document.write( "Question 186542: Solve the System.\r
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document.write( "2x+3y+2z=1
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document.write( "2x-3y+2z=-1
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document.write( "4x+3y-2z=4\r
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document.write( "x=
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document.write( "y=
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document.write( "z= \n" );
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Algebra.Com's Answer #139798 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! 2x + 3y + 2z = 1 \n" ); document.write( "2x - 3y + 2z =-1 \n" ); document.write( "4x + 3y - 2z = 4 \n" ); document.write( ": \n" ); document.write( "Use elimination with subtraction on 1st two equations \n" ); document.write( "2x + 3y + 2z = 1 \n" ); document.write( "2x - 3y + 2z =-1 \n" ); document.write( "------------------ \n" ); document.write( "0x + 6y + 0z = 2 \n" ); document.write( "6y = 2 \n" ); document.write( "y = \n" ); document.write( ": \n" ); document.write( "Add the last two equations, eliminate y & z, find x \n" ); document.write( "2x - 3y + 2z =-1 \n" ); document.write( "4x + 3y - 2z = 4 \n" ); document.write( "------------------ \n" ); document.write( "6x + 0y + 0z = 3 \n" ); document.write( "6x = 3 \n" ); document.write( "x = \n" ); document.write( ": \n" ); document.write( "Using the 1st equation, substitute for x & y, to find z \n" ); document.write( "2( \n" ); document.write( "1 + 1 + 2z = 1 \n" ); document.write( "2z = 1 - 2 \n" ); document.write( "z = \n" ); document.write( "; \n" ); document.write( "; \n" ); document.write( "You can check the solutions in the 2nd and 3rd equation \n" ); document.write( " |