document.write( "Question 697072: I know a priest who is 44 years old. Sometime ago, he was twice as old as I. In the following year he was one and a half times older than I.
\n" ); document.write( "How old am I now? \n" ); document.write( "
Algebra.Com's Answer #429648 by ankor@dixie-net.com(22740)\"\" \"About 
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I know a priest who is 44 years old.
\n" ); document.write( " Sometime ago, he was twice as old as I.
\n" ); document.write( " In the following year he was one and a half times older than I.
\n" ); document.write( " How old am I now?
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\n" ); document.write( "We can approach this with this relationship:
\n" ); document.write( "let p = priest's age some time ago
\n" ); document.write( "let a = your age some time ago
\n" ); document.write( "Find the age in which p is twice a, and one year later p is 1.5a
\n" ); document.write( ":
\n" ); document.write( "2a = p
\n" ); document.write( "and one year later
\n" ); document.write( "1.5(a+1) = p + 1
\n" ); document.write( "1.5a + 1.5 = p + 1
\n" ); document.write( "1.5a = p + 1 - 1.5
\n" ); document.write( "1.5a = p - .5
\n" ); document.write( "replace p with 2a
\n" ); document.write( "1.5a = 2a - .5
\n" ); document.write( "1.5a - 2a = -.5
\n" ); document.write( "-.5a = -.5
\n" ); document.write( "a = 1 year old some time ago
\n" ); document.write( "and
\n" ); document.write( "p = 2 yrs old some time ago
\n" ); document.write( "then one year later:
\n" ); document.write( "a = 2, then p = 3, we know that 1.5(2) = 3
\n" ); document.write( "there is only one year difference, therefore when the priest is 44 you are 43\r
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