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)\"\" \"About 
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\n" ); document.write( "With formal algebra... if it is required....

\n" ); document.write( "x = # dimes
\n" ); document.write( "x+8 = # nickels

\n" ); document.write( "\"10x%2B5%28x%2B8%29+=+145\"

\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

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