
Victoria V. answered 01/25/19
Math Teacher: 20 Yrs Teaching/Tutoring CALC 1, PRECALC, ALG 2, TRIG
You have known for a long time that distance = rate * time (the rate is the speed).
If you solve this for time, you find that time = distance/rate
Let t = the time it took to fly there at two different speeds, and it will be in hours.
This way our rate (speed) can be in miles per hour.
Let r = faster rate, and r-90 = slower rate (Converted 1.5 mi/min * 60 min/hour = 90 miles/hour)
So t = 340/r + 850/(r-90)
Going only the slower rate (r-90), the time required would have been t (our original time) + 0.1 (6 min = 0.1 hours) and that would be the whole 1190/(r-90)
So we now have 2 equations that can both be simplified to = t (our original time)
t = 340/r + 850/(r-90)
t + 0.1 = 1190/(r-90), solving this for "t", our second equation becomes
t = 1190/(r-90) - 0.1
Since we have two expressions that both = t, they equal each other so we can set up this equation:
340/r + 850/(r-90) = 1190/(r-90) - 0.1
Getting a common denominator [ r(r-90) ] on the left, we have
(1190r -30600)
--------------------
r ( r-90)
Getting a common denominator on the right we get
1280 - 0.1 r
----------------
r - 90
Need the right to have the same denominator as the right, so multiply top and bottom by "r"
Now the right side has
1280 r - 0.1 r2
------------------------
r ( r - 90 )
Now that the left and right sides both have the same denominator, we can just deal with the numerators.
(or put another way, multiply both sides by "r ( r - 90 )" and all that will be left are the numerators.)
1190r - 30600 = 1280r - 0.1 r2
To solve a quadratic, put everything on one side so other side = 0
0.1 r2 - 90 r - 30600 = 0
I multiplied everything by 10 so I did not have to deal with the 0.1.
Now I have r2 - 900 r - 30600 = 0
This solves to:
r = 1163 miles per hour or r = -263 miles per hour. Generally we think of speed as being positive, so
the faster speed was 1163 miles per hour. Subtract 90 miles per hour to find that the slower speed was
1073 miles per hour.