document.write( "Question 1132202: Four members of one family have a combined age of 135 years.\r
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document.write( "Jeff is twice as old as his daughter Ann, while Trudy is 5 years younger than her husband, Jeff.\r
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document.write( "Harry is a third of his mother Trudy's age.\r
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document.write( "How old are each of the four members of this family? \n" );
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Algebra.Com's Answer #749159 by addingup(3677)![]() ![]() You can put this solution on YOUR website! J + A + T + H = 135 \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \n" ); document.write( "J = 2A \n" ); document.write( "T = J - 5 = 2A - 5 \n" ); document.write( "H = 1/3T = 1/3(2A - 5) \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \n" ); document.write( "A + 2A + 2A - 5 + 1/3(2A - 5) = 135 \n" ); document.write( "5A - 5 + 2/3A - 1 2/3 = 135 \n" ); document.write( "5 2/3A - 6 2/3 = 135 \n" ); document.write( "5 2/3A = 141 2/3 \n" ); document.write( "A = 141 2/3/ 5 2/3 = 25 \n" ); document.write( "Ann is 25 years old. \n" ); document.write( "Jeff is 25 x 2 = 50 \n" ); document.write( "Trudy is 2(25) - 5 = 45 \n" ); document.write( "Harry is 1/3(45) = 15 \n" ); document.write( ". \n" ); document.write( "The sum of their ages: \n" ); document.write( "25 + 50 + 45 + 15 = 135 Correct \n" ); document.write( " \n" ); document.write( " |