document.write( "Question 336059: a father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. the present age of the father is. \n" ); document.write( "
Algebra.Com's Answer #241004 by mananth(16946)![]() ![]() You can put this solution on YOUR website! let son be x years \n" ); document.write( "father = 2x \n" ); document.write( ".. \n" ); document.write( "20 years ago \n" ); document.write( "son = x-20 \n" ); document.write( "father = 2x-20 \n" ); document.write( ".. \n" ); document.write( "2x-20=12*(x-20) \n" ); document.write( "2x-20 = 12x -240 \n" ); document.write( "add 20 to both sides \n" ); document.write( "2x = 12x-240+20 \n" ); document.write( "2x=12x-220 \n" ); document.write( "add -12x to both sides \n" ); document.write( "2x-12x= 12x-12x-220 \n" ); document.write( "-12x+2x= -240+20 \n" ); document.write( "-10x= -220 \n" ); document.write( "divide by -10 \n" ); document.write( "x= -220/-10 \n" ); document.write( "x=22 son's age \n" ); document.write( "Father's age = 2*22 = 44 years \n" ); document.write( " \n" ); document.write( " |