Answer :
Certainly! Let's solve the given system of linear equations step-by-step. The system of equations is:
[tex]\[ \left\{ \begin{align} 6x + y - z &= 14 \\ 4x + 2y - z &= 13 \\ x + 2y + 4z &= 12 \end{align} \right. \][/tex]
Here's how we would solve it:
1. Label the equations for reference:
[tex]\[ \begin{align*} (1) \quad 6x + y - z &= 14 \\ (2) \quad 4x + 2y - z &= 13 \\ (3) \quad x + 2y + 4z &= 12 \end{align*} \][/tex]
2. First, we will eliminate [tex]\( z \)[/tex] from the first two equations:
Subtract equation (2) from equation (1):
[tex]\[ (6x + y - z) - (4x + 2y - z) = 14 - 13 \][/tex]
[tex]\[ 6x + y - z - 4x - 2y + z = 1 \][/tex]
[tex]\[ 2x - y = 1 \quad \Rightarrow \quad (4) \][/tex]
3. Next, we will eliminate [tex]\( y \)[/tex] from equations (3) and (2):
Multiply equation (3) by 2 and subtract equation (2) from it:
[tex]\[ 2(x + 2y + 4z) - (4x + 2y - z) = 2 \cdot 12 - 13 \][/tex]
[tex]\[ 2x + 4y + 8z - 4x - 2y + z = 24 - 13 \][/tex]
[tex]\[ -2x + 2y + 9z = 11 \quad \Rightarrow \quad (5) \][/tex]
4. Now solve equations (4) and (5) together:
Equation (4):
[tex]\[ 2x - y = 1 \][/tex]
Equation (5):
[tex]\[ -2x + 2y + 9z = 11 \][/tex]
Add equations (4) and (5) to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x - y) + (-2x + 2y + 9z) = 1 + 11 \][/tex]
[tex]\[ y + 9z = 12 \quad \Rightarrow \quad (6) \][/tex]
5. Solve for [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex] using Equation (6):
[tex]\[ y = 12 - 9z \quad \Rightarrow \quad (7) \][/tex]
6. Substitute [tex]\( y \)[/tex] from Equation (7) into Equation (4):
[tex]\[ 2x - (12 - 9z) = 1 \][/tex]
[tex]\[ 2x - 12 + 9z = 1 \][/tex]
[tex]\[ 2x + 9z = 13 \][/tex]
[tex]\[ x = \frac{13 - 9z}{2} \quad \Rightarrow \quad (8) \][/tex]
7. Substitute [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex] into one of the original equations, say Equation (3):
[tex]\[ x + 2y + 4z = 12 \][/tex]
Substitute [tex]\( x \)[/tex] from Equation (8) and [tex]\( y \)[/tex] from Equation (7):
[tex]\[ \left(\frac{13 - 9z}{2}\right) + 2(12 - 9z) + 4z = 12 \][/tex]
[tex]\[ \left(\frac{13 - 9z}{2}\right) + 24 - 18z + 4z = 12 \][/tex]
[tex]\[ \frac{13 - 9z}{2} + 24 - 14z = 12 \][/tex]
Multiply everything by 2 to clear the fraction:
[tex]\[ 13 - 9z + 48 - 28z = 24 \][/tex]
Combine like terms:
[tex]\[ 61 - 37z = 24 \][/tex]
[tex]\[ 37z = 37 \][/tex]
[tex]\[ z = 1 \][/tex]
8. Now, substitute [tex]\( z = 1 \)[/tex] back into Equation (7) to find [tex]\( y \)[/tex]:
[tex]\[ y = 12 - 9(1) = 3 \][/tex]
9. Finally, substitute [tex]\( z = 1 \)[/tex] into Equation (8) to find [tex]\( x \)[/tex]:
[tex]\[ x = \frac{13 - 9(1)}{2} = \frac{13 - 9}{2} = 2 \][/tex]
The solution to the system of equations is:
[tex]\[ (x, y, z) = (2, 3, 1) \][/tex]
So the values are:
