Elimination Method - 3 Variables

Solve System of Equations - Elimination Method - 3 Variables - How it Works - Video

Example 1

Example 1:

Here we have 3 variables. One way to solve it is with the elimination method; another is graphing; another is with matrices.

We are going to use the elimination with these examples.

We have the equations: 4p - 4q + 4r = 4, 4p + q - 2r = 5, -3p - 3q - 4r = -16. When we solve for the system, we have to use all 3 equations.

In order to add/subtract we need the coefficient in front the variable that we want to eliminate to be the same or opposite so 4x and 4x or 4x and -4x for example.

In this example we are going to eliminate the variable r. The 1st equation has 4r and the 2nd equation has -2r so we need to multiple the 2nd equation by 2 so the coefficient in front of r is the same/opposite.

After multiply the 2nd equation by 2, we get 8p + 2q - 4r = 10. Now we can add the first equation, 4p - 4q + 4r = -4, with the new second equation together since the coefficients in front of r are opposite 4 and -4.

4p + 8p = 12p            -4q + 2q = -2q            4r + -4r = 0r            -4 + 10 = 6

So our result is 12p - 2q + 0r = 6, which becomes 12p - 2q = 6.

No we have used the 1st and 2nd equation. We have to use the 3rd equation. You can pick if you want to use the 1st or the 2nd equation. 

Since we are eliminating the variable, r, let's use the first equation and the third equation since they have the same/opposite coefficient, 4r and -4r. Let's add the 1st equation and the 3rd equation.

4p + -3p = 12p            -4q + -3q = -2q            4r + -4r = 0r            -4 + -16 = -20

So our result is 12p - 2q + 0r = 6, which becomes p - 7q = 20.

Now, we have eliminated the r's. We are left with the two equations : 12p - 2q = 6 and p - 7q = 20.

We have the two equations : 12p - 2q = 6 and p - 7q = 20. We can use elimination again so we are left with one variable. Now we can eliminate either the p or the q. If we eliminate the p, then we only have to multiply one equation. If we eliminate the q, we have to multiply two equations. So, let's multiple by 12 to the second equation so we only have to multiply one equation.

Now we have 12p - 2q = 6 and p - 7q = -20. After multiply the 2nd equation by 2, we get 12p - 84q = -20. Now we have coefficients that are the same/opposite, 12p and 12p. Now, we can subtract the two equations and the result is 

12p - 12p = 0p            -2q - -84q = 82q 6 - -240 = 246

So our result is 0p +82q = 246, which becomes 82q = 246. After dividing both sides by 82, we find q to equal 3.

Now that we know q = 3.

We can find p using either one of the equations that have two variables: 12p - 2q = 6 or  p - 7q = -20. Let's use p - 7q = -20 since it has smaller numbers. Now let's find p by substituting 3 for q.

Now that we know q = 3 and p = 1. We can find r. Once again we can use any of the 3 equations. Let's use 4p + q - 2r = 5 since it has smaller numbers. Now let's find p by substituting 3 for q and 1 for p.

Now that we know q = 3, p = 1, and r = 1. We have our answer (1, 3, 1).

Example 2

Example 2:

Here we have 3 variables. One way to solve it is with the elimination method; another is graphing; another is with matrices.

We are going to use the elimination with these examples.

We have the equations: 2p + 2q + 2r = -3, 2p - q - 5r = 1, 4p + 4q + 4r = -8. When we solve for the system, we have to use all 3 equations.

In order to add/subtract we need the coefficient in front the variable that we want to eliminate to be the same or opposite so 4x and 4x or 4x and -4x for example.

Before doing any math, you should look at the equations closely to see if any or multiples of each other. In this example the first equation and the third equation look like they are multiples of each other.

In this example we are going to eliminate the variable p. The 1st equation has 2p and the 2nd equation has 4p so we need to multiple the 1st equation by 2 so the coefficient in front of p is the same/opposite.

After multiply the 1st equation by 2, we get 4p + 4q + 4r = -6. Now we can subtract the 3rd equation, 4p + 4q + 4r = -8, from the new 1st equation since the coefficients in front of p are opposite 4 and 4.

4p - 4p = 0p            4q - 4q = 0q            4r - 4r = 0r           6 - -8 = 14

So our result is 0p + 0q + 0r = 14, which becomes 0 = 14. Since 0 does not equal 14, our answer is no solution because the 3 lines never intersect. 

Example 2 - Method 2:

Here is what happens if you pick the two other equations and do not notice the 1st and 3rd equation are multiples of each other.

