document.write( "Question 519525: There is a number of $2 and $5 coins. It is known that 1/3 of the number of $2 coins is greater than that of $5 coins by 1. If the total value of the coins is $39, (a) find the number of $2 coins. (b）find the total value of $5 coins \n" ); document.write( "
Algebra.Com's Answer #345669 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
$2 coins = x
\n" ); document.write( "$5 coins y\r
\n" ); document.write( "\n" ); document.write( "x/3=y+1
\n" ); document.write( "x=3(y+1)
\n" ); document.write( "x=3y+3\r
\n" ); document.write( "\n" ); document.write( "2x+5y=39\r
\n" ); document.write( "\n" ); document.write( "1 x -3 y = 3 .............1
\n" ); document.write( "2 x + 5 y = 39 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by 5
\n" ); document.write( "Multiply (2) by 3
\n" ); document.write( "5 x -15 y = 15
\n" ); document.write( "6 x 15 y = 117
\n" ); document.write( "Add the two equations
\n" ); document.write( "11 x = 132
\n" ); document.write( "/ 11
\n" ); document.write( "x = 12.00 12
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1 x + -3 y = 3
\n" ); document.write( "12 + -3 y = 3
\n" ); document.write( " -3 y = 3 -12
\n" ); document.write( " -3 y = -9
\n" ); document.write( " y = 3.00
\n" ); document.write( "$ 2 coins 12 numbers
\n" ); document.write( "$5 coins 3 numbers \r
\n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );