document.write( "Question 1132373: in 3 years angelo will be 3x as old as his sister Precy, a year ago angelo was 7 times as old as precy how old are they now \n" ); document.write( "
Algebra.Com's Answer #749442 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "in 3 years angelo will be 3x as old as his sister Precy:

\n" ); document.write( "\"a%2B3+=+3%28p%2B3%29\"
\n" ); document.write( "\"a%2B3+=+3p%2B9\"
\n" ); document.write( "\"a+=+3p%2B6\" (1)

\n" ); document.write( "a year ago angelo was 7 times as old as precy:

\n" ); document.write( "\"a-1+=+7%28p-1%29\"
\n" ); document.write( "\"a-1+=+7p-7\"
\n" ); document.write( "\"a+=+7p-6\" (2)

\n" ); document.write( "Set the two expressions for a equal to each other and solve for p:

\n" ); document.write( "\"3p%2B6+=+7p-6\"
\n" ); document.write( "\"12+=+4p\"
\n" ); document.write( "\"p+=+3\"

\n" ); document.write( "Precy is 3 years old; 1 year ago he was 2. 1 year ago Angelo was 7 times as old as precy, so he was 14 1 year ago; so he is now 15.

\n" ); document.write( "ANSWERS: Precy is 3; Angelo is 15.

\n" ); document.write( "If this were a problem on a math competition test where the speed of solving is important, the problem can be solved much faster by logical trial and error.

\n" ); document.write( "(1) Choose an age for Precy 1 year ago and make Angelo's age 7 times that number;
\n" ); document.write( "(2) Add 4 years to each of those ages to get their ages 3 years from now and see if those ages have Angelo 3 times as old a Precy.

\n" ); document.write( "Precy 1 year ago = 1; Angelo 1 year ago = 7; Precy 3 years from now 5, Angelo 3 years from now 11. No; 11 is not 3 times 5.

\n" ); document.write( "Precy 1 year ago = 2; Angelo 1 year ago = 14; Precy 3 years from now 6, Angelo3 years from now 18. YES! 18 is 3 times 6. Their ages 1 year ago were 2 and 14; their ages now are 3 and 15.

\n" ); document.write( "That solution takes a long time to write out; but the mental calculations are quick and easy.
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