Question 927849
basic equation is rt = d, where r = rate and t = time and d = distance.
since d = 36, this equation becomes:
rt = 36
if you increase r by 10 and you decrease t by 18, then you get the equation of:
(r+10) * (t-18) = 36
this is because the distance stays the same.


since rt = 36 and (r+10) * (t-18) = 36, and 36 = 36, you can set rt equal to (r+10) * (t-18) to get:


rt = (r+10) * (t-18)


simplify this equation to get:


rt = rt - 18r + 10t - 180


subtract rt from both sides of this equation to get:


0 = -18r + 10t - 180


solve this equation for t and you will get:


t = 1.8r + 18


go back to the first equation of rt = 36 and replace t with 1.8r + 18 to get:


r * (1.8r + 18) = 36


simplify this to get:


1.8r^2 + 18r = 36


subtract 36 from both sides of this equation to get:


1.8r^2 + 18r - 36 = 0


divide both sides of this equation by 1.8 to get:


r^2 + 10r - 20 = 0


factor using the quadratic formula to get:


r = -5 plus or minus 3 * sqrt(5)


-5 minus 3 * sqrt(5) is negative.
since rate can't be negative, this solution is no good.


-5 plus 3*sqrt(5 is positive.
this solution for r is potentially good.


in the equation of rt = 36, replace r with -5 plus 3*sqrt(5) and solve for t to get:


t = 36 / (-5 + 3*sqrt(5))
this solution for t is potentially good.


r = -5 + 3*sqrt(5) = 1.708203932


t = 36 / (-5 + 3 * sqrt(5)) = 21.07476708


confirm these solutions are good by replacing r and t in the original two equations to see that they are both true.


r*t becomes 1.708203932 * 21.07476708 which is equal to 36.


(r+10) * (t-18) becomes 11.70820393 * 3.074767078 which is equal to 36.


the solution is confirmed to be good in both original equations and so the values of r and t are considered to be good.