document.write( "Question 1079856: Fathers is three years older than mother
\n" ); document.write( "Brother is two years older than sister
\n" ); document.write( "Their ages added up is 73
\n" ); document.write( "Four years ago their ages added up to 58
\n" ); document.write( "Find the ages of each person in the family \n" ); document.write( "

Algebra.Com's Answer #694153 by MathTherapy(10552) $\"\"$ $\"About$

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Fathers is three years older than mother
\n" ); document.write( "Brother is two years older than sister
\n" ); document.write( "Their ages added up is 73
\n" ); document.write( "Four years ago their ages added up to 58
\n" ); document.write( "Find the ages of each person in the family
\n" ); document.write( "

If the sum of the 4 people's ages is 73, then 4 years ago, the sum would've been: 73 - 4(4), or 73 - 16 = 57, unless of course,
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document.write( "the youngest person was not born 4 years ago. That's what it is!\r
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document.write( "Let the father’s age, be F, and brother’s, B
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document.write( "Then mother’s is: F – 3, and sister’s is: B – 2
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document.write( "We then get: F + B + F – 3 + B – 2 = 73
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document.write( "2F + 2B – 5 = 73
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document.write( "2F + 2B = 78 ------- eq (i)\r
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document.write( "The youngest is the sister, so it appears that 4 years ago, she was not born so we get:
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document.write( "F – 4 + F – 3 – 4 + B – 4 = 58
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document.write( "2F + B – 15 = 58
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document.write( "2F + B = 73 ------- eq (ii)
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document.write( "B = 78 – 73 ------ Subtracting eq (ii) from eq (i)
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document.write( "B, or \r
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document.write( "2F + 5 = 73 ------- Substituting 5 for B in eq (ii)
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document.write( "2F = 73 – 5
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document.write( "2F = 68
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document.write( "F, or 
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