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solve 8x-6z=38, 2x+5y+3z=5, x+10y-4z=8
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Looks like eliminating y from the last two equations will give you a system of two equations in two unknowns. @shizuko
4x - 3z = 19, 2x + 5y + 3z = 5; x + 10y - 4z = 8; [4x + 10y + 6z = 10 ] - [x+10y-4z = 8] => 3x + 10z = 2; 12x - 9z = 57 ; 12x + 40z = 8; => z = -1 ; x = 4; y = 0 ;
@mathmate two unknowns?
x = 4 , y = 0, z = -1
IF you eliminate y from the last two equations. 4x+10y+6z=10 (2 times second equation) x+10y-4z=8 (third equation) Subtract: 3x+10z=2 ....(4) 4x-3z=19 ...(1A, half of (1)) Now you have a system of TWO equations to solve for x and z.
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ohh
no metal?
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