Question 129756
Let his first speed be represented by x and his second speed be represented by y. Then recall that
the distance covered is found by multiplying speed times the time spent traveling at that speed.
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In the first case the motorist drove 2 hours at the first speed and 3 hours at the second speed.
So the distance he traveled at the first speed is x times the 2 hours at that speed and 
is 2x km. And the distance traveled at the second speed is the rate y times the 3 hours that
he traveled at that speed or 3y km. Add these two distances together and set them equal to the 
252 km he covered:
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2x + 3y = 252
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Similarly, had he traveled 4 hours at the first speed he would have gone a distance of 4x km.
And had he then driven for 1 hour at the second speed he would have traveled 1*y or y km. Adding
these two distances together are to result in a total distance of 244 km. In equation form this
is:
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4x + y = 244 km
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So our equation set is:
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2x + 3y = 252 and
4x + y = 244
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Multiply the top equation (all terms on both sides) by 2 and the equation set then becomes:
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4x + 6y = 504
4x + y = 244
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Subtract these two equations in vertical columns. This will eliminate the 4x terms in each
equation and you are left with:
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5y = 260
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Solve for y, the second speed, by dividing both sides of this equation by 5 and you get that
the second speed is:
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y = 260/5 = 52 km per hour
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You can now solve for x, the first speed, by returning to either of the original equations and
substituting 52 for y. Let's use the equation:
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4x + y = 244
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Substitute 52 for y and you have:
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4x + 52 = 244
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Get rid of the 52 on the left side by subtracting 52 from both sides to reduce the equation to:
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4x = 192
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Solve for x by dividing both sides by 4 and you have:
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x = 192/4 = 48
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So we know that the first speed that he traveled at in both trips was 48 km per hour, and the
second part of both trips he traveled at 52 km per hour.
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Hope this helps you to understand the problem.
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