document.write( "Question 1156188: Mike has $1.45 in dimes and nickels. If he has 8 more nickels than dimes, how many of each coin does he have?
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Algebra.Com's Answer #778869 by greenestamps(13203)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "With formal algebra... if it is required.... \n" ); document.write( "x = # dimes \n" ); document.write( "x+8 = # nickels \n" ); document.write( " \n" ); document.write( "Solve using basic algebra; I leave that to you. \n" ); document.write( "Informally.... \n" ); document.write( "Count the \"extra\" 8 nickels first. That is 40 cents; what is left is equal numbers of dimes and nickels, with a value of 145-40 = 105 cents. \n" ); document.write( "The value of one dime and one nickel is 15 cents; to make the remaining 105 cents, the number of dimes and nickels has to be 105/15 = 7. \n" ); document.write( "ANSWER: 7 dimes and 7+8=15 nickels \n" ); document.write( "CHECK: 7(10)+15(5) = 70+75 = 145 \n" ); document.write( " \n" ); document.write( " |