document.write( "Question 13789: X is 6 years younger than Y, and the average of their ages is twice X age 5 years ago, how old are they? \n" ); document.write( "

Algebra.Com's Answer #6945 by LilSkittleMd(119) $\"\"$ $\"About$

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First set up equations
\n" ); document.write( "x=y-6
\n" ); document.write( "(x+y)/2=2x-5
\n" ); document.write( "Substitute \"y-6\" where any x's are
\n" ); document.write( "(y-6+y)/2=2(y-6)-5
\n" ); document.write( "Combine like terms,distribute
\n" ); document.write( "2y-6/2=2y-12-5
\n" ); document.write( "Factor 2y-6 to cancel the 2 so there is no fraction
\n" ); document.write( "2(y-3)/2=2y-17
\n" ); document.write( "The two 2's cancel each other out so you are left with
\n" ); document.write( "y-3=2y-17
\n" ); document.write( "subtract y from both sides
\n" ); document.write( "-3=y-17
\n" ); document.write( "add 17 to both sides
\n" ); document.write( "14=y
\n" ); document.write( "x=14-6
\n" ); document.write( "x=8
\n" ); document.write( "X is 8 years old, Y is 14 years old \n" ); document.write( "

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