document.write( "Question 161939: A year ago, a mother was eight times as old as her daughter. Now, her age is the square of her daughter's age. How old are they now? \n" ); document.write( "
Algebra.Com's Answer #119331 by Alan3354(69443)\"\" \"About 
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A year ago, a mother was eight times as old as her daughter. Now, her age is the square of her daughter's age. How old are they now?
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\n" ); document.write( "M-1 = 8*(D-1) One year ago
\n" ); document.write( "M = D^2 Now
\n" ); document.write( "Sub for M in 1st eqn
\n" ); document.write( "D^2 - 1 = 8*(D-1)
\n" ); document.write( "D^2 - 1 = 8D-8
\n" ); document.write( "D^2 - 8D + 7 = 0
\n" ); document.write( "(D-7)*(D-1) = 0
\n" ); document.write( "D = 7, D = 1
\n" ); document.write( "The D=1 solution isn't reasonable, as one year ago the Daughter would be zero.
\n" ); document.write( "So the daughter is 7, and the mother is 49.
\n" ); document.write( "One year ago, they were 6 and 48.
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