document.write( "Question 498187: Two years ago, a man was eight times as old as his son. In three years time, he will be four and a half times his son's age then. Find their present ages. \n" ); document.write( "
Algebra.Com's Answer #336936 by ptaylor(2198)\"\" \"About 
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Let x=son's present age
\n" ); document.write( "And let y father's present age
\n" ); document.write( "x-2=son's age 2 yrs ago; x+3=son's age 3 yrs hence
\n" ); document.write( "y-2=fathers age 2 yrs ago; y+3=father's age 3 yrs hence
\n" ); document.write( "Now we are told the following:
\n" ); document.write( "y-2=8(x-2) simplify
\n" ); document.write( "y-2=8x-16
\n" ); document.write( "y-8x=-14-------------------eq1
\n" ); document.write( "and
\n" ); document.write( "y+3=4.5(x+3)simplify
\n" ); document.write( "y+3=4.5x+13.5
\n" ); document.write( "y-4.5x=10.5----------------eq2
\n" ); document.write( "subtract eq1 from eq2
\n" ); document.write( "3.5x=24.5
\n" ); document.write( "x=7 yrs old-----------------present age of son
\n" ); document.write( "from eq1
\n" ); document.write( "y-56=-14
\n" ); document.write( "y=42---------------present age of father
\n" ); document.write( "CK
\n" ); document.write( "two years ago:
\n" ); document.write( "42-2=8*5
\n" ); document.write( "40=40
\n" ); document.write( "3 years from present:
\n" ); document.write( "42+3=(4.5)*10
\n" ); document.write( "45=45\r
\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor\r
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