document.write( "Question 242785: paddy is three times as old as Siobhana. Seven years ago, he was five times as old as Sibohana.How old is Paddy now? \n" ); document.write( "
Algebra.Com's Answer #177748 by oberobic(2304)![]() ![]() ![]() You can put this solution on YOUR website! Start by defining your terms. \n" ); document.write( "p = Paddy's age \n" ); document.write( "s = Sibohana's age \n" ); document.write( "Now we set up the givens: \n" ); document.write( "p = 3s OR Paddy is 3 times Siobhana's age \n" ); document.write( "p -7 = 5(s-7) OR 7 years ago, Paddy was 5 times Siobhana's age \n" ); document.write( "Notice that we show BOTH p-7 and s-7 as their ages 7 years ago. \n" ); document.write( "Substituting the value of p from the first equation into the second: \n" ); document.write( "3s -7 = 5(s - 7) \n" ); document.write( "Multiplying through the right side: \n" ); document.write( "3s -7 = 5s - 35 \n" ); document.write( "Subtracting 3s from both sides: \n" ); document.write( "-7 = 2s - 35 \n" ); document.write( "Adding 35 to both sides: \n" ); document.write( "-7 + 35 = 28 = 2s \n" ); document.write( "Dividing both sides by 2: \n" ); document.write( "14 = s. Siobhana is 14. \n" ); document.write( "We know from the setup that: \n" ); document.write( "p = 3s, so p = 3(14) = 42. Paddy is 42. \n" ); document.write( "Check your work! \n" ); document.write( "Substitute the values into the second equation. \n" ); document.write( "p - 7 = 5(s - 7) \n" ); document.write( "42 - 7 = 5(14 -7) \n" ); document.write( "35 = 5(7), which is true. \n" ); document.write( " |