Question 266485
Let x=amount of time it takes the old computer working alone to do the payroll
Then the old computer works at the rate of 1/x of the payroll per hour
We are told that the new computer working alone can do the payroll in x-1 hours
Then the new computer works at the rate of 1/(x-1) of the payroll per hour
Together the two computers work at the rate of 1/x + 1/(x-1) of the payroll per hour
We are also told that when both computers are used, they can do the job in 3 hours.
So, together the computers work at the rate of 1/3 of the payroll per hour
So, our equation to solve is:
1/x + 1/(x-1)=1/3 multiply each term by 3*x*(x-1)
3(x-1)+3x=x(x-1) simplify
3x-3+3x=x^2-x or
6x-3=x^2-x subtract 6x from and add 3 to both sides
6x-6x-3+3=x^2-x-6x+3  collect like terms
x^2-7x+3=0-----------Quadratic in standard form; solve using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (7+- sqrt( 49-12 ))/(2) }}} 
{{{x = (7+- sqrt(37))/(2) }}}
{{{x = (7+-6.1)/(2) }}}
{{{x = (7+6.1)/(2) }}}
{{{x = 6.55hr }}}------------------time it takes the old computer to do the payroll working alone
{{{x-1=6.55-1=5.55hr }}}-----time it would take the new computer working alone

Another possible solution is:
{{{x = (7-6.1)/(2) }}}
{{{x = 0.45hr}}}---- time it would take the old computer to do do the payroll working alone ---NO GOOD!!!! the new computer can do the job in 1 hour less time and that would put us into negative time

CK
(1/6.55)+1/(5.55)=1/3
0.15267+0.18018=0.33333333
0.333~~~~~~0.333


Hope this helps-----ptaylor