document.write( "Question 1079508: Al is ten years older than Jerry. At one point Al was three times Jerry's age. How old was Jerry when Al three times older than him?
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Algebra.Com's Answer #693833 by ankor@dixie-net.com(22740)\"\" \"About 
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Al is ten years older than Jerry.
\n" ); document.write( "a = j + 10
\n" ); document.write( " At one point Al was three times Jerry's age.
\n" ); document.write( "let y = \"the point\"
\n" ); document.write( "a - y = 3(j-y)
\n" ); document.write( "a = 3j - 3y + y
\n" ); document.write( "a = 3j - 2y
\n" ); document.write( "replace a with (j+10)
\n" ); document.write( "j + 10 = 3j - 2y
\n" ); document.write( "10 = 3j - j - 2y
\n" ); document.write( "10 = 2j - 2y
\n" ); document.write( "simplify, divide by 2
\n" ); document.write( "5 = j - y
\n" ); document.write( "We can write it:
\n" ); document.write( "y = j - 5
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\n" ); document.write( "The problem with this problem is there are many solutions:
\n" ); document.write( "let a = 40 yrs old, then we know j = 30 yrs, find y
\n" ); document.write( "y = 30 - 5
\n" ); document.write( "y = 25 yrs ago Al(40) was 3 times Jerry (30)
\n" ); document.write( "check this
\n" ); document.write( "40 - 25 = 3(30-25)
\n" ); document.write( "15 = 3(5)
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\n" ); document.write( "If a = 20, then j = 10, find y
\n" ); document.write( "y = 10 - 5
\n" ); document.write( "y = 5
\n" ); document.write( "20 - 5 = 3(10 - 5)
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\n" ); document.write( "How about a = 27, then j = 17 and y = 12
\n" ); document.write( "27 - 12 = 3(17 - 12)\r
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