document.write( "Question 986193: There was a time when a man was twice the age of his wife but the next year he was only 1.5 years older. If he is 44 how old is his wife? \n" ); document.write( "
Algebra.Com's Answer #607006 by macston(5194)\"\" \"About 
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\n" ); document.write( "x=husband's age when he was twice wife's age; y=wife.s age when husband was twice her age;
\n" ); document.write( "H=husband's current age; W=wife's current age
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\n" ); document.write( "x=2y
\n" ); document.write( "x+1=1.5(y+1)
\n" ); document.write( "2y+1=1.5y+1.5
\n" ); document.write( "0.5y=0.5
\n" ); document.write( "y=1 Wife was one year old when husband was twice her age.
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\n" ); document.write( "x=2y
\n" ); document.write( "x=2(1)=2 Husband was 2 years old when he was twice wife's age.
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\n" ); document.write( "x-y=2-1=1 year Husband is one year older than wife.
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\n" ); document.write( "H-1 year=W
\n" ); document.write( "44 years - 1 year=W
\n" ); document.write( "43 years=W
\n" ); document.write( "ANSWER: Wife is 43 years old.
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