SOLUTION: I need help!! - A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I need help!! - A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h      Log On


   



Question 174973: I need help!!
- A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
I tried
x+y=5
100x+90=470
100x+100y=500
100x+90y=470
= 10y = 30
x=2 y=3
but I only got half marks. Where did I go wrong and what was the answer?

Found 2 solutions by Electrified_Levi, stanbon:
Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
.
I need help!!
- A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
We will start fresh
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First we need to know a formula
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+%28rate%29%28time%29=+distance+, or with abbreviations, +r%29%28t%29+=+d+
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A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
This problem is a little tricky, since the car is going two speeds, instead of just one, we also have two different times.
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This will mean we will have two unknowns, we can find the other unknowns with the information we have
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A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car travelled at 100 km/h. For the rest of the trip, it travelled at 90 km/h. How far did the car travel at each speed?
.
Remember that +rate%28time%29+=+distance+, we don't know the time or distance for either one of the speeds, but we know the total distance and total time
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We can say that the time it took when the car went 90 km/h, the time it took would be +x+
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We can say that how long it took when the car went 100 km/h, the time it took would be +5+-+x+, ( if you subtract the other time, from the total time, you will find the second unknown time )
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It is the same concept with the distances
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one distance would be +y+
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the second unknown distance would be +470+-+y+
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We can now find equations to this problem
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+r%28t%29+=+d+
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Equation 1
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if the car went 100 km/h ( rate )
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if the car took +5+-+x+ hours ( time ), ( doesn't matter which time we could use "x" if we wanted, but we will use this one )
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These two multiplied together will equal a distance +470+-+y+ , ( once again it doen't matter which distance, we could use "y" if we wanted )
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Now we would put the words into a formula,
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+rate%28time%29=+distance+ = +100%285+-+x%29+=+470+-+y+, this is the first equation
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Now we will use the other distance, time, and rate, to find the second equation
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If the car went 90 km/h ( rate )
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If the car took +x+ hours ( time )
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If the car drove +y+ km ( distance )
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Now replace "r", "t" and "d" with the other veriables
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+r%28t%29+=+d+ = +90%28x%29+=+y+, this is the second equation
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Now we can put these two equations side by side
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+system+%28+100%285+-+x%29+=+470+-+y%2C+90%28x%29+=+y%29+
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To make it easier, we will simplify the two equations
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Equation 1, +100%285+-+x%29+=+470+-+y+, we can use the distribution to simplify this equation
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+100%285+-+x%29+=+470+-+y+ = +highlight%28100%29%28highlight%285%29+-+x%29+=+470+-+y+ = +highlight%28100%29%285+-+highlight%28x%29%29+=+470+-+y+
(remember the signs
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+100%285%29+-+100%28x%29+=+470+-+y+ = +500+-+100x+=+470+-+y+, we don't have to move anything right now, so this is the simplified form
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Equation 2, we will just multiply the "90" and "x"
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+90%28x%29+=+y+ = +90x+=+y+, this is simplified
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Now we can start solving for the two unknowns
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+system+%28+500+-+100x+=+470+-+y%2C+90x+=+y%29+
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This is a system of equations, the easiest way to solve this system is by substitution
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Since +y+=+90x+, we can just simply replace "y" in the first equation with "90x"
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+500+-+100x+=+470+-+y+ = +500+-+100x+=+470+-+%2890x%29+
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+500+-+100x+=+470+-+%2890x%29+ = +500+-+100x+=+470+-+90x+, now we can just solve for "x", we will move (-100x) to the right side
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+500+-+100x+=+470+-+90x+ = +500+-+100x+%2B+100x+=+470+-+90x+%2B+100x+ = +500+=+470+%2B+10x+, now we will move "470" to the left side
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+500+=+470+%2B+10x+ = +500+-+470+=+470+-+470+%2B+10x+ = +30+=+10x+, to find "x" we will divide each side by "10"
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+30+=+10x+ = +30%2F10+=+10x%2F10+ = +3+=+highlight%2810%29x%2Fhighlight%2810%29+ = +3+=+cross%2810%29x%2Fcross%2810%29+ = +3+=+x+, +x+=+3+
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(remember "x" is the time in hours), we can replace "x" with "3" in the second equation, to find "y"
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+90x+=+y+ = +90%283%29+=+y+ = +270+=+y+, +y+=+270+, we can check our two answers by replacing "x" and "y" in the first equation
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x = 3
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y = 270
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+500+-+100x+=+470+-+y+ = +500+-+100%283%29+=+470+-+%28270%29+ = +500+-+300+=+470+-+270+ = +200+=+200+ (True)
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Here are the simplified equations with the numbers
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+system+%28+100%285+-+3%29+=+470+-+270%2C+90%283%29+=+270%29+
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+system+%28+100%282%29+=+200%2C+90%283%29+=+270%29+
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If we look at the formula, +rate%28time%29+=+distance+
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+system+%28+100%282%29+=+200%2C+90%283%29+=+270%29+
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This means that the car went 200 km at 100 km/h, and it went 100 km/h for 2 hours
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The car went 270 km at 90 km/h, and it went 90 km/h for 3 hours
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The distances add up to the total distance, as well as the time with the total time
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The car went 200 km at 100 km/h
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The car went 270 km at 90 km/h
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Hope I helped, Levi

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A car travelled 470 km from Sudbury to Brandford in 5 hours. For part of the trip, the car traveled at 100 km/h. For the rest of the trip it traveled at 90 km/h. How far did the car travel at each speed?
--------------------------
100 kph section DATA:
Distance = x km ; rate = 100 kph ; time = d/r = x/100 hrs.
--------------------
90 kph section DATA:
Distance = 470 - x km ; rate = 90 kph ; time = d/r = (470-x)/90 hrs
==========================
EQUATION:
time + time = 5 hrs
x/100 + (470-x)/90 = 5
90x + 100(470-x) = 450*100
90x + 100*470 - 100x = 450*100
-10x = -20*100
x = 200 km (distance traveled at 100 kph)
470-x = 270 km (distance traveled at 90 kph)
================================================
Cheers,
Stan H.