document.write( "Question 795430: This year, Jake is 5 years older than his sister. Three years ago, Jake was twice his sister's age. How old is Jake's sister? \n" ); document.write( "
Algebra.Com's Answer #480968 by DrBeeee(684)\"\" \"About 
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Let j = Jake's current age
\n" ); document.write( "Let s = his sisters current age
\n" ); document.write( "The first statement gives us
\n" ); document.write( "(1) j = s + 5
\n" ); document.write( "The second statement about 3 years ago we have
\n" ); document.write( "(2) j = 2*s, but this is 3 years ago , so we have
\n" ); document.write( "(3) j-3 = 2*(s-3) OK? or
\n" ); document.write( "(4) j = 2*s -6 +3 or
\n" ); document.write( "(5) j = 2*s -3
\n" ); document.write( "Now solve (1) and (5) simultaneously.
\n" ); document.write( "Since j = j we can set (1) = (5) to get
\n" ); document.write( "(6) s + 5 = 2*s - 3 or
\n" ); document.write( "(7) s = 8
\n" ); document.write( "From (1) we get
\n" ); document.write( "(8) j = 8 + 5 or
\n" ); document.write( "(9) j = 13
\n" ); document.write( "Let's check this using (3).
\n" ); document.write( "Is (13 - 3 = 2*(8 - 3))?
\n" ); document.write( "Is (10 = 2*5)?
\n" ); document.write( "Is (10 = 10)? Yes
\n" ); document.write( "Answer: This year Jake's sister is 8 years old. \n" ); document.write( "
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