document.write( "Question 627671: a woman is 4 times older than her daughter, 6 years ago the product of their ages was 136. find their present age \n" ); document.write( "
Algebra.Com's Answer #395057 by Maths68(1474)![]() ![]() You can put this solution on YOUR website! Let \n" ); document.write( "Present age of woman = w years old \n" ); document.write( "Present age of daughter = d years old\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "woman is 4 times older than her daughter \n" ); document.write( "w=4d............(1) \n" ); document.write( "6 years ago \n" ); document.write( "woman was = w-6 years old \n" ); document.write( "daughter was = d-6 \n" ); document.write( "then \n" ); document.write( "product of their ages was 136 \n" ); document.write( "(w-6)(d-6)=136 \n" ); document.write( "substitute the value of from (1) to above equation \n" ); document.write( "(4d-6)(d-6)=136 \n" ); document.write( "4d^2-24d-6d+36=136 \n" ); document.write( "4d^2-30d+36=136 \n" ); document.write( "2(2d^2-15d+18)=136 \n" ); document.write( "Divide by 2 both sides \n" ); document.write( "2(2d^2-15d+18)/2=136/2 \n" ); document.write( "2d^2-15d+18=68 \n" ); document.write( "2d^2-15d+18-68=0 \n" ); document.write( "2d^2-15d-50=0 \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "d=10 or d=-2.5 (unacceptable) \n" ); document.write( "so \n" ); document.write( "d=10 \n" ); document.write( "Put the value of d in (1) \n" ); document.write( "w=4d \n" ); document.write( "w=4(10) \n" ); document.write( "w=40\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Present age of woman = w = 40 years old \n" ); document.write( "Present age of daughter = d = 10 years old \n" ); document.write( " \n" ); document.write( " |