document.write( "Question 245670: Four years ago, Katie was twice as old as Anne was then. In 6 years, Anne will be the same age that katie is now. How old is each now? \n" ); document.write( "
Algebra.Com's Answer #179505 by College Student(505)\"\" \"About 
You can put this solution on YOUR website!
Four years ago:
\n" ); document.write( "Anne = x
\n" ); document.write( "Katie = 2x
\n" ); document.write( ".
\n" ); document.write( "Today (four years later)
\n" ); document.write( "Anne = x+4
\n" ); document.write( "Katie = 2x+4
\n" ); document.write( ".
\n" ); document.write( "In six years Anne will be the same age Katie is now. So our equation becomes:
\n" ); document.write( "\"x%2B4%2B6=2x%2B4\"
\n" ); document.write( "Combine like terms and solve for x to find today's Anne's age.
\n" ); document.write( "I'll let you take it from here. :)
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