document.write( "Question 1203174: Six years ago, Lydia was three times as old as Keiko. In three years time, Lydia will be twice as old as Keiko. What are their present ages? \n" ); document.write( "
Algebra.Com's Answer #838539 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "Six years ago, Lydia was three times as old as Keiko. In three years time, Lydia will be twice as old as Keiko. What are their present ages?\r\n" );
document.write( "\r\n" );
document.write( "Let Lydia and Keiko's ages be L and K, respectively\r\n" );
document.write( "Then we get L - 6 = 3(K - 6)____L - 6 = 3K - 18___L - 3K = - 12 ----- eq (i)\r\n" );
document.write( "Also, L + 3 = 2(K + 3)_____L + 3 = 2K + 6_____L - 2K = 3 ----- eq(ii)\r\n" );
document.write( "Subtracting eq (i) from eq (ii), we find that Keiko, or K = 12 + 3 = 15 years-old\r\n" );
document.write( "\r\n" );
document.write( "Substituting 15 for K in eq (ii), we get: L - 2(15) = 3 \r\n" );
document.write( "                                             L - 30 = 3\r\n" );
document.write( "                                       Lydia, or L = 3 + 30 = 33 years-old
\n" ); document.write( "
\n" );