text: i need help with math? solve the following system x-2y+3z+=9 -x+3y  =-4 2x-5y+5z=17 (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)  what 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.  there 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:  we have 3 equations, all of which are assumed to be true for the "solution" values of x, y, and z:  (1): x - 2y + 3z = 9 (2): -x + 3y = -4 (3): 2x - 5y + 5z = 17  now 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:  -x+3y =-4 -x = -4 - 3y x = 3y + 4 (2a): x = 3y + 4  <<< x defined in terms of y  -x+3y =-4 3y = -4 + x y = (x-4)/3 (2b): y = (x-4)/3 <<< y defined in terms of x  equation 2a is definitely cleaner, so substitute that into the other two equations (solving or in this case simplifying by substitution):  (1): x-2y+3z+=9  <<<< original equation 3y + 4 - 2y + 3z = 9 <<<< substitute "3y+4" for x y + 3z + 4 = 9  <<<< combine like terms y + 3z = 5 <<<< remove 4 from each side of the equation (1a): y + 3z = 5  (3): 2x-5y+5z=17  <<<< original equation 2(3y + 4) - 5y + 5z = 17 <<<< substitute "3y+4" for x 6y + 8 -5y + 5z = 17 <<<< distribute the multiplication y + 5z + 8 = 17  <<<< combine like terms y + 5z = 9 <<<< remove 8 from each side of the equation (3a): y + 5z = 9  now, what happens if we subtract all of the terms from equation 1a from all of the terms in equation 3a?  . (y + 5z = 9) - (y + 3z = 5) ------------------------ . (y-y) + (5z-3z) = (9-5)  this simplifies to: 2z = 4  therefore z = 4/2  z = 2  <<<< solution for z  great!  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:  (1): x-2y+3z=9 3y + 4 - 2y + 3(2) = 9 3y - 2y + 4 + 6 = 9 y + 10 = 9 y = 9 - 10 y = -1 <<<< solution for y  finally, substitute that value for y back into equation 2 in order to solve for x:  (2): -x + 3y = -4 -x +3(-1) = -4 -x -3 = -4 -x = -1 x = 1 <<<< solution for x  now 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:  (1): x - 2y + 3z = 9 1 - 2(-1) + 3(2) = 9 ? 1 - (-2) + 6 = 9? 1 + 2 + 6 = 9 ? 9 = 9 ? *** yes ***  (2): -x + 3y = -4 -1 + 3(-1) = -4 ? -1 + (-3) = -4 ? -1 - 3 = -4 ? -4 = -4 ? *** yes ***  (3): 2x - 5y + 5z = 17 2(1) - 5(-1) + 5(2) = 17 ? 2 - (-5) + 10 = 17 ? 2 + 5 + 10 = 17 ? 17 = 17 ? *** yes ***  the values check out for all of the equations, so that is the correct solution.