Question 245715
Let x = # hours it takes Pipe A to drain the tank working alone
Let x + 5 = # hours it takes Pipe B to drain the tank working alone
Let 1/x = portion of the job Pipe A can perform in one hour 
Let 1/(x+5)= portion of the job Pipe B can perform in one hour


Set up the equation:
Here it is in words:

(portion of the job Pipe A can perform in one hour + portion of the job Pipe B can perform in one hour) x # hours to perform the job = 100% of the job or 1 job completed


And Algebraically:
Solve for x:
(1/x + 1(x+5))(6) = 1
{{{((x+5)/((x)(x+5)) + x/((x)(x+5)))(6) = 1}}}
{{{(2x+5/(x^2 +5x))(6) = 1}}}
{{{(12x + 30)/(x^2 + 5x) = 1}}}
{{{12x + 30 = x^2 + 5x}}]
{{{x^2 - 7x - 30 = 0}}}
{{{(x - 10)(x + 3) = 0}}}
{{{x = 10}}} 
x = -3  Reject this answer (cannot have a negative value for time)

It takes Pipe A 10 hours to drain the tank working alone