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)![]() ![]() You can put this solution on YOUR website! 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( "-------------------- \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. \n" ); document.write( " \n" ); document.write( " |