SOLUTION: A man walks a certain distance due North and then the same distance plus a further 7km due East. If the final distance from the starting point is 17km, find the distances he walks

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Question 530736: A man walks a certain distance due North and then the same distance plus a further 7km due East. If the final distance from the starting point is 17km, find the distances he walks North and East.
Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!
He walks x km north and x+7 km east.
At this point he is 17 km from the starting point.
This is the measure of the hypotenuse of a right triangle.
.
c^2 = a^2 + b^2
.
17^2 = x^2 + (x+7)^2
.
289 = x^2 + x^2 + 14x + 49
.
2x^2 +14x - 240 = 0
.
x^2 +7x -120 = 0
.
(x+15)(x-8) = 0
.
x = -15 or 8.
.
You cannot walk a negative distance, so x = 8 is the solution.
.
Always check your solution.
He walks 8 km north and then 8+7 = 15 km east.
8^2 + 15^2 = 64 + 225 = 289
sqrt(289) = 17
Correct.
.
Done.

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