document.write( "Question 132068: The sum of Joe's age and Mary's age is 60 years. Ten years ago, Joe was 3 times as old as Mary was then. How old is each now? \n" ); document.write( "
Algebra.Com's Answer #96447 by solver91311(24713)\"\" \"About 
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Joe is x years old
\n" ); document.write( "Mary is y years old
\n" ); document.write( "We are given that \"x%2By=60\"\r
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\n" ); document.write( "\n" ); document.write( "Ten years ago, Joe was x - 10 years old and Mary was y - 10 years old.
\n" ); document.write( "We are given that \"x-10=3%28y-10%29\"\r
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\n" ); document.write( "\n" ); document.write( "Put the second relationship into standard form (\"ax%2Bby=c\")
\n" ); document.write( "\"x-10=3%28y-10%29\"
\n" ); document.write( "\"x-10=3y-30\"
\n" ); document.write( "\"x-3y=-30%2B10\"
\n" ); document.write( "\"x-3y=-20\"\r
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\n" ); document.write( "\n" ); document.write( "Now we have two equations in standard form:
\n" ); document.write( "\"x%2By=60\"
\n" ); document.write( "\"x-3y=-20\"\r
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\n" ); document.write( "\n" ); document.write( "Multiply the second equation by -1:
\n" ); document.write( "\"-x%2B3y=20\"\r
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\n" ); document.write( "\n" ); document.write( "Add the first equation to the second, term by term:
\n" ); document.write( "\"0x%2B4y=80\"\r
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\n" ); document.write( "\n" ); document.write( "Divide by the coefficient on y:
\n" ); document.write( "\"y=20\"\r
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\n" ); document.write( "\n" ); document.write( "Therefore Mary is 20 years old now. That means that Joe is 40 years old now (60 - 20). Ten years ago, Joe would have been 30 and Mary would have been 10, and 30 is 3 times 10. Answer checks.
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