Let the slower speed be x mph.
The faster speed is therefore x+10 mph
You know the distance and the speeds, write expressions for the each time..
t = d/s
t_1 = 280/x (longer time) " "and t_2 = 280/(x+10) (shorter time)
The difference between the two times was 1 " hour"
t_1 - t_2 = 1
280/x - 280/(x+10) = 1" "larr xx LCM to cancel the denominators
(color(blue)(x(x+10)xx280))/x - (color(blue)(x(x+10)xx280))/(x+10) = color(blue)(x(x+10)xx)1
280(x+10) -280x = x(x+10)
280x+2800 -280x = x^2+10x" "larr
Now you have a quadratic equation, make it equal to 0.
0 = x^2 +10x -2800
This expression does not have factors.
Completing the square gives:
x^2 +10x +25 = 2800+25
(x+5)^2 = 2825
x +5 = sqrt2825" " only the positive root is valid
x = sqrt2825 -5
x = 48.1507 mph - this is the slower speed
x +10 = 58.1507 mph -this is the faster speed
t_1 = 280/48.1507 = 5.815 hrs
t_2 = 280/58.1507 = 4.815 hrs
"ave speed" = ("total distance")/("total time")
"ave speed" = 560/10.63
"ave speed" = 52.68mph
An expression for the ave speed would be
"ave speed" = 560/(t_1+t_2)
