document.write( "Question 502348: five years ago chloe was twice as old as danny. seven years hence, three times chloe's age will be four times danny's age .ow old each now? \n" ); document.write( "
Algebra.Com's Answer #338816 by Maths68(1474)\"\" \"About 
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Let
\n" ); document.write( "Present age of Chloe = c
\n" ); document.write( "Present age of Danny = d\r
\n" ); document.write( "\n" ); document.write( "Given
\n" ); document.write( "Five years ago means we have to subtract 5 from their present age
\n" ); document.write( "c-5=d-5
\n" ); document.write( "then
\n" ); document.write( "Chloe was twice as old as Danny
\n" ); document.write( "(c-5)=2(d-5)
\n" ); document.write( "c-5=2d-10
\n" ); document.write( "c=2d-10+5
\n" ); document.write( "c=2d-5...............(1)
\n" ); document.write( "Seven years hence means we have to add 6 in their present age
\n" ); document.write( "c+7=d+7
\n" ); document.write( "then
\n" ); document.write( "Three times Chloe's age will be four times Danny's age
\n" ); document.write( "3(c+7)=4(d+7)
\n" ); document.write( "3c+21=4d+28
\n" ); document.write( "3c=4d+28-21
\n" ); document.write( "3c=4d+7..............(2)
\n" ); document.write( "Put the value of c from (1) to (2)
\n" ); document.write( "3c=4d+7
\n" ); document.write( "3(2d-5)=4d+7
\n" ); document.write( "6d-15=4d+7
\n" ); document.write( "6d=4d+7+15
\n" ); document.write( "6d-4d=22
\n" ); document.write( "2d=22
\n" ); document.write( "divide by 2 both sides of the above equation
\n" ); document.write( "2d/2=22/2
\n" ); document.write( "d=11
\n" ); document.write( "Put the value of d in (1)
\n" ); document.write( "c=2d-5...............(1)
\n" ); document.write( "c=2(11)-5
\n" ); document.write( "c=22-5
\n" ); document.write( "c=17\r
\n" ); document.write( "\n" ); document.write( "Present age of Chloe = c = 17 years
\n" ); document.write( "Present age of Danny = d = 11 years\r
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