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)\"\" \"About 
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
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