iaesthetic90
Answered

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Domingo picks up 5 coins from his collection of quarters and nickel. Is it possible for him to pick up exactly $0.50
a. write an equation
b. solve the equation
c. does the equation have a solution?
d. does the word problem have a solution?
PLEASE HURRY GUYS ILL GIVE YOU BRAINLIEST!!!!

Sagot :

evgeniylevi

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.

sqdancefan

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

__

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

__

c)

The equations have a solution, but the values are not integers. There is no integer solution.

__

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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