document.write( "Question 881039: a father is twice as old as his son. 12 years ago the father was 4 times as old as the son was then.
\n" );
document.write( "Determine their present age.
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #531941 by algebrahouse.com(1659)![]() ![]() You can put this solution on YOUR website! x = son's age \n" ); document.write( "2x = father's age {he is twice as old as his son}\r \n" ); document.write( "\n" ); document.write( "x - 12 = son's age 12 years ago \n" ); document.write( "2x - 12 = father's age 12 years ago\r \n" ); document.write( "\n" ); document.write( "2x - 12 = 4(x - 12) {twelve years ago the father was 4 times his son} \n" ); document.write( "2x - 12 = 4x - 48 {used distributive property} \n" ); document.write( "-12 = 2x - 48 {subtracted 2x from each side} \n" ); document.write( "36 = 2x {added 48 to each side} \n" ); document.write( "x = 18 {divided each side by 2} \n" ); document.write( "2x = 36 {substituted 18, in for x, into 2x}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "son is 18 \n" ); document.write( "father is 36 \n" ); document.write( " For more help from me, visit: www.algebrahouse.com \n" ); document.write( " |