document.write( "Question 38940: Allen is 3 times as old as Betty. Catherine is 5 yrs younger than Allen. 3 yrs ago sum of all ages is 56. What is each of their present age? \n" ); document.write( "
Algebra.Com's Answer #24386 by Earlsdon(6294)![]() ![]() ![]() You can put this solution on YOUR website! Start by letting: \n" ); document.write( "A = Allen's present age. \n" ); document.write( "B = Betty's present age. \n" ); document.write( "C = Catherine's present age.\r \n" ); document.write( "\n" ); document.write( "From the problem desciption: \n" ); document.write( "1) A = 3B (Allen is 3 times as old as Betty) \n" ); document.write( "2) C = A-5 (Catherine is 5 years younger than Allen). \n" ); document.write( "3) (A-3)+(B-3)+(C-3) = 56 (Three years ago, the sum of their ages was 56).\r \n" ); document.write( "\n" ); document.write( "Simplify equation 3) \n" ); document.write( "A+B+C-9 = 56 Add 9 to both sides. \n" ); document.write( "A+B+C = 65 Substitute equation 1) for A. \n" ); document.write( "3B+B+C = 65 Rewrite equation 2 as: C = 3B-5 and substitute for C. \n" ); document.write( "3B+B+3B-5 = 65 Simplify. \n" ); document.write( "7B-5 = 65 Add 5 to both sides \n" ); document.write( "7B = 70 Divide both sides by 7. \n" ); document.write( "B = 10 This is Betty's age. \n" ); document.write( "A = 3B = 3(10) = 30 This is Allen's age. \n" ); document.write( "C = A-5 = 30-5 = 25 This is Catherine's age. \n" ); document.write( " |