document.write( "Question 933313: if sum of age of three brothers are 42 and before 10 years ratio of their ages are 1:2:3 then what is the age of younger brother.? \n" ); document.write( "
Algebra.Com's Answer #566777 by ankor@dixie-net.com(22740)\"\" \"About 
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before 10 years ratio of their ages are 1:2:3
\n" ); document.write( " let x = the multiplier ten yrs ago
\n" ); document.write( " then 10 yrs ago, their ages:
\n" ); document.write( " 1x, 2x, 3x
\n" ); document.write( "Now
\n" ); document.write( "if sum of age of three brothers are 42
\n" ); document.write( "(x+10) + (2x+10) + (3x+ 10) = 42
\n" ); document.write( "6x + 30 = 42
\n" ); document.write( "6x = 42 - 30
\n" ); document.write( "6x = 12
\n" ); document.write( "x = 12/6
\n" ); document.write( "x = 2 is the multiplier
\n" ); document.write( ":
\n" ); document.write( "then what is the age of younger brother.?
\n" ); document.write( " 2 + 10 = 12 yrs is the age of the younger brother
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "see if the adds up, 10 to their ages 2, 4, 6
\n" ); document.write( "12 + 14 + 16 = 42 \n" ); document.write( "
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