Alan

Your second equation is x - 2y = 9

If we set a = x - 3 then by adding 3 to each side we have x = a + 3

If we set b = y + 3 then by subtracting 3 from each side we have y = b - 3

Now we replace x and y in x - 2y = 9,

so (a + 3) - 2(b - 3) = 9

or a + 3 -2b + 6 = 9

or a - 2b + 9 = 9

or a - 2b = 0

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As far as I can see the only force that the car applies to the track at point A is the normal force.  

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Assuming conservation of energy:

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You could re-write the second equation as  x-3 -2(y+3) = 0

Let a = x-3,   b = y+3

Now your two equations become

a² + b² = 5

a - 2b = 0  so a = 2b

so

(2b)² + b² = 5  or b² = 1  so  b = ±1  so  a = ±2

Therefore we must have x = a+3 = ±2 + 3  and  y = b-3 = ±1 - 3

or x = 5 and 1  with  y = -2 and -4

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Like this:

T is tension in string; N is normal reaction force; F is force of frictional resistance.

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Melody, the first term in the third line of your derivation should be 0.2X not 0.23X.  

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The prime factors of 80 are 2, 2, 2, 2, 5, so any combination of these multiplied together is a factor of 80.  This means 2, 4 and 5 are factors.  3 is a prime number, so the first non-prime, non-factor is 6.

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You have left out a "y" that multiplies the 350.

(25 - 7y)² +y² = 25

25² - 2*25*7*y + 7²y² + y² = 25

625 - 350y + 50y² = 25

50y² - 350y + 600 = 0

y² - 7y + 12 = 0

(y - 3)(y - 4) = 0

y  = 3 or y = 4

so x = 25 - 7*3 or x = 4  when y = 3

and x = 25 -7*4 or x = -3  when y = 4 (as Heureka illustrates).

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Melody, your 2nd answer is correct!

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