document.write( "Question 233484: A mother is five times as old as her son. If twice the age of the son is 3 years less than half of the mother, how old are they? \n" ); document.write( "
Algebra.Com's Answer #172309 by Theo(13342)\"\" \"About 
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let m = the mother's age.
\n" ); document.write( "let s = the son's age.\r
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\n" ); document.write( "\n" ); document.write( "m = 5*s\r
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\n" ); document.write( "\n" ); document.write( "Twice the age of the son is 3 years less than half of the mother's age.\r
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\n" ); document.write( "\n" ); document.write( "2*s = (1/2)*m - 3\r
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\n" ); document.write( "\n" ); document.write( "replace m from the first equation with 5*s in the second equation to get:\r
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\n" ); document.write( "\n" ); document.write( "2*s = (1/2)*5*s - 3\r
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\n" ); document.write( "\n" ); document.write( "multiply both sides of the equation by 2 to get:\r
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\n" ); document.write( "\n" ); document.write( "4*s = 5*s - 6\r
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\n" ); document.write( "\n" ); document.write( "subtract 4*s from both sides of the equation and add 6 to both sides of the equation to get:\r
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\n" ); document.write( "\n" ); document.write( "s = 6\r
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\n" ); document.write( "\n" ); document.write( "since m = 5*s, this makes m = 30\r
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\n" ); document.write( "\n" ); document.write( "2 * 6 = 12\r
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\n" ); document.write( "\n" ); document.write( "30/2 = 15 - 3 = 12\r
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\n" ); document.write( "\n" ); document.write( "12 is twice the age of the son.
\n" ); document.write( "12 is half the mother's age minus 3\r
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\n" ); document.write( "\n" ); document.write( "since they're the same, looks like we're good.\r
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\n" ); document.write( "\n" ); document.write( "mother is 30 years old.
\n" ); document.write( "son is 6 years old.\r
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