document.write( "Question 535595: Nina is 6 years older than Francine is. Three years ago, Nina was four times as old as Francine was then. What is the present age of each? \n" ); document.write( "
Algebra.Com's Answer #351955 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let x=Francine's age now; x-3=Francine's age 3 yrs ago \n" ); document.write( "Then x+6=Nina's age now; x+6-3=x+3=Nina's age 3 yrs ago \n" ); document.write( "Soooo, our equation to solve is: \n" ); document.write( "x+3=4(x-3) get rid of parens \n" ); document.write( "x+3=4x-12 \n" ); document.write( "subtract 3 and also 4x from each side \n" ); document.write( "x-4x+3-3=4x-4x-12-3 \n" ); document.write( "-3x=-15 \n" ); document.write( "x=5----Francine's age now \n" ); document.write( "x+6=5+6=11 Nina's age now\r \n" ); document.write( "\n" ); document.write( "CK \n" ); document.write( "\"Nina is 6 yrs older than Francine is\"---11-5=6---OK \n" ); document.write( "\"3 Yrs ago, Nina was 4 times as old as Francine\"---(11-3)=4(5-3)--OK\r \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " \n" ); document.write( " |