Determine the values for x,y and z in the following system: 2x+3y-z=5 ; 4x-y-z=-1 ; x+4y+z=12.

Answer:x = 1y = 2z = 3Step-by-step explanation:Given:2x + 3y - z = 5orz = 2x + 3y - 5     ............(1) 4x - y - z = -1       ..........(2)x + 4y + z = 12 ...........(3)substituting 1 in 2, we get4x - y - ( 2x + 3y - 5 ) = -1  or 4x - y - 2x - 3y + 5 = -12x - 4y + 6 = 0orx - 2y + 3 = 0 orx = 2y - 3substituting x in equation 1 , we getz = 2( 2y - 3 ) + 3y - 5orz = 4y - 6 + 3y - 5 z = 7y - 11substituting x and z in 3 we get( 2y - 3 ) + 4y + ( 7y - 11 ) = 12or-14 + 13y = 12ory = 2substituting y in z = 7y - 11, we getz = 7 × 2 - 11orz = 3substituting y in x = 2y - 3we getx = 2 × 2 - 3orx = 1thus,x = 1y = 2z = 3