SOLUTION: A coin bank for decoration contained 1570. There were twice as many 5 php coins as 10 php coins and 14 less 1 php coins as 5 php coins. How many of each kind were there?

Question 1064611: A coin bank for decoration contained 1570. There were twice as many 5 php coins as 10 php coins and 14 less 1 php coins as 5 php coins. How many of each kind were there?
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
X-number of 1 php coins
X-number of 5 php coins
Z-number of 10 php coins
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1.
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2.
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3.
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From eq. 2,

So substituting into 1,

So then,

and

Answer by ikleyn(52879) (Show Source): You can put this solution on YOUR website!
.
A coin bank for decoration contained 1570. There were twice as many 5 php coins as 10 php coins and 14 less 1 php coins as 5 php coins.
How many of each kind were there?
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Let T be the number of 10 php coins. Then the number of 5 php coins is 2T, and the number of 1 php coins is (2T-14), according to the condition. The value equation is 10T + 5*(2T) + (2T - 14) = 1570. 10T + 10T + 2T - 14 = 1570, 22T = 1570 + 14, 22T = 1584 ---> T = = 72. 72 of 10-php coins. Now you calculate the rest.

The lesson to learn from this solution: This problem is for ONE unknown.

Not for 3 unknowns and even not for 2. Only for one.

Therefore, the solution MUST be explained adequately.

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As long I live, NEVER saw the coin problem for 3 unknowns.

Coin problems are intended, as a rule, for young students, who even don't know about systems of equations.

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