Solve the System of Linear Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ x – y + z = 4 \quad (1) \]

\[ x + y + z = 2 \quad (2) \]

\[ 2x + y – 3z = 0 \quad (3) \]

Solution

Step 1: Express x in Terms of y and z

From equation (1):

\[ x = 4 + y – z \quad (4) \]

Step 2: Substitute x in Equation (2)

Substitute equation (4) into equation (2):

\[ (4 + y – z) + y + z = 2 \]

\[ 4 + 2y = 2 \]

\[ 2y = -2 \]

\[ y = -1 \quad (5) \]

Step 3: Substitute x and y in Equation (3)

Substitute \(x = 4 + y – z\) and \(y = -1\) into equation (3):

\[ 2(4 – 1 – z) – 1 – 3z = 0 \]

\[ 2(3 – z) – 1 – 3z = 0 \]

\[ 6 – 2z – 1 – 3z = 0 \]

\[ 5 – 5z = 0 \]

\[ z = 1 \quad (6) \]

Step 4: Find the Value of x

Substitute \(y = -1\) and \(z = 1\) into equation (4):

\[ x = 4 – 1 – 1 = 2 \]

Conclusion

The solution of the given system of equations is:

\[ x = 2,\quad y = -1,\quad z = 1 \]

\[ \therefore \quad \text{The solution is } (2,\; -1,\; 1). \]