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Sagot :
Try this option:
1) if the number of nickel coins is 'n', of quarter coins is 'q', then according to the condition 5 coins is: n+q=5;
2) total sum is 50 cents can be written as 5n+25q=50;
3) Using the two equations it is possible to make up the next system:
[tex]\left \{ {{n+q=5} \atop {5n+25q=50}} \right. \ => \ \left \{ {{n=3.75} \atop {q=1.25}} \right.[/tex]
Finally,
a. see item 3 above;
b. see item 3 above;
c. one solution, see item 3 above;
d. no, it is not possible to have 1.25 and 3.75 coins, n&q∈Z.
Answer:
a) system of equations: q+n=5; 25q+5n=50
b) by elimination, (q, n) = (1.25, 3.75)
c) no integer solution
d) no solution
Step-by-step explanation:
a)
Let q and n represent numbers of quarters and nickels, respectively. The problem tells us two desired relations:
q + n = 5 . . . . . . . . . number of coins
25q +5n = 50 . . . . value in cents
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b)
Subtracting 5 times the first equation from the second gives ...
(25q +5n) -5(q +n) = (50) -5(5)
20q = 25
q = 25/20 = 1.25
n = 5 -1.25 = 3.75
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c)
The equations have a solution, but the values are not integers. There is no integer solution.
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d)
There is no solution to the word problem that involves integer numbers of coins.
_____
Additional comment
The problem is not as hard as it is made out to be. Each coin is an odd number of cents. An odd number of coins will total an odd number of cents, so cannot total 50 cents (an even number).
There cannot be a solution to this problem.
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