hard fraction

**andyboy179** · November 13th 2010, 11:53 AM

i really don't understand this! it's just not making sense!

**Ackbeet** · November 13th 2010, 12:01 PM

Ok, let's start here: forget the rest of this thread for now. Focus on what I'm about to tell you.

Would you agree that for any real number $a,$ it is true that $a\cdot 1=a?$

That is, the number $1$ is the multiplicative identity. I can multiply something, anything, by $1$ and not change the something. Right?

Examples: $3\cdot 1=3;\;6\cdot 1=6;\; x\cdot 1=x;\; (x-5)\cdot 1=x-5;\;(\text{stick figure})\cdot 1=\text{stick figure}.$

Make sense?

**andyboy179** · November 13th 2010, 12:14 PM

yep

**Ackbeet** · November 13th 2010, 12:17 PM

Excellent. And, of course, since multiplication by real numbers is commutative, I could just as easily have written, for all the previous examples, the following:

$1\cdot 3=3;\;1\cdot 6=6;\; 1\cdot x=x;\; 1\cdot(x-5)=x-5;\;1\cdot (\text{stick figure})=\text{stick figure}.$

Still with me?

**andyboy179** · November 13th 2010, 12:22 PM

yep

**Ackbeet** · November 13th 2010, 12:52 PM

Fantastic. Now, with any of these examples, I could simply divide one side of them by the "something", and get one. So, from the examples

$1\cdot 3=3;\;1\cdot 6=6;\; 1\cdot x=x;\; 1\cdot(x-5)=x-5;\;1\cdot (\text{stick figure})=\text{stick figure},$

I could have

$1=\dfrac{3}{3};\;1=\dfrac{6}{6};\;1=\dfrac{x}{x}\; (\text{for}\;x\not=0);\;1=\dfrac{x-5}{x-5}\;(\text{for}\;x\not=5);\;1=\dfrac{\text{stick figure}}{\text{stick figure}}\;(\text{for}\;\text{stick figure}\not=0).$

So you would agree, then, that if I have any nonzero number whatsoever, call it $\mathfrak{R},$ I have

$1=\dfrac{\mathfrak{R}}{\mathfrak{R}}.$

All of these fractions of "something over itself" are all just fancy ways of writing the number $1.$

Make sense?

**andyboy179** · November 13th 2010, 12:55 PM

it kind of makes sense, where would i apply this to my question?

**Ackbeet** · November 13th 2010, 04:50 PM

...where would i apply this to my question?

We're getting there. I just want to make sure you understand every step of the way. The next step is this: if I can multiply any number by 1 and get the number back again, it follows that I can multiply any fraction by 1 and get the fraction back again. Examples:

$\dfrac{2}{3}\cdot 1=\dfrac{2}{3};\;\dfrac{x}{6}\cdot 1=\dfrac{x}{6}.$

And if I can multiply any fraction by 1, it follows that I can multiply any fraction by one of these "fancy" ways of writing 1. Examples:

$\dfrac{2}{3}\cdot\dfrac{3}{3}=\dfrac{2}{3};\;\dfra c{x}{6}\cdot\dfrac{x-5}{x-5}=\dfrac{x}{6}\;(\text{if}\;x\not=5),$ and so on.

Do you follow?

So what's the point of all this? The point is, that if I can multiply by any of these fancy ways of writing 1, then I can get fractions to have practically any denominator I want, simply by multiplying by "what's missing". So, let's say I have the fraction

$\dfrac{2}{3},$

but I want the denominator to be 36, AND I don't want to change the fraction's value. Well, what do I have to multiply 3 by to get 36? 12. So I can multiply by 12/12 (which is equal to 1), and not change the value of the fraction. I get

$\dfrac{2}{3}=\dfrac{2}{3}\cdot\dfrac{12}{12}=\dfra c{24}{36}.$

I now have the denominator I want without changing the fraction. I could always cancel the 12's to get back to where I started. But now, if I wanted to add this fraction to another fraction with a 36 in the denominator, I could do that. You can ONLY add fractions when they have the same denominator (the common denominator). If you do have a common denominator, then adding fractions works like this:

$\dfrac{a}{c}+\dfrac{b}{c}=\dfrac{a+b}{c}.$

The new numerator is the sum of the old numerators, and the new denominator is the old common denominator.

Make sense?

All of this doesn't change when you throw in variables. If I have the fraction

$\dfrac{x^{2}-4}{x-3},$

but I want the denominator $(x-3)(x+5),$ then I'd have to multiply by a fancy way of writing 1 that looks like $(x+5)/(x+5).$

That is,

$\dfrac{x^{2}-4}{x-3}=\dfrac{x^{2}-4}{x-3}\cdot\dfrac{x+5}{x+5}=\dfrac{(x^{2}-4)(x+5)}{(x-3)(x+5)}.$

Do you follow?

**andyboy179** · November 14th 2010, 02:48 AM

Yes it's starting to make sense.

**HallsofIvy** · November 14th 2010, 05:12 AM

No. The two denominators are 2 and x+ 2. Since they have no factors in common, their least common multiple is just the product of the two expressions: 2(x+ 2)= 2x+ 4. You learned how to add fractions years ago- if 1/2+ 2/3= 3/6+ 4/6= 7/6. It's the same here.

Rather than add the fractions on the left you can just multiply both sides by that least common multiple:
$(2x+4)\frac{x-1}{2}- (2x+4)\frac{5}{x+2}= (2x+4)(1)$ .

**andyboy179** · November 23rd 2010, 06:29 AM

hello again!

we have been working a bit on these sort of questions in class.

so far i have got x^2+x-2-8/ 2x+4!

from that point im stuck, do i simplify it so i have, x^2+x-10= 2x+4 ??

**Ackbeet** · November 23rd 2010, 06:38 AM

Excellent. Everything looks good so far. Remember the general principles of solving for a variable: get everything that has that variable over to one side of the equation. How could you do that?

**andyboy179** · November 23rd 2010, 06:49 AM

take away 2x and 4 from both sides to leave you with. x^2-x-14=0
then do a quadratic equation!
for the quadratic equation would i do:
$hard fraction-quadratic-equation.jpg$
and then the rest?

**Ackbeet** · November 23rd 2010, 06:52 AM

Everything is correct up to and including your use of the quadratic formula. However, you simplified incorrectly from the quadratic formula. What's inside the square root is, I grant you, 57, but there's more to the expression than that. You should have two real solutions!

**andyboy179** · November 23rd 2010, 06:56 AM

i would do this wouldn't i:
$hard fraction-quadratic-equations-2.jpg$

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