document.write( "Question 206804: Half of Henry's age added to 1/3 of Daisy's age is 11 years. Six years from now the sum of their ages will be 40 years how old is each? \n" ); document.write( "
Algebra.Com's Answer #156340 by profemmanuel2q@yahoo.com(15)\"\" \"About 
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Half of Henry's age added to 1/3 of Daisy's age is 11 years. Six years from now the sum of their ages will be 40 years how old is each?\r
\n" ); document.write( "\n" ); document.write( "Let Henry's age be x and Daisy's age be y
\n" ); document.write( "Half of Henry's age added to 1/3 of Daisy's age is 11 years can be expressed as 1/2x + 1/3y = 11 ---- Eqn 1 Multiply eQN - 1 both 6 to cancel the denominator
\n" ); document.write( "3x + 2y = 66 ---- 1
\n" ); document.write( "Six years from now the sum of their ages will be 40 years
\n" ); document.write( "Henry will be (x + 6) and Daisy will (y + 6), Their sum will be x + 6 + y + 6 = 40
\n" ); document.write( "x + y = 40 - 6 - 6
\n" ); document.write( "x + y = 28 --- Eqn 2
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\n" ); document.write( "Joining Eqn 1 and Eqn. 2
\n" ); document.write( "3x + 2y = 66 ---- 1
\n" ); document.write( "x + y = 28 --- Eqn 2
\n" ); document.write( "Multiply Eqn 2 by 2
\n" ); document.write( "3x + 2y = 66 ---- 1
\n" ); document.write( "2x + 2y = 56 ---- 2
\n" ); document.write( "Eqn 1 - Eqn 2
\n" ); document.write( "x = 10
\n" ); document.write( "Put x = 10 into Eqn. 1
\n" ); document.write( "3(10) + 2y = 66
\n" ); document.write( "30 + 2y = 66
\n" ); document.write( "2y = 66 - 30
\n" ); document.write( "2y = 36
\n" ); document.write( "Divide both sides by 2
\n" ); document.write( "y = 18\r
\n" ); document.write( "\n" ); document.write( "Hence, Henry is 10 years and Daisy is 18 years
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