SOLUTION: Stuart and Dave each had some money. If Stuart spends 3 times the amount Dave spends each day, Stuart would have left $280 by the time Dave spends all his money. If Stuart spends 1

Question 1208421: Stuart and Dave each had some money. If Stuart spends 3 times the amount Dave spends each day, Stuart would have left $280 by the time Dave spends all his money. If Stuart spends 1/3 the amount that Dave spends each day, Stuart would have $568 by the time Dave spends all his money. How much does each of them have?
Answer by ikleyn(52800) (Show Source): You can put this solution on YOUR website!
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Stuart and Dave each had some money. If Stuart spends 3 times the amount Dave spends each day,
Stuart would have left $280 by the time Dave spends all his money.
If Stuart spends 1/3 the amount that Dave spends each day, Stuart would have $568 by the time
Dave spends all his money. How much does each of them have?
~~~~~~~~~~~~~~~~~~~~~

Let S be initial amount of Stuart's money; D is the initial amount of Dave's money.


Let x (dollars per day) be the regular rate at which Dave spends his money each day.
Then in the first scenario, the rate at which Stuart spends his money is 3x dollars per day.


For the first scenario, we have these equations 

    S - 3x*d   = 280,     (1)  

    D - x*d =  0,         (2)

where d is the number of days in the first scenario.


We can exclude the unknown d from equations (1) and (2).  For it, multiply equation (2) by 3 (both sides).
Keep equation (1) as is.  You will get

     S - 3x*d = 280,     (1')

    3D - 3x*d =   0.     (2')


Now from eq.(1') subtract equation (2').  You will get

    S - 3D = 280.    (3)


In the second scenario, Dave spends x dollars per day;  Stuart spends    dollars per day.

For the second scenario, the equations are

     = 568,    (4)

    D - x*d = 0.                (5)


Multiply equation (4) by 3; keep equation (5) as is.

    3S - x*d = 1704,            (4')

    D - x*d = 0.                (5')


Subtract eq.(5') from eq.(4').  You will get

    3S - D = 1704.              (6)


Thus we have two equations (3) and (6)

    S  - 3D = 280,              (3)

    3S -  D = 1704.             (6)


From (3), express  S = 3D + 280  and substitute it into (6)

    3(3D+280) - D = 1704,

    9D + 840 - D = 1704,

    9D - D = 1704 - 840,

       8D  =    864,

        D  =    864/8 = 108.


Now from (3)  S = 280 + 3*D = 280 + 3*108 = 604.


ANSWER.  Stuart has $604;  Dave has $108.

Solved.

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