To solve a system of three equations with three variables, we can use the Addition/Subtraction method. This method involves eliminating one variable at a time by adding or subtracting the equations in the system. Let’s go through the steps using two pairs of equations.

Step 1: Choose two pairs of equations

Let’s say we have the following system of equations:

Equation 1: 2x + 3y – z = 5

Equation 2: x – 2y + 3z = 4

Equation 3: 3x + y + 2z = 1

We can choose any two pairs of equations to start with. For this example, let’s choose equations 1 and 2 as the first pair and equations 1 and 3 as the second pair.

Step 2: Eliminate the same variable from each pair

First Pair: Equations 1 and 2

To eliminate a variable, we need to manipulate the equations so that the coefficients of the variable are opposites. In this case, let’s eliminate the variable x.

Multiply equation 2 by 2:

2(x – 2y + 3z) = 2(4)

2x – 4y + 6z = 8

Now, add equation 1 and the modified equation 2:

(2x + 3y – z) + (2x – 4y + 6z) = 5 + 8

4x – y + 5z = 13

Second Pair: Equations 1 and 3

To eliminate a variable, we need to manipulate the equations so that the coefficients of the variable are opposites. In this case, let’s eliminate the variable x.

Multiply equation 3 by -1:

-1(3x + y + 2z) = -1(1)

-3x – y – 2z = -1

Now, add equation 1 and the modified equation 3:

(2x + 3y – z) + (-3x – y – 2z) = 5 + (-1)

-x + 2y – 3z = 4

Step 3: Solve the system of the two new equations

Now we have a new system of two equations with two variables:

Equation 4: 4x – y + 5z = 13

Equation 5: -x + 2y – 3z = 4

We can solve this system using the Addition/Subtraction method. Let’s eliminate the variable y by multiplying equation 4 by 2 and equation 5 by 1:

2(4x – y + 5z) = 2(13)

-2x + 4y – 6z = 8

Now, add equation 4 and the modified equation 5:

(4x – y + 5z) + (-2x + 4y – 6z) = 13 + 8

2x – z = 21

Step 4: Substitute the solution back into one of the original equations

Now we have a new equation:

Equation 6: 2x – z = 21

We can substitute the value of x into equation 6. Let’s say x = 5 (just an example):

2(5) – z = 21

10 – z = 21

-z = 21 – 10

-z = 11

Z = -11

Step 5: Solve for the third variable

Now we know that z = -11. We can substitute this value back into equation 6 to solve for x:

2x – (-11) = 21

2x + 11 = 21

2x = 21 – 11

2x = 10

X = 5

We can substitute the values of x and z back into one of the original equations to solve for y. Let’s use equation 1:

2(5) + 3y – (-11) = 5

10 + 3y + 11 = 5

3y = 5 – 10 – 11

3y = -16

Y = -16/3

Therefore, the solution to the system of equations is x = 5, y = -16/3, and z = -11.

By following these steps, we can solve a system of three equations with three variables using the Addition/Subtraction method.