​Using Elimination

Suppose we are given the following system:​

3x + y - 2z = 10​

​6x - 2y + z = -2

​x + 4y + 3z = 7

​

​These systems can have one single solution (a point), infinitely many solutions (a line), or no solution (no common intersection).

​Using Elimination

​3x + y - 2z = 10​

​6x - 2y + z = -2

​x + 4y + 3z = 7

​

​We can solve for the solution by using elimination. First, we need to choose which variable to eliminate. Let's eliminate z for this example.

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

We should label each equation to keep up with all of our pieces. Let's multiply (b) by 2 so the coefficient of z is the same absolute value as the coefficient of z in (a).

​(6x - 2y + z = -2) x 2 -> 12x - 4y + 2z = -4

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Now, the coefficients of our z values in (a) and (b) match, so we can add them to eliminate z (we add because they have opposite signs already).

​(a) 3x + y - 2z = 10

(b) +(12x - 4y + 2z = -4)

​(a') 15x - 3y = 6

​We'll call our resulting equation (a') and set it aside for now.

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Let's now eliminate the same variable using a different combination of the three equations. Let's use (b) and (c). First, we have to multiply (b) by 3 so that the coefficients match up:

(b) (​6x - 2y + z = -2) x 3 -> 18x - 6y + 3z = -6

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Now, we can eliminate z by subtracting (c) from (b):

(b) 18x - 6y + 3z = -6

​(c) -( x + 4y + 3z = 7)

​(b') 17x - 10y = -13

​We will call this resulting equation (b') and move forward.

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Now, we can use (a') and (b') to eliminate yet another variable. ​Let's eliminate y. But first, we have to match the coefficient of y in each equation. The least common multiple of 3 and 10 is 30. So, we multiply (a') by 10 and (b') by 3.

​(a') (15x - 3y = 6​) x 10 -> 150x - 30y = 60

(b') (17x - 10y = -13) x 3 -> 51x - 30y = -39

​

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Since the signs of y are the same, we will have to subtract the equations:

(a') 150x - 30y = 60

(b') -(51x - 30y = -39)

​ 99x = 99

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Now, we can solve for x:

​ 99x = 99

​ 99 99

​ x = 1

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

We can then use this value of x in either (a') or (b') to solve for the value of the other variable in that equation:

​17x - 10y = -13

​17(1) - 10y = -13

​17 - 10y = -13

​-17 -17

​ -10y = -30

​ -10 -10

​ y = 3

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

Then, we can plug in our values for x and y into any of the first three equations to solve for z:

​(a) 3x + y - 2z = 10

​ 3(1) + 3 - 2z = 10

​ 6 - 2z = 10

​ -6 -6

​ -2z = 4

​ -2 -2

​ z = -2

​Using Elimination

(a)​ 3x + y - 2z = 10​

(b) ​6x - 2y + z = -2

(c) ​x + 4y + 3z = 7

​

So, the solution to this system is (1, 3, -2).