document.write( "Question 358078: hello, we are working with linear equations. i am trying to solve 3 equations with 3 variables each x,y, and z but the substitution method is not working for me. here are equations:
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document.write( "x - y + 3z = 8
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document.write( "3x + y - 2z = -2
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document.write( "2x + 4y + z = 0
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document.write( "I don't really want answers i need to see how. thank you \n" );
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Algebra.Com's Answer #255609 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! There's more than one way. I'll use elimination. \n" ); document.write( "Look for coefficients that make it easiest to eliminate one of the 3 variables. \n" ); document.write( "x - y + 3z = 8 \n" ); document.write( "3x + y - 2z = -2 \n" ); document.write( "2x + 4y + z = 0 \n" ); document.write( "----------- \n" ); document.write( "1*y is in the 1st and 2nd, so eliminate it first \n" ); document.write( "x - y + 3z = 8 \n" ); document.write( "3x + y - 2z = -2 \n" ); document.write( "------------------ Add \n" ); document.write( "4x + z = 6 \n" ); document.write( "Multiply eqn 1 by 4 \n" ); document.write( "4x - 4y +12z = 32 \n" ); document.write( "2x + 4y + z = 0 \n" ); document.write( "----------------- Add \n" ); document.write( "6x + 13z = 32 \n" ); document.write( "Now the x's can be eliminated \n" ); document.write( "Multiply 4x + z = 6 by 3 and the other by 2 \n" ); document.write( "12x + 3z = 18 \n" ); document.write( "12x +26z = 64 \n" ); document.write( "-------------- Subtract \n" ); document.write( "-23z = -46 \n" ); document.write( "z = 2 \n" ); document.write( "------ \n" ); document.write( "Sub for z in 4x + z = 6 \n" ); document.write( "--> x = 1 \n" ); document.write( "----------- \n" ); document.write( "Sub for x and z in one of the original eqns to find y. \n" ); document.write( " \n" ); document.write( " |