Question 251971
x = rate of the first car.
x+10 = rate of the second car.
h = time in hours.


general formula is rate * time = distance


first car:


x*h = 60


second car:


(x+10)*h = 90


solve for one of the variables in the first equation in terms of the other variable and substitute in the second equation.


x*h = 60


solve for h to get:


h = 60/x


substitute in the second equation to get:


(x+10)*h = 90 becomes:


(x+10)*(60/x) = 90


multiply both sides of the equation by x to get:


(x+10)*60 = 90*x


simplify by removing parentheses to get:


60*x + 600 = 90*x


subtract 60*x from both sides of the equation to get:


600 = 90*x - 60*x = 30*x


divide both sides of the equation by 30 to get:


x = 20


use the value of x to solve for h in the first equation.


x*h = 60 becomes:


20*h = 60


divide both sides of the equation by 20 to get:


h = 60/20 = 3


use value of x and h in the first equation to get:


x*h = 60 becomes:


20*3 = 60 which is true.


use value of x and h in the second equation to get:


(x+10)*h = 90 becomes:


(20+10)*3 = 90 


combine like terms to get:


30*3 = 90 which is also true.


answer is:


first car travels at 20 miles per hour.


second car travels at 30 miles per hour.