SOLUTION: A person traveling 210 km by car drove the first part at 100 kph and the rest at 80 kph. If the trip took 2.25 hours to complete, what distance was traveled at 100 kph? Write equat

Question 926751: A person traveling 210 km by car drove the first part at 100 kph and the rest at 80 kph. If the trip took 2.25 hours to complete, what distance was traveled at 100 kph? Write equation and solve.
Found 2 solutions by TimothyLamb, MathTherapy:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
x = distance driven at 100 kph
y = time driven at 100 kph
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s = d/t
t = d/s
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y = x/100
2.25 - y = (210 - x)/80
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100y = x
80*2.25 - 80y = 210 - x
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put the system of linear equations into standard form
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100y = x
180 - 80y = 210 - x
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x - 100y = 0
x + 180 - 80y = 210
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x - 100y = 0
x - 80y = 30
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copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
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solution:
x = distance driven at 100 kph = 150 km
y = time driven at 100 kph = 1.5 hours
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Solve systems of linear equations up to 6-equations 6-variables:
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Answer by MathTherapy(10551) (Show Source): You can put this solution on YOUR website!

A person traveling 210 km by car drove the first part at 100 kph and the rest at 80 kph. If the trip took 2.25 hours to complete, what distance was traveled at 100 kph? Write equation and solve.

Let distance covered at 100 km/h be D
Then distance covered at 80 km/h = 210 - D
Time taken to cover distance, travelling at 100 km/h (1st part of trip) =
Time taken to cover distance, travelling at 80 km/h (2nd part of trip) =
Since total time taken for entire trip was 2.25 hours, then we can say that:
------ 2.25 hours converted to , or hours
4D + 5(210 - D) = 9(100) ------- Multiplying by LCD, 400
4D + 1,050 - 5D = 900
4D - 5D = 900 - 1,050
- D = - 150
D, or distance traveled at 100 km/h = , or km
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