text: (there is a stray "+" before the "=" in the first equation -- I'm going to assume it is just a typo and should not be there)\n\nWhat you have is a system of simultaneous equations.  The assumption is that for some value of each of the variables (x, y, and z) all of these equations will be true -- so that this combination of values is the "solution" to the system of equations.\n\nThere are two basic techniques used to solve such a system.  The first is through substitution, the second through addition.  Sometimes, the best approach is a combination of both:\n\nWe have 3 equations, all of which are assumed to be true for the "solution" values of x, y, and z:\n\n(1): x - 2y + 3z = 9\n(2): -x + 3y = -4\n(3): 2x - 5y + 5z = 17\n\nNow if we look at equation 2, we see it only involves 2 of the 3 variables, x and y.  This means we can re-write it to express each variable in terms of the other:\n\n-x+3y =-4\n-x = -4 - 3y\nx = 3y + 4\n(2a): x = 3y + 4  <<< x defined in terms of y\n\n-x+3y =-4\n3y = -4 + x\ny = (x-4)/3\n(2b): y = (x-4)/3 <<< y defined in terms of x\n\nEquation 2a is definitely cleaner, so SUBSTITUTE that into the other two equations (solving or in this case simplifying by substitution):\n\n(1): x-2y+3z+=9  <<<< Original equation\n3y + 4 - 2y + 3z = 9 <<<< substitute "3y+4" for x\ny + 3z + 4 = 9  <<<< combine like terms\ny + 3z = 5 <<<< remove 4 from each side of the equation\n(1a): y + 3z = 5\n\n(3): 2x-5y+5z=17  <<<< Original equation\n2(3y + 4) - 5y + 5z = 17 <<<< substitute "3y+4" for x\n6y + 8 -5y + 5z = 17 <<<< distribute the multiplication\ny + 5z + 8 = 17  <<<< combine like terms\ny + 5z = 9 <<<< remove 8 from each side of the equation\n(3a): y + 5z = 9\n\nNow, what happens if we subtract ALL of the terms from equation 1a from ALL of the terms in equation 3a?\n\n. (y + 5z = 9)\n- (y + 3z = 5)\n------------------------\n. (y-y) + (5z-3z) = (9-5)\n\nThis simplifies to:\n2z = 4\n\nTherefore z = 4/2\n\nz = 2  <<<< solution for z\n\nGreat!  Now we have one of the three values solved.  To get the others, first SUBSTITUTE that value AND the definition of x in terms of y (equation 2a) back into either equation 1 or equation 3:\n\n(1): x-2y+3z=9\n3y + 4 - 2y + 3(2) = 9\n3y - 2y + 4 + 6 = 9\ny + 10 = 9\ny = 9 - 10\ny = -1 <<<< solution for y\n\nFinally, substitute that value for y back into equation 2 in order to solve for x:\n\n(2): -x + 3y = -4\n-x +3(-1) = -4\n-x -3 = -4\n-x = -1\nx = 1 <<<< solution for x\n\nNow we have values for x (1), y (-1), and z (2).  All we need to do is check our answers to make sure that these values hold true in all three original equations:\n\n(1): x - 2y + 3z = 9\n1 - 2(-1) + 3(2) = 9 ?\n1 - (-2) + 6 = 9?\n1 + 2 + 6 = 9 ?\n9 = 9 ? *** YES ***\n\n(2): -x + 3y = -4\n-1 + 3(-1) = -4 ?\n-1 + (-3) = -4 ?\n-1 - 3 = -4 ?\n-4 = -4 ? *** YES ***\n\n(3): 2x - 5y + 5z = 17\n2(1) - 5(-1) + 5(2) = 17 ?\n2 - (-5) + 10 = 17 ?\n2 + 5 + 10 = 17 ?\n17 = 17 ? *** YES ***\n\nThe values check out for all of the equations, so that is the correct solution.