# Algebraic fractions

• May 12th 2008, 10:09 PM
Turple
Algebraic fractions
I'm just confused about a couple of rules. For example if I have multiplied a fraction out to get 9(x+4)-2(x+2) does the 2 in the second bracket get multiplied by -2 or 2?

Also if I have this: 2(x+4)(x+2) is it only the x+4 which is doubled? Or do you work out the quadratic and multiply the whole thing by two?

Thanks
• May 12th 2008, 10:21 PM
Moo
Hello,

Quote:

Originally Posted by Turple
I'm just confused about a couple of rules. For example if I have multiplied a fraction out to get 9(x+4)-2(x+2) does the 2 in the second bracket get multiplied by -2 or 2?

-2 ;)

Quote:

Also if I have this: 2(x+4)(x+2) is it only the x+4 which is doubled? Or do you work out the quadratic and multiply the whole thing by two?

Thanks
2(x+4)(x+2)=[2(x+4)](x+2)=2[(x+4)(x+2)]=(x+4)[2(x+2)]

So it's both of them (Wink)
• May 12th 2008, 11:57 PM
earboth
Quote:

Originally Posted by Turple
...

I also can't understand how to work out these fractions:

2x ...... 9
---- - ---- = 2
x+4 ........ x+2

...

I'm going to show you the first example step by step. The following examples can be solved in exactly the same way:

1. Domain: The denominator must be unequal to zero. Therefore

$\displaystyle D=\mathbb{R}\setminus \{-4, -2\}$

2. The common denominator of the fractions is $\displaystyle (x+4)(x+2)$. Change the fractions so they have both the same denominator:

$\displaystyle \frac{2x (x+2)}{(x+4)(x+2)} - \frac{9 (x+4)}{(x+4)(x+2)} = 2$

3. Multiply both sides of the equation by the common denominator to get rid of the fractions:

$\displaystyle 2x(x+2)-9(x+4)=2(x+4)(x+2)$

4. Expand all brackets:

$\displaystyle 2x^2+4x-9x-36=2(x^2+6x+8)$

5. Collect like terms:

$\displaystyle 2x^2-5x-36=2x^2+12x+16$

$\displaystyle -17x=52$

6. Divide by the leading factor of x:

$\displaystyle \boxed{x=-\frac{52}{17}}$