Question 767571: John has 21 paper bills consisting of $20, $50, $100. He has twice as many $50 bills as $20 bills and has a total amount of 1380.
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = number of $20 bills
2x = number of $50 bills {he has twice as many $50 bills as $20 bills}
21 - 3x = number of $100 bills {total number of bills minus $20 and $50 bills}
20x + 50(2x) + 100(21 - 3x) = 1380 {value of bill times number of bills equals total value}
20x + 100x + 2100 - 300x = 1380 {used distributive property}
-180x + 2100 = 1380 {combined like terms}
-180x = -720 {subtracted 2100 from each side}
x = 4 $20 bills {divided each side by -180}
2x = 8 $50 bills {substituted 4, in for x, into 2x}
21 - 3x = 9 $100 bills {substituted 4, in for x, into 21 - 3x}
John has 4 $20 bills, 8 $50 bills, and 9 $100 bills
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