Question 1190240
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The setup is easy....<br>
Let x = # of quarters
Then 71-x = # of pennies<br>
Total value of the coins in cents: 25(x)+1(71-x) = 24x+71<br>
Number of quarters if he had 3 times as many: 3x
Number of pennies if he had half as many: (1/2)(71-x)<br>
Total value of the new numbers of coins: 25(3x)+1((1/2)(71-x))<br>
The new total value is 1522 cents more than the original:<br>
{{{25(3x)+1((1/2)(71-x))=24x+71+1522}}}<br>
Unfortunately, solving that equation gives a non-integer value for x, which is impossible.<br>
The problem is faulty as given.  Correct the statement of the problem and re-post.<br>
Or take the correct numbers in the problem and set up and solve the problem yourself using the method shown.<br>