Question 225733
Assume the problem is:
{{{(5pq^2)/(12r^2s)}}}
----------
{{{(20q^5)/(3r^3s^4)}}}
invert the dividing fraction & multiply
{{{(5pq^2)/(12r^2s)}}} * {{{(3r^3s^4)/(20q^5)}}}
we can cancel across the "*" sign
cancel q^2 into q^5
{{{(5p)/(12r^2s)}}} * {{{(3r^3s^4)/(20q^3)}}}
: 
Cancel r^2 into r^3
{{{(5p)/(12s)}}} * {{{(3rs^4)/(20q^3)}}}
:
Cancel s in s^4
{{{(5p)/(12)}}} * {{{(3rs^3)/(20q^3)}}}
:
cancel 3 into 12
{{{(5p)/4}}} * {{{(rs^3)/(20q^3)}}}
:
Cancel 5 into 20; then multiply
{{{p/4}}} * {{{(rs^3)/(4q^3)}}} = {{{(prs^3)/(16q^3)}}}
:
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/(y-2)}}} = {{{7/(y+1)}}}
:
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/(y-2)}}} = (y+1)(y-2)*{{{7/(y+1)}}}
;
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