apparently use for solving equations.

Apologies to ryanf1 if I caused offence.

That's really nice of you. Tempers can often be too easily roused on the internet, so it's lovely to see a poster as polite as you.

Two fractions to be added:
a/b + c/d

You multiply together the two denominators.
You multiply the numerator a of the first by denominator d of the second.
You multiply the numerator c of the second by denominator b of the second.
So you get (ad + bc)/bd

The two steps in italics can be described as cross-multiplying because you're multiplying across the numerator/denominator divide and across the addition symbol.

"Cross-multiplying" isn't like addition or integration - it's not a formal mathematical operation. It's a verbal description of what you're doing.

**Timbuk2**

Cross-multiplying isn't really possible if it's not an equation. People often do the cross-multiply thing, and then multiply the two denominators, but that's not really cross-multiplying, that's just allowing the fractions to be added. It can be confusing for people to grasp this, as when there is an equals sign in the middle you don't have to multiply the denominators.

For example
[latex]\displaystyle \frac{a}{b} + \frac{c}{d}[/latex]
This would be the easiest way to approach it, rather than that incorrect cross-multiplying (it's not an equation!)
[latex]\displaystyle (\frac{a}{b}\times\frac{d}{d}) + (\frac{c}{d}\times\frac{b}{b})[/latex]
I'm only showing it in steps like this to make it clear. As you are multiplying by d/d or b/b, you are not changing the value of the fraction. If you were to cross-multiply, you would be (providing you didn't multiply out the denominators - a step that people often forget if they try to cross-multiply)
[latex]\displaystyle \frac{ad + cb}{bd}[/latex]

dustintheturkey

Algebra is very simple.

how do you add 1/4 to 1/2 ?
you add 1/4 to 2/4 because you are adding quarters.
one quarter + two quarters is three quarters just like adding apples.

But how to add 1/2 plus 1/3 ?

This is what your teachers don't tell you
MULTIPLY THE TWO FRACTIONS BY 1
sound stupid? no more than Pat Kenny.

multiply 1/2 by 3/3 and 1/3 by 2/2 so you are adding sixths, just like apples again.
When you have x's, just imagine it's a number (which it is, but what we dont know yet)
and make the denominators the same.

So here's what you do....

multiply 1/(x+1) by (2x-1)/(2x-1)

Now you are combining 1/{(x+1)(2x-1)}'s

{(2x-1)/(2x-1)}{1/(x+1)} +3/{(2x-1)(x+1)}

Basically, you can think of it as.... when the denominators are the same,
just combine (add or subtract) the numerators.

So the answer is, if you write that down {2x-1+3}/{(2x-1)(x+1)}
which is (2x+2)/{(2x-1)(x+1)}, equals 2(x+1)/{(2x-1)(x+1)}

and (x+1)/(x+1) = 1 so the answer is 2/(2x-1)

What you are taught usually is cross-multiplication or whatever,
a dumbed-down version of straightforward thinking.

Us turkeys had no schooling, so we're still smart.