Algebra.Com's Answer #705441 by MathTherapy(10552)  You can put this solution on YOUR website! Can someone help me? \n" );
document.write( "I need to write the final answer as an ordered triple (x,y,z) \n" );
document.write( "Solve the system for x,y,z \n" );
document.write( "2x+3y-z=4 \n" );
document.write( "3x-y+2z=5 \n" );
document.write( "x-4y+3z=1 \n" );
document.write( "I tried it, but ended up weird. \n" );
document.write( "Thank you. \n" );
document.write( "2x + 3y - z = 4 -------- eq (i) \n" );
document.write( "3x - y + 2z = 5 -------- eq (ii) \n" );
document.write( " x - 4y + 3z = 1 -------- eq (iii)\r \n" );
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document.write( "I can clearly see what you're saying! \n" );
document.write( "The EASIEST method to be used here is ELIMINATION, as follows:\r \n" );
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document.write( "1) Multiply eq (i) by 2 to get: 4x + 6y - 2z = 8 ---- eq (iv) \n" );
document.write( "2) Add eqs (ii) & (iv) to get: 7x + 5y = 13 --------- eq (v) \r \n" );
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document.write( "1) Multiply eq (i) by 3 to get: 6x + 9y - 3z = 12 --- eq (vi) \n" );
document.write( "2) Add eqs (iii) & (vi) to get: 7x + 5y = 13 -------- eq (vii) \r \n" );
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document.write( "Do you notice anything? If not, it's clear that eqs (v) & (vii) are the same \n" );
document.write( "Therefore, there are an INFINITE number of solutions to this system!\r \n" );
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document.write( "If you want to see what I'm talking about, substitute a value for x in eq (v) or (vii), and then solve for y. You will then have values for x and y. \n" );
document.write( "Substitute these 2 values in any of the 3 ORIGINAL equations (I would choose eq 1 since that equation has a z-value with a coefficient of - 1), and then solve for z. \n" );
document.write( "You now have values for x, y, and z. \n" );
document.write( "Substitute these 3 values into all 3 ORIGINAL equations, and you will see that these values satisfy all 3. \n" );
document.write( "You can follow the same process if you so choose to use a different value for x. \n" );
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