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Title: Algebra Word Problem - Ages Post by TruthlessHero on Apr 21st, 2007, 10:23pm The ages of Old and Young total 48. Old is twice as old as Young was when Old was half as old as Young will be when Young is three times as old as Old was when Old was three times as old as Young. Anyone? |
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Title: Re: Algebra Word Problem - Ages Post by Eigenray on Apr 22nd, 2007, 10:34am Break it down into [hide]a system of 5 equations in 5 unknowns[/hide]: [hide]Old is twice is old as Young was T1 years ago. T1 years ago, Old was half as old as Young will be T2 years from now. In T2 years, Young will be three times as old as Old was T3 years ago. T3 years ago, Old was three times as old as Young.[/hide] |
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Title: Re: Algebra Word Problem - Ages Post by TruthlessHero on Apr 22nd, 2007, 10:53am That's a little more complex than I ended up doing it. [hide] O + Y = 48 O = 48 - Y Y = 48 - O g = age gap = O - Y = O - (48 - O) = O - 48 + O = 2O - 48 x = number of years since Old was three times as old as Young >>Old is twice as old as Young was when Old was half as old as Young will be when Young is three times as old as (O-x = 3(Y-x)). Solve for x, to be used later: O-x = 3(Y-x) O-x = 3Y - 3x 48 - y - x = 3Y - 3x 48 + 2x = 4Y x = 2Y-24 >>Old is twice as old as Young was when Old was half as old as 9(Y-x). >>Old is twice as old as Young was when Old was (9(Y-x))/2. >>Old is twice as old as (9(Y-x))/2 -g. >>Old is 2((9(Y-x))/2 -g). Replace x with 2Y-24, Y with 48 - O, and g with 2O - 48. O = 9(Y-x) - 2g O = 9(Y - 2Y + 24) - 2g = 9(24 - Y) -2g O = 9(24 - (48 - O)) - 2g = 9(24 - 48 + O) - 2g = 9(O-24) - 2g O = 9O - 216 - 2g 8O = 216 + 2g = 236 + 2(2O - 48) = 4O + 216 - 96 4O = 120 O = 30 Y = 18 >>Old is 30. Young is 18.[/hide] I guess mine is lengthier, but it worked :P. |
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