SOLUTION: Assistance is needed and greatly appreciated. 5pq^2/ 12r^2s / 20q^5/ 3r^3s4 so far I got 100pq^7/ 36r^5s^5 (not sure if I am on the right path. Also... 6 + 2/ y -2 = 7

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Assistance is needed and greatly appreciated. 5pq^2/ 12r^2s / 20q^5/ 3r^3s4 so far I got 100pq^7/ 36r^5s^5 (not sure if I am on the right path. Also... 6 + 2/ y -2 = 7      Log On


   

Question 225733: Assistance is needed and greatly appreciated.
5pq^2/ 12r^2s / 20q^5/ 3r^3s4
so far I got 100pq^7/ 36r^5s^5 (not sure if I am on the right path.
Also...
6 + 2/ y -2 = 7/ y + 1 (not sure how to start to solve)
Please help!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Assume the problem is:
%285pq%5E2%29%2F%2812r%5E2s%29
----------
%2820q%5E5%29%2F%283r%5E3s%5E4%29
invert the dividing fraction & multiply
%285pq%5E2%29%2F%2812r%5E2s%29 * %283r%5E3s%5E4%29%2F%2820q%5E5%29
we can cancel across the "*" sign
cancel q^2 into q^5
%285p%29%2F%2812r%5E2s%29 * %283r%5E3s%5E4%29%2F%2820q%5E3%29
:
Cancel r^2 into r^3
%285p%29%2F%2812s%29 * %283rs%5E4%29%2F%2820q%5E3%29
:
Cancel s in s^4
%285p%29%2F%2812%29 * %283rs%5E3%29%2F%2820q%5E3%29
:
cancel 3 into 12
%285p%29%2F4 * %28rs%5E3%29%2F%2820q%5E3%29
:
Cancel 5 into 20; then multiply
p%2F4 * %28rs%5E3%29%2F%284q%5E3%29 = %28prs%5E3%29%2F%2816q%5E3%29
:
There is a lot of chances for dumb errors here, check each step and confirm that I did not make a math mistake.
:
;
Assume this problem is:
6 + 2%2F%28y-2%29 = 7%2F%28y%2B1%29
:
the first thing we want to do is get rid of those annoying denominators.
Multiply each term by (y+1)(y-2) to do this:
:
6((y+1)(y-2)) + (y+1)(y-2)*2%2F%28y-2%29 = (y+1)(y-2)*7%2F%28y%2B1%29
;
FOIL, then cancel the denominators
6(y^2 - y - 2) + 2(y+1) = 7(y-2)
:
Get rid of the brackes
6y^2 - 6y - 12 + 2y + 2 = 7y - 14
:
Combine like terms on the left
6y^2 - 6y + 2y - 7y - 12 + 2 + 14 = 0
:
6y^2 - 11y + 4 = 0; a quadratic equation
:
Fortunately this will factor, otherwise use the quadratic formula
(3y - 4)(2y - 1) = 0
Two solutions
3y = 4
y = 4/3
and
2y = 1
y = 1/2
:
Check each solution in original equation