document.write( "Question 543176: The age of a person is twice the sum of the ages of his two sons and five years ago his age was thrice the sum of their ages.find his present age. \n" ); document.write( "
Algebra.Com's Answer #355184 by JBarnum(2146)\"\" \"About 
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f=father a=first son b=second son
\n" ); document.write( "\"f=2%28a%2Bb%29\" this says father is 2 times the age of his two sons combined.
\n" ); document.write( "\"f-5=3%28%28a-5%29%2B%28b-5%29%29\" 5 years ago means you have to subtract 5 years from the father and 2 sons
\n" ); document.write( "\"2%28a%2Bb%29-5=3%28%28a-5%29%2B%28b-5%29%29\"
\n" ); document.write( "\"2a%2B2b-5=3a-15%2B3b-15\"
\n" ); document.write( "\"-5=a%2Bb-30\"
\n" ); document.write( "\"0=a%2Bb-25\"
\n" ); document.write( "\"-a=b-25\"
\n" ); document.write( "\"a=25-b\" this tells us both a and b are less than 25 yrs old
\n" ); document.write( "take this and substitute it for a in the first equation
\n" ); document.write( ".
\n" ); document.write( "\"f=2%28a%2Bb%29\"
\n" ); document.write( "\"f=2%28%2825-b%29%2Bb%29\" b will cancel out
\n" ); document.write( "\"f=2%2825%29\"
\n" ); document.write( "\"f=50\"
\n" ); document.write( ".
\n" ); document.write( "to check use the other equation
\n" ); document.write( "\"f-5=3%28%28a-5%29%2B%28b-5%29%29\" sub f and a
\n" ); document.write( "\"50-5=3%28%28%2825-b%29-5%29%2B%28b-5%29%29\"
\n" ); document.write( "\"45=3%2815%29\"
\n" ); document.write( "\"45=45\"
\n" ); document.write( "correct
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