reviewing restrictions

[yfelix45]

hello,

this is another review question that I would help with.

Simplify: 9x/ x^2 -5x - 6 - 6x/ x^2 - x - 12 and state any restrictions on the variables.

my solution so far:

9x/(x-3)(x-2) - 6x/(x-4)(x+3)

9x(x-3) - 6x(x-4) / (x-3)(x-2)(x-4)(x+3)

9x^2 - 27x - 6x^2 - 24x / (x-3)(x-2)(x-4)

Have I done this right so far? How can I simplify this better?

Thanks
yfelix45

 

[pappus]

hello,

this is another review question that I would help with.

Simplify: 9x/ (x^2 -5x - 6) - 6x/ (x^2 - x - 12) <--- use brackets!
and state any restrictions on the variables.

my solution so far:

9x/(x-3)(x-2) <-- that's wrong; see below - 6x/(x-4)(x+3)

9x(x-3) - 6x(x-4) / (x-3)(x-2)(x-4)(x+3)

9x^2 - 27x - 6x^2 - 24x / (x-3)(x-2)(x-4)

Have I done this right so far? <-- no
How can I simplify this better?

Thanks
yfelix45

1. Since the factors of the denominators are coprime the common denominator is (x-4)(x-3)(x-6)(x+1)

2. Re-arrange the quotients so that both fractions have the same denominator:

\(\displaystyle \displaystyle{\frac{9x}{(x-6)(x+1)}-\frac{6x}{(x+3)(x-4)} = \frac{9x(x+3)(x-4)}{(x-6)(x+1)(x+3)(x-4)}-\frac{6x(x-6)(x+1)}{(x-6)(x+1)(x+3)(x-4)}}\)

\(\displaystyle \displaystyle{ = \frac{9x(x+3)(x-4)-6x(x-6)(x+1)}{(x-6)(x+1)(x+3)(x-4)}}\)

3. Expand the brackets in the numerator, collect like terms, factor out 3x.

4. To get the restrictions determine those values of x which yields zero in the denominator. (Remember: A product equals zero if one of the factors is zero)

 

[yfelix45]

thank you

hello


Thank you pappus for answering my question.

To denis : I quite sorry for not placing brackets in part of my solution. The question that I presented to the forum was posted exactly the way it was presented to me on paper. I did not mean to offend you!!!!!!!!!!!!!!!

thanks

yfelix45

 

[mmm4444bot]

The question that I presented to the forum was posted exactly the way it was presented to me on paper.

Who cares whether somebody else wrote it wrongly?

We would like to know whether you understand why the grouping symbols are required on these boards. Do you?

PS: Nobody at this end is offended. :cool: