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Question 511488: a laborer was employed to work for a daily wage of $15 and paid $2 per day of board. He was idle on certain days and received no wage for that day time, but was charged for food every day. His net pay at the end of the period is $382. Had his daily wage been $18 and his board $3, the net pay would have been $438. How may days did he work and how many days was he idle?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! income = 15 -2 on days worked
income = -2 on days not worked
x = days worked
y = days not worked
income = 13x -2y
income = 382
13x -2y = 382
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The other option says for the same number of days worked and not worked, so these are the same x and y, he would have earned:
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15x -3y = 438
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Set up the simultaneous equations
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13x -2y = 382
15x -3y = 438
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multiply the first by 3 and the second by 2, then subtract the second from the first
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39x -6y = 1146
30x -6y = 876
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9x +0y = 270
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x = 30
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substitute to find y
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13x -2y = 382
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13(30) -2y = 382
390 -2y = 382
-2y = -8
y = 4
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check the answers to see if these values are correct
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15(30) -3(4) = ??
450 - 12 = 438
correct
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Answer: He worked 30 days and did not work 4 days.
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Done.
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