Solve the System of Linear Equations by the Substitution Method

Video Explanation

Question

Solve the following system of equations:

\[ x – y + z = 4 \quad (1) \]

\[ x – 2y + 3z = 9 \quad (2) \]

\[ 2x + y + 3z = 1 \quad (3) \]

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):

\[ (4 + y – z) – 2y + 3z = 9 \]

\[ 4 – y + 2z = 9 \]

\[ -y + 2z = 5 \quad (5) \]

Step 3: Substitute x in Equation (3)

Substitute equation (4) into equation (3):

\[ 2(4 + y – z) + y + 3z = 1 \]

\[ 8 + 2y – 2z + y + 3z = 1 \]

\[ 3y + z = -7 \quad (6) \]

Step 4: Solve Equations (5) and (6)

From equation (5):

\[ y = 2z – 5 \quad (7) \]

Substitute equation (7) into equation (6):

\[ 3(2z – 5) + z = -7 \]

\[ 6z – 15 + z = -7 \]

\[ 7z = 8 \]

\[ z = \frac{8}{7} \]

Step 5: Find the Value of y

Substitute \(z = \frac{8}{7}\) into equation (7):

\[ y = 2\left(\frac{8}{7}\right) – 5 \]

\[ y = \frac{16}{7} – \frac{35}{7} = -\frac{19}{7} \]

Step 6: Find the Value of x

Substitute \(y = -\frac{19}{7}\) and \(z = \frac{8}{7}\) into equation (4):

\[ x = 4 – \frac{19}{7} – \frac{8}{7} \]

\[ x = \frac{28 – 27}{7} = \frac{1}{7} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{7},\quad y = -\frac{19}{7},\quad z = \frac{8}{7} \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{7},\; -\frac{19}{7},\; \frac{8}{7}\right). \]