1. ## Re: Algebraic fraction?

Originally Posted by Ashir
Why did you do that?
Where did x+1 come from?
Why both numerator and denominator?
Why can't we just treat it as a normal fraction?
Why wasn't my method correct? Or was it?
What do you mean 'foil'?
What do you mean by distribute?
1.) The $\displaystyle x+1$ comes from:

$\displaystyle \frac{(x+1)(x-1)}{x-1}=x+1$

More simply, we see that the denominator in the first term needs the factor $\displaystyle x+1$ to be equal to the denominator in the second term.

2.) You have to multiply both the numerator and the denominator by the same thing so that in effect you are multiplying the term by 1, and so its value remains unchanged.

3.) What do you mean by normal fraction?

4.) No, your method was not correct.

5.) FOIL is an acronym used as a mnemonic device for multiplying binomials. If you have:

$\displaystyle (a+b)(c+d)$ you can find the product from the sum of:

F: product of "first" terms or $\displaystyle ac$
O: product of "outer" terms or $\displaystyle ad$
I: product of "inner" terms or $\displaystyle bc$
L: product of "last" terms or $\displaystyle bd$

Hence:

$\displaystyle (a+b)(c+d)=ac+ad+bc+bd$

6.) To distribute essentially means:

$\displaystyle a(b+c)=ab+ac$

The term on the outside is "distributed" to the terms on the inside.

2. ## Re: Algebraic fraction?

Does it make any sense now?

3. ## Re: Algebraic fraction?

Not at all I'm afraid. My teacher can do it much easier but she only has so much time

4. ## Re: Algebraic fraction?

I can guarantee you that your teacher is finding the lowest common denominator, then combining the terms and simplifying.

The only thing easier is explaining the process in person rather than online.

5. ## Re: Algebraic fraction?

Originally Posted by MarkFL2
I can guarantee you that your teacher is finding the lowest common denominator, then combining the terms and simplifying.

The only thing easier is explaining the process in person rather than online.
On the nose my friend...

6. ## Re: Algebraic fraction?

Originally Posted by Ashir
2x/x-1 - 7x-3/x^2-1
$\displaystyle \frac{2x}{x-1}-\frac{7x-3}{x^2-1}$

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

$\displaystyle \frac{2x^2-5x+3}{x^2-1}$

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

7. ## Re: Algebraic fraction?

Originally Posted by Plato
$\displaystyle \frac{2x}{x-1}-\frac{7x-3}{x^2-1}$

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

$\displaystyle \frac{2x^2-5x+3}{x^2-1}$

$\displaystyle \frac{(2x+1)(x-3)}{x^2-1}$
Almost got it. Why did x^2-1 remain the denominator and why did 2x become 2x(x+1)

8. ## Re: Algebraic fraction?

Because, like I told you you multiply both the numerator and denominator by $\displaystyle x+1$...
$\displaystyle x^2-1$ remained in the denominator because it is the least common denominator...
Not to be rude, but you need to brush up on your basic fraction mathematics. You can't do this level if you don't have a firm grasp on fractions with just numbers...
Tell me what these would equal.

$\displaystyle 1)\frac{8}{5}+\frac{3}{7}=$

$\displaystyle 2)\frac{1}{6}-\frac{7}{9}=$

$\displaystyle 3)\frac{5}{8}\times\frac{2}{5}=$

$\displaystyle 4)\frac{3}{7}\div\frac{9}{13}=$

9. ## Re: Algebraic fraction?

Originally Posted by alane1994
Because, like I told you you multiply both the numerator and denominator by $\displaystyle x+1$...
$\displaystyle x^2-1$ remained in the denominator because it is the least common denominator...
Not to be rude, but you need to brush up on your basic fraction mathematics. You can't do this level if you don't have a firm grasp on fractions with just numbers...
Tell me what these would equal.

$\displaystyle 1)\frac{8}{5}+\frac{3}{7}=$

$\displaystyle 2)\frac{1}{6}-\frac{7}{9}=$

$\displaystyle 3)\frac{5}{8}\times\frac{2}{5}=$

$\displaystyle 4)\frac{3}{7}\div\frac{9}{13}=$
I'll just do the first one to show I can, don't have time to do the rest.
56/35 + 15/35 = 71/35 = 2 1/35

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