how do you use eliminator method or substitution method for three linear questions? like in this problem... 4x+3y+z=-5 2x-3y-z=23 5x+y+5z=0
add equations 1 and equation 2 to eliminate z and y 4x + 3x + z = -5 2x =3y - z = 23 -------------- 6x = 18 then x = 3 substitute into all equations and you have 12 + 3y + z = -5 3y + z = -17 (4) 6 - 3y - z = 23 -3y - z = 17 (5) 15 + y + 5z = 0 y + 5z = -15 (6) Equation (5) + 3 x equation (6) to eliminate y and find z -3y - z = 17 3y + 15z = -45 ---------------- 14z = - 28 z = -2 substitute x = 3 and z = -2 into the 1st equation to find y 12 + 3y - 2 = -5 3y = -15 y = -5 so the solution is x = 3, y = -5 and z = -2 hope this helps
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