document.write( "Question 201804: the father's age 10 years ago was 35 years more than twice his son's age. After how many years from now will the father be
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Algebra.Com's Answer #152128 by RAY100(1637)\"\" \"About 
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Let f = father's age 10 yrs ago, and s= son's age 10 yrs ago
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\n" ); document.write( "f = 2s +35,,,,,we can continue to use variable symbols, but it is much clearer to use arbitrary numbers.\r
\n" ); document.write( "\n" ); document.write( "Assume son is 10 at this point,,f= 2(10) +35 =55,,,all 10 yrs ago
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\n" ); document.write( "Adding 10 to both, today they are f=65,,,s=20
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\n" ); document.write( "father will be double if,,, (65+x) = 2(20+x)
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\n" ); document.write( "65+x = 40 +2x
\n" ); document.write( "25 = x,,,,,,,,,that is 25 yrs hence, father is twice age of son
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\n" ); document.write( "checking f= 65 + 25 =90,,,,,son = 20 +25 =45,,,90/45=2,,,ok
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\n" ); document.write( "father will be tripled if ,,,(65+y) = 3 (20 +y)
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\n" ); document.write( "65+y =60 +3y
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\n" ); document.write( "5= 2y
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\n" ); document.write( "2.5 = y,,,,or 2.5 yrs hence father will be triple sons age
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\n" ); document.write( "checking 65 + 2.5 =67.5,,,20+2.5 =22.5,,,,67.5/22.5 =3,,,,ok
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