document.write( "Question 683983: if a boys age and his fathers age amount together to 24 years. fourth part of the product of their ages exceeds the boys age by 9 years.find how old they are. \n" ); document.write( "
Algebra.Com's Answer #423847 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
boy's age =x
\n" ); document.write( "father's age = y\r
\n" ); document.write( "\n" ); document.write( "x+y=24\r
\n" ); document.write( "\n" ); document.write( "(xy)/4 -x=9
\n" ); document.write( "multiply equation by 4
\n" ); document.write( "xy-4x=36
\n" ); document.write( "x(y-4)=36
\n" ); document.write( "x=36/(y-4)\r
\n" ); document.write( "\n" ); document.write( "Now x+y = 24
\n" ); document.write( "substitute x in the equation\r
\n" ); document.write( "\n" ); document.write( "36/(y-4) +y = 24\r
\n" ); document.write( "\n" ); document.write( "multiply equation by (y-4)
\n" ); document.write( "36+y(y-4)=24(y-4)\r
\n" ); document.write( "\n" ); document.write( "36+y^2-4y=24y-96\r
\n" ); document.write( "\n" ); document.write( "y^2-4y-24y+36+96=0\r
\n" ); document.write( "\n" ); document.write( "y^2-28y+132=0\r
\n" ); document.write( "\n" ); document.write( "y^2-22y-6y+132=0\r
\n" ); document.write( "\n" ); document.write( "y(y-22)-6(y-22)=0
\n" ); document.write( "(y-22)(y-6)=0\r
\n" ); document.write( "\n" ); document.write( "y=22 OR 6\r
\n" ); document.write( "\n" ); document.write( "Father's age = 22\r
\n" ); document.write( "\n" ); document.write( "Son's age = 2 (because x+y=24)
\n" ); document.write( "
\n" );