In this example, notice that each of our variables, x, y, and z, appears in all three equations. To solve this system, we use the addition method. In other words, let’s start with our first two equations, x + y + z = 4, and x – y + z = 2. Notice that if we add these equations together, the +y and –y will cancel out, and we have 2x + 2z = 6. So, in our new equation, 2x + 2z = 6, we’ve eliminated the variable y. Unfortunately, we still haven’t solved for any of our variables. However, if we can create another equation with just x and z in it, then we’ll have a system of equations in two variables, which we can use to solve for x and z. To create another equation with just x and z in it, we need to eliminate y. We can’t add the first and second equations together, because we’ve already done that. However, notice that if we add the first and third equations together, the first equation has a +y and the third equation has a –y, so we’ll be able to eliminate the y. So we have our first equation, x + y + z = 4, and our third equation, x – y – z = 0, and adding them together, notice that the +y – y cancels out, and, as a bonus, the +z – z also cancels out, so we have 2x = 4, and dividing both sides by 2, x = 2. Now, since we know that x = 2, notice that if we plug a 2 in for x in the equation that we created earlier, we can solve for z. And we have 2(2) + 2z = 6, or 4 + 2z = 6, and subtracting 4 from both sides, we have 2z = 2, and dividing both sides by 2, z = 1. So x = 2 and z = 1, and to find the value of y, we simply plug our values of x and z into any of the equations in the original system. Let’s use the first equation, x + y + z = 4. Since x = 2 and z = 1, we plug a 2 in for x and a 1 in for z, and we have 2 + y + 1 = 4, or 3 + y = 4, and subtracting 3 from both sides, y = 1. So x = 2, y = 1, and z = 1, and finally, we write our answer as the ordered triple, x, y, z, or (2, 1, 1).