document.write( "Question 999539: It says to solve each system by substitution and elimination but I don't understand on how to do it.
\n" ); document.write( "Y=-3x+4
\n" ); document.write( "Y=3x-2 \n" ); document.write( "
Algebra.Com's Answer #617142 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
3 x + 1 y = 4 .............1
\n" ); document.write( "Total value
\n" ); document.write( "-3 x + 1 y = -2 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by -1
\n" ); document.write( "Multiply (2) by 1
\n" ); document.write( "-3 x -1 y = -4
\n" ); document.write( "-3 x + 1 y = -2
\n" ); document.write( "Add the two equations
\n" ); document.write( "-6 x = -6
\n" ); document.write( "/ -6
\n" ); document.write( "x = 1
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "3 x + 1 y = 4
\n" ); document.write( "3 + y = 4
\n" ); document.write( " y = 4 -3
\n" ); document.write( " y = 1
\n" ); document.write( " y = 1
\n" ); document.write( "x= 1
\n" ); document.write( "y= 1 \r
\n" ); document.write( "\n" ); document.write( "SUBSTITUTION METHOD
\n" ); document.write( "
\n" ); document.write( " 3 x + 1 y = 4 .............1
\n" ); document.write( " 1 y = 4 + -3 x
\n" ); document.write( " / 1 =
\n" ); document.write( " y = 4 + -3 x
\n" ); document.write( " -3 x + 1 y = -2 .....................2
\n" ); document.write( " Substitute y in (2)
\n" ); document.write( " -3 x + 1 ( 4 + -3 x ) = -2
\n" ); document.write( " -3 x + 4 + -3 x = -2
\n" ); document.write( " -3 x -3 x = -6
\n" ); document.write( " -6 x = -6
\n" ); document.write( " / -6
\n" ); document.write( " x= 1
\n" ); document.write( " Plug the value of x in (1)
\n" ); document.write( " 3 x + 1 y = 4
\n" ); document.write( " 3 * 1 + 1 y = 4
\n" ); document.write( " 3 + 1 y = 4
\n" ); document.write( " 1 y = 1
\n" ); document.write( " / 1
\n" ); document.write( " y= 1
\n" ); document.write( " m.ananth@hotmail.ca
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " \n" ); document.write( "
\n" );