Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. How do you determine speed of the vehicle?

Let #s# be the speed (in miles/hour) that Steve was traveling for #h# hours to cover #200# miles.

We are told that if he had traveled at a speed of #(s+10)# miles/hour that it would have taken him #(h-1)# hours to cover the #200# miles.

Since distance travelled # = #speed # xx # time

#color(white)("XXX")200 = sh#
#color(white)("XXXXXXXXXXX")rarr color(blue)(h)=color(green)(200/s)#
and
#color(white)("XXX")200=(s+10)(color(blue)(h)-1))#

So we have
#color(white)("XXX")200=(s+10)(200/s-1)#

#color(white)("XXXXXX")=s(200/s-1)+10(200/s-1)#

#color(white)("XXXXXX")=200-s+2000/s-10#

#rArrcolor(white)("XXX")s-2000/s+10=0#

#rarrcolor(white)("XXX")s^2+10s-2000=0#

Using the quadratic formula
#color(white)("XXX")s=(-10+-sqrt(10^2-4(1)(-2000)))/(2(1))#

#color(white)("XXXX")=(-10+-sqrt(8100))/2#

#color(white)("XXXX")=(-10+-90)/2#

#s=-25# or #s=40#

Since the speed must be non-negative, #s=-25# is an extraneous solution.

There are 2 different scenarios described here, Write an expression for the speed of each of them.

The difference between the times would be 1 hour. This allows us to make an equation.

Let the slower speed be #x# miles per hour.
The faster speed is #x +10# miles per hour.

#time ="distance"/"speed"#

At the slower speed, the time, #T_1 = 200/x # hours

At the faster speed, the time, #T_2 = 200/(x+10) #hours

(#T_1 "will be more than " T_2# because if we drive at slower speed, the journey will take longer.)
The difference between the two times is 1 hour.

#T_1 - T_2 = 1#

#200/x -200/(x+10) = 1 " now solve the equation"#

Multiply each term by #color(red)(x(x+10))#

#(color(red)(x(x+10))xx200)/x -(color(red)(x(x+10))xx200)/(x+10) = color(red)(x(x+10))xx1#

#(cancelx(x+10)xx200)/cancelx -(xcancel((x+10))xx200)/cancel((x+10)) = x(x+10)xx1#

#200x+2000 -200x =x^2+10x#

#x^2 +10x -2000 = 0#

Find factors of 2000 which differ by 10.

The factors must be quite close to #sqrt2000#, because there is a very small difference between them.

We find #40xx50 = 2000#

#(x-40)(x+50) = 0#

#x = 40 or x = -50, # (reject -50)

The slower speed is 40mph, the faster speed is 50 mph.