find the quotient of each of the following

Question

state the non-permissible value(s)

 

a. 2x/6x+6 ÷ x+4/3+3x

 

 

b.9-x²/8x÷x-3/8x

 

 

c. x(x+4)/x+1÷6x²+24x/2

 

 

I am having trouble understanding how to do these type of questions

Answer 1

Hello, Layne

 

a) Lets factor 6x+6= 6(x+1)   and 3+3x= 3(x+1).  When you divide flip the second fraction upside down (reciprocal). So, a becomes

 

(2)/[6(x+1)] * 3(x+1)/(x+4)        Cancel out (x+1)

2/6 * 3/(x+4)                              2/6 becomes 1/3

1/3 *3/(x+4)                               cancel out the 3

       1/(x+4)                                Should be the final answer

 

b)  9-x^2.  You can factor out a negative 1.  So, it becomes -1(x^2-9).  x^2-9 is the difference of perfect squares.  It becomes (x-3)(x+3).. don't forget -1.   We are ready.  Remember, when you divide fractions you multiply by the reciprocal of the second fraction.  So, we have

 

[(-1)(x-3)(x+3)]/(8x)  *  (8x)/(x+3)      cancel out 8x and (x+3) which is the same as (3+x)

 

Final answer (-1)(x-3) or -x+3, if distribute.

 

c) For this one, we need to factor  6x from 6X^2+24x= 6x(x+4)... and again turn the second fraction upside down.  We have:

 

[x(x+4)](x+1) * 2/[(6x)(x+4)]   cancel out (x+4)

 

x/(x+1) * 2/(6x)                      cancel out the x's

1/(x+1)* 2/6                           reduce 2/6 to 1/3

1/(x+1) * 1/3                          multiply

   1/[3(x+1)]

 

There you have it Layne... Please rate my answer :)

 

D. Y. Taylor