document.write( "Question 469233: How do I solve:\r
\n" ); document.write( "\n" ); document.write( "x+y+z=4
\n" ); document.write( "x+y-z=6
\n" ); document.write( "2x-3y+z=-1\r
\n" ); document.write( "\n" ); document.write( "Thank you for your time.
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Algebra.Com's Answer #321920 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" ); document.write( "Hi,
\n" ); document.write( "Different ways to solve this system. EQs lend to elimination, substitution methods
\n" ); document.write( "x+y+z=4
\n" ); document.write( "x+y-z=6\r
\n" ); document.write( "\n" ); document.write( " x +y+z=4
\n" ); document.write( "-x -y+z =-6 |multiplying 2nd Eq thru by -1 to eliminate both x and y
\n" ); document.write( " 2z = -2 z = -1\r
\n" ); document.write( "\n" ); document.write( "2x-3y+(-1)=-1 |substituting z = -1
\n" ); document.write( " x+y+(-1)=4
\n" ); document.write( "therefore:
\n" ); document.write( "2x-3y = 0
\n" ); document.write( " x+y = 5\r
\n" ); document.write( "\n" ); document.write( "2x-3y = 0
\n" ); document.write( "-2x-2y = -10 Multiplying 2 EQ thru by -2 to eliminate x
\n" ); document.write( " -5y = -10 y = 2 and x = 3 (x + y = 5)\r
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