In this example we are going to eliminate the variable p. Luckily for us the Eq-1 and Eq-2 both have 2 has the coefficient so we can just subtract the two equations.

After subtracting the 1st equation, 2p + 2q + 2r = -3,  by the 2nd equation, 2p - q - 5r = 1, since the coefficients in front of p are the same 2 and 2.

2p - 2p = 0p            2q - -q = 3q            2r - -5r = 7r           -3 - 1 = -4

So our result is 0p + 3q + 7r = -4 which becomes 3q + 7r = -4.

Since we used the 1st equation and the 2nd equation, we have to used the 3rd equation now. Remember we need to have the coefficient in front of p to be the same/opposite. This time we used the 2nd equation, but you can use the 1st equation as well.

After multiply the 1st equation by 2, we get 4p - 2q - 10r = 2. Now we can subtract the new 2nd equation, 2p - 4q + 6r = 8, by the 3rd equation since the coefficients in front of p are the same 4 and 4.

4p - 4p = 0p            -2q - 4q = -6q            -10r - 4r = -14r           2 - -8 = 10

So our result is 0p - 6q - 14r = 10 which becomes -6q - 14r = 10.

Now if we have multiply equation 1 by so we can match the coefficients in front of q. In this case one will be +6 and the other will be -6.

After multiplying eq-1 by 2, the result is 6q+ 14r = -48. Now we can add the two new equations, 6q + 14r = -8 and -6q - 14r = 10, our result is 0 = 2. 

So the end result is the same, it just a few extra steps along the way to find the same conclusion.

Example 3

Example 3 - Method 1:

Here we have 3 variables. One way to solve it is with the elimination method; another is graphing; another is with matrices.

We are going to use the elimination with these examples.

We have the equations: 2p - 4q + 6r = 8, 2p + 3q - 2r = -4, p - 2q + 3r = 4. When we solve for the system, we have to use all 3 equations.

In order to add/subtract we need the coefficient in front the variable that we want to eliminate to be the same or opposite so 4x and 4x or 4x and -4x for example.

Before doing any math, you should look at the equations closely to see if any or multiples of each other. In this example the first equation and the third equation look like they are multiples of each other.

In this example we are going to eliminate the variable p. The 1st equation has 2p and the 2nd equation has p so we need to multiple the 3rd equation by 2 so the coefficient in front of p is the same/opposite.

After multiply the 3rd equation by 2, we get 2p - 4q + 6r = 8. Now we can subtract the 1st equation, 2p - 4q + 6r = 8, by the new 3rd equation since the coefficients in front of p are the same 2 and 2.

2p - 2p = 0p            4q - 4q = 0q            6r - 6r = 0r           8 - 8 = 0

So our result is 0p + 0q + 0r = 0 which becomes 0 = 0. Since 0 does equal 0, our answer is infinity or all real numbers because the 2 lines coincide. 

Example 3 - Method 2:

Here is what happens if you pick the two other equations and do not notice the 1st and 3rd equation are multiples of each other.

In this example we are going to eliminate the variable p. Luckily for us the Eq-1 and Eq-2 both have 2 has the coefficient so we can just subtract the two equations.

After subtracting the 1st equation, 2p - 4q + 6r = 8,  by the 2nd equation, 2p + 3q - 2r = -4, since the coefficients in front of p are the same 2 and 2.

2p - 2p = 0p            -4q - 3q = -7q            6r - -2r = 8r           8 - -4 = 12

So our result is 0p - 7q + 8r = 12 which becomes -7q + 8r = 12.

Since we used the 1st equation and the 2nd equation, we have to used the 3rd equation now. Remember we need to have the coefficient in front of p to be the same/opposite. After multiply the 3rd equation by 2, we get 2p - 4q + 6r = 8. Now we can subtract the 2nd equation, 2p - 4q + 6r = 8, by the new 3rd equation since the coefficients in front of p are the same 2 and 2.

2p - 2p = 0p            3q - -4q = 0q            -2r - 6r = -8r           -4 - 8 = -12

So our result is 0p + 7q - 8r = -12 which becomes 7q - 8r = -12.

Now if we add the two new equations, -7q + 8r = 12 and 7q - 8r = -12, our result is 0 = 0. So the end result is the same, it just a few extra steps along the way to find the same conclusion.

Now if we add the two new equations, -7q + 8r = 12 and 7q - 8r = -12, our result is 0 = 0. So the end result is the same, it just a few extra steps along the way to find the same conclusion.

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