SOLUTION: Please help me solve this word problem: Chip and Layla found a hole in the fence and decided to make a run for it at the same time. Chip headed north trotting at x miles per hou

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Question 131749: Please help me solve this word problem:
Chip and Layla found a hole in the fence and decided to make a run for it at the same time. Chip headed north trotting at x miles per hour. Layla ran east, running 2 miles an hour faster than Chip. In 2 hours, they were 20 miles apart (as the crow flies- think hypotenuse). How fast was each dog going?
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
C: x mph
L: x + 2 mph

d=rt

So the distance Chip ran was 2x (2 hours at x miles per hour), the distance Layla ran was 2(x + 2) (2 hours at x + 2 miles per hour). Since their directions of travel were 90 degrees apart, the square of Chip's distance plus the square of Layla's distance equals the square of the straight line distance between the two.

x%5E2%2B%28x%2B2%29%5E2=20%5E2

x%5E2%2Bx%5E2%2B4x%2B4=400

2x%5E2%2B4x-396=0

x%5E2%2B2x-198=0

Use the quadratic formula on this beast.

x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A1%2A%28-198%29+%29%29%2F%282%2A1%29+

x+=+%28-2+%2B-+sqrt%28+796+%29%29%2F2+

x=%28-2+%2B+2%2Asqrt%28199%29%29%2F2 or x=%28-2+-+2%2Asqrt%28199%29%29%2F2

x=-1+%2B++sqrt%28199%29 or x=-1+-+sqrt%28199%29. Second root is < 0, therefore extraneous.

So Chip was going -1%2Bsqrt%28199%29 mph, and Layla managed 1%2Bsqrt%28199%29, a little over 13 and a little over 15 mph respectively.


Of course, I don't believe it for a minute. Some breeds of dog can manage 42 mph in a short sprint, but I would be surprised to learn that there is a breed that can sustain 15 or even 13 mph for 2 hours.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

SOLVER91311'S SOLUTION IS INCORRECT BECAUSE HE
FORGOT TO INCLUDE THE "2 HOURS" IN HIS EQUATION.

Correct solution:



Let R = Chip's northward rate in mi/h

Chip's time is 2 hours

Using DISTANCE = RATE x TIME,

Distance of Chip's northward path = (R)(2) = 2R mi

>>...Layla ran east, running 2 miles an hour faster than Chip...<<

Therefore Layla's eastward rate is R+2

Using DISTANCE = RATE x TIME,

Distance of Layla's eastward path = (R+2)2 = 2(R+2) = 2R+4 mi.

We'll write those distances on the appropriate sides of the 
triangle.



So we use the Pythagorean theorem:

                   a² + b² = c²

           (2R)² + (2R+4)² = 20²

        4R² + (2R+4)(2R+4) = 400

4R² + (4R² + 8R + 8R + 16) = 400

    4R² + (4R² + 16R + 16) = 400

      4R² + 4R² + 16R + 16 = 400

            8R² + 16R + 16 = 400

           8R² + 16R - 384 = 0

Divide every term thru by 8

              R² + 2R - 48 = 0

            (R - 6)(R + 8) = 0

Solutions are R = 6 and R = -8

We discard the negative solution.

Solution:  Chip's rate = R = 6 mi/h
           Layla's rate = R+2 = 6+2 = 8 mi/h

Edwin