1. Re: Yet another fractions problem

Originally Posted by uperkurk
For example you multiplied the denominators to get $8(x-3)^2$ are there certain rules you have to follow to multiply these like that? Because looking at it I would have done $x-3\times4\times2(x-3)$ which is ofcourse wrong. How did you know in which order to multiply these denominators by?
Order? In multiplication? You should not care about order in multiplication since A·B = B·A for any 2 numbers!
So A·B·C·6·Q·77= 77·6·C·A·B·Q = B·Q·77·C·A·6 = ...,etc.

Also when you want to multiply x-3 with 4 with 2(x-3) the result: x-3·4·2(x-3) is not correct since in that way you do not multiply x-3 with 4 but you multiply only 3 with it. Remember that 5-3·2 = 5-6 = -1 and NOT 5-3·2 = 2·2 = 4
So you should put brackets to x-3 to make it be multiplied with 4 so the right way to write it, is (x-3)·4·2(x-3)

Lets just assume that the denominators are now $x-1 , 4$ and $x+2$ can we simply just multiply all these together?
Yes. (x-1)·4·(x+2) or 4(x-1)(x+2) is the correct product which happens to be the LCM also(for unknown variables).

2. Re: Yet another fractions problem

To see the difference between using the LCM in the denominator, and just using the product, let's look at a simple example:

$\frac{1}{6} + \frac{3}{8} = x$.

Now LCM(6,8) = (6*8)/GCD(6,8) = 6*8/2 = 24. So we can multiply everything by 24:

$24\left(\frac{1}{6} + \frac{3}{8}\right) = \frac{24}{6} + \frac{72}{8} = 4 + 9 = 13 = 24x$.

Thus $x = \frac{13}{24}$.

If we use 6*8 = 48 instead, we get:

$48\left(\frac{1}{6} + \frac{3}{8}\right) = \frac{48}{6} + \frac{144}{8} = 8 + 18 = 26 = 48x$.

Thus $x = \frac{26}{48}$, and we have to divide top and bottom by 2, to get our original answer. So we get "one more step", and some larger numbers to work with, but it still works.

Remember, even a complicated expression like:

$\sqrt{3x^2 + 2y - 15}$ boils down to some number, once we know what x and y are. So we can treat things as "chunks" until we have to combine them in some way (like multiplying or adding together), or break them down (by factoring, for example).

Of course, using the LCM often results in "simpler" expressions, so if the LCM isn't TOO hard to figure out, it makes sense to use it. In the previous problem we worked out, the LCM is 4(x - 3) = 4x - 12 (it might be smarter to leave it factored to recognize when we can cancel, though), which would have saved us the steps of dividing by 2 and x - 3 after we "cleared the denominators".

It's OK to start a problem, and make some progress, and then say: "now I'm stumped". That is how you LEARN what gives you trouble, and how you can finally master the skills you need to solve problems.

Expressions like:

$\frac{2x+3}{x^2 - 2}$ are called "rational expressions", and it turns out that the same rules for manipulating rational numbers (ordinary fractions) work for them, too. Let me state these basic rules, so you can commit them to memory:

For all quantities A,B and C:

(A+B)+C = A+(B+C) = A+B+C (all groupings of SUMS are on equal footing).

For all quantities A and B:

A+B = B+A

There is a quantity called 0 for which:

A+0 = 0+A = A, for any quantity A.

For every quantity A, there is another quantity (denoted -A) such that:

A+(-A) = -A+A = 0 (where 0 is the expression from the previous law).

Associativity of Multiplication:

For all quantities A,B and C:

(A*B)*C = A*(B*C) = A*B*C (note this only works with "all multiplies" or "all sums"...it is NOT true that (A+B)*C = A+(B*C), for example: (3+4)*5 = 35, while 3+(4*5) = 23).

Commutativity of Multiplication:

For all quantities A,B:

A*B = B*A

Existence of a Multiplicative Identity:

There exists a quantity, usually denoted 1, such that, for all quantities A:

A*1 = 1*A = A

Existence of Multiplicative Inverses:

For all quantities A except 0, there exists a quantity B (denoted by 1/A, typically) such that:

A*(1/A) = (1/A)*A = 1.

For all quantities A,B and C:

1) A*(B+C) = (A*B) + (A*C)

2) (A+B)*C = (A*C) + (B*C)

There is no arguing with any of these rules, they are "the law" and must be obeyed (or there are severe penalties). There is one final rule, which may SEEM obvious, but later on turns out to be necessary:

1 ≠ 0.

Any set of expressions or quantities which obeys all these rules is called a FIELD, which is the usual setting for doing algebra. Examples include:

Rational numbers
Real numbers
Rational functions in one or more variables
Complex numbers
Algebraic numbers (sometimes we need to use square roots, or higher roots).

You seem to have particular trouble with understanding the distributive law (often used in "collecting like terms"). I would practice using it some more.

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