Lesson Alternative Way of Clearing Fractions with Simple Denominators

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This Lesson (Alternative Way of Clearing Fractions with Simple Denominators) was created by by boilpoil(127)

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About boilpoil: I'm a student that likes maths... I can teach some basic operations though

Hello everyone, this is my first lesson, I hope you will find this helpful :)

So, first of all. Let me explain what is this.

This will explain an alternate way of clearing annoying fractions out of your way, by just simply multiplying! After reading this, it should be more comfortable for you to solve something like {{{(1/3)((1/4)(7y-5)-3+3/4)=3/2-(y+4)/(4)}}} or {{{(3/2)((2/3)(x/4-1)-2)-x=2}}}.

I will now explain the most important contents you must know: Terms, LCM and denominator
1) Terms: For simple understanding, I would say whenever you see a '+' or '-', anything that follows it, along with that symbol, is a term. Remember, multiplication and division are NOT terms! If you messed this up, this method will not work for you, also even if the fraction contains + and -, the fraction itself along with the symbol following it still counts as 1 term only.
1.1) E.g. of terms: {{{1}}} is a term, {{{3x}}} is a term, {{{8xyz}}} is a term, {{{9x(3/7)}}} is a term, {{{(28-x)((x-3)^(x+1)/9)}}} is a term. etc.
2) LCM: I think you know this very well, Least Common Multiple, which is a number that is a multiple for 2 or several numbers and is already as small as it can.
2.1) E.g. of LCM: LCM of 5 and 6 is 30, LCM for 8 and 16 is 16, LCM for 10, 15 and 20 is 60. etc.
3) Denominator: I hope you still remember that the number below the fraction line is denominator.

And, I will start explaining how to use this method. P.S. I hope you read the title, "with simple denominators", if you are having some denominators like x+5+2y or such you may get yourself in trouble with this method because it just makes the equation more complicated

Step 1): As you are viewing this, you must be stuck with fractions. So, we need to look up at our fraction and find it's denominator. Make sure that you have ALL fraction's denominators, then find it's LCM.
Step 2): Now you've got it's LCM, I hope you still remember what is term. If you forgot, look back right now. If you remember, good, now multiply the LCM we just found to ALL terms, every single term in the equation must be multiplyed, please, don't think it's too hard work, and do it.
Step 3): Good, you have finished clearing the denominators? After that you should multiply the quotient(if applicable) into the numerator of all fractions.

It's harder to understand to just talk, but do nothing. Let's view some examples.

A) {{{60/x=3}}} --- Not such a hard equation?
{{{60/x-3=0}}} --- Not a clever move though, but this can apply to our clearing fractions method.
{{{x(60/x)-3(x)=0(x)}}} --- Remember, ALL terms must be multiplyed by the denominators LCM, in this case, 'x'.
{{{1(60)-3x=0}}} --- Easy-to-fail mistakes: Remember 0 multiplyed to anything is 0, do NOT write x, according to my experience this is a common way people fails even in simple algebras.
{{{60-3x=0}}}
{{{-3x=-60}}} --- Small Reminder: The -3x is not 3x, it follows the symbol in front of it, don't erase it, at all. This is another easy-to-fail mistake, but not that common.
{{{x=20}}} --- Small Reminder: You are dividing -60 by '-'3, don't forget about symbol.

Great, but is this correct? Let's do some checking (which I seldom do :P)

Checking:
60/x=3 and x=20
60/20=3
3=3

Let's look at another more complicated example, which has some more easy-to-fail common mistakes.

B) {{{(5/3)-(y+1)/4=2/3}}}
{{{12(5/3)-(12)(y+1)/4=(12)(2/3)}}} --- Remember: It's LCM of all fractions.
{{{4(5)-3(y+1)=4(2)}}} --- Don't skip process! According to my experience, it is the easiest to fail mistake around the '-3(y+1)' because if you skip this step you will be easily eliminated because you might think of '3(y+1)' instead. And, {{{12/4}}} = 3, {{{12/3}}} = 4, you have to multiply to quotient back to the fraction's numerators.
{{{20-3y-3=8}}} --- Must remember you are multiplying '-3' into (y+1) but not '3'.
{{{20-3y=11}}} --- 8+3=11.
{{{-3y=11-20}}} --- Just never forget about negative symbols.
{{{-3y=-9}}} --- 11-20=-9
{{{y=3}}} --- Remember you're dividing -9 by -3.

More complicated means easier to fail, so let's double check.

Checking:
{{{(5/3)-(y+1)/4=2/3}}} and y=3
{{{(5/3)-(3+1)/4=2/3}}}
{{{(5/3)-4/4=2/3}}}
{{{(5/3)-1=2/3}}}
{{{(5/3)-(3/3)=(2/3)}}} --- Strategy: 1 = 2/2, 3/3, 4/4 or anything with same numerator and denominator.
{{{2/3=2/3}}}

I hope you all enjoy my lesson, see you soon :)