Algebra fractions, a little scary?

Question

The /// is just there because I am trying to separate the numbers they keep running together.

x +6 /// 3x-12

_____ . _____

x^2-16 /// 3x+18

Answer 1

http://www.themathpage.com/Alg/multiply-...

http://www.mathsrevision.net/gcse/pages....

Answer 2

x+6  ///      3x-12

_____  . ______

x^2-16 ///    3x+18            

you simplify it so to cancel like terms

x+6   ///       3(x-4)

_____ . ________

x^2-16  ///    3(x+6)

you cancel both 3s because (3/3=1) and you're left with:

x+6   ///     (x-4)

_____ . _______

x^2-16  ///    (x+6)

since you are multiplying both sides, you can cancel both (x+6)

and you're left with:

(x-4)

_____

x^2 -16

you have to factor out your denominator, and it will look like this:

(x-4)

___________

(x+4) (x-4)

now cancel both (x-4) because it is identical, and you have:



answer:

1

_______

x+4

or

1/ (x+4)

algebra is easy as long as you've familiarized yourself with its rules. don't hesitate to read its rules of addition, subtraction, multiplication and division. remember them while solving and you won't have a problem. you can do it! :)

Answer 3

= ([x + 6]/[x² - 16])([3x - 12]/[3x + 18])

= ([x + 6]/[{x + 4}{x - 4}])([3{x - 4}]/[3{x + 6}]) cancel out 3

= ([x + 6]/[{x + 4}{x - 4}])([x - 4]/[x + 6]) cancel out x + 6

= (1/[{x + 4}{x - 4})([x - 4]) cancel out x - 4

= 1/(x + 4)

Answer: 1/(x + 4)

Answer 4

(x+6)/(x^2 -16) * (3x-12)/(3x+18)

= (x+6)/[(x-4)*(x+4)] * [3(x-4)]/[3(x+6)]

= 1/(x+4)  

Answer 5

3x - 12  =   3(x-4)

3x+18  = 3(x+6)

X^2  - 16   = (x+4)(x-4)

so we get

1/(x+4)

Answer 6

3X-12=3(X-4)

X^2-16=(X+4)(X-4)

3X+18=3(X+6)

PUT AND Simplify by eliminationg facors

1/(x+4)   is the ans

Answer 7

they're right  

Answer 8

(3x+18)(x+6) - (x^2-12)(3x-12)

(3x^2+18x+18x+108) - (3x^3-12x^2-36x+144)

3x^3+15x^2+54x-36