document.write( "Question 611044: Ten years from now, A will be twice as old as B. Five years ago, A was three times as old as B. What are the present ages of A and B?
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Algebra.Com's Answer #384860 by Earlsdon(6294)\"\" \"About 
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You can express these two situations algebraically:
\n" ); document.write( "A+10 = 2(B+10) \"Ten years from now (A+10), A will be twice as old as B (will be ten years from now) 2(B+10)\" and
\n" ); document.write( "A-5 = 3(B-5) \"Five years ago (A-5), A was three times as old as B (was five years ago), (3(B-5))\" Solve this for A to get:
\n" ); document.write( "A = 3(B-5)+5 and simplify to get:
\n" ); document.write( "A = 3B-15+5 or
\n" ); document.write( "A = 3B-10 Now substitute into the first equation:
\n" ); document.write( "(3B-10)+10 = 2(B+10) Simplify and solve for B.
\n" ); document.write( "3B =2B+20
\n" ); document.write( "\"highlight%28B+=+20%29\" and...
\n" ); document.write( "A = 3B-10
\n" ); document.write( "A = 3(20)-10
\n" ); document.write( "A = 60-10
\n" ); document.write( "\"highlight%28A+=+50%29\"
\n" ); document.write( "Check:
\n" ); document.write( "A+10 = 2(B+10)
\n" ); document.write( "50+10 = 2(20+10)
\n" ); document.write( "60 = 2(30)
\n" ); document.write( "60 = 60 and...
\n" ); document.write( "A-5 = 3(B-5)
\n" ); document.write( "50-5 = 3(20-5)
\n" ); document.write( "45 = 3(15)
\n" ); document.write( "45 = 45
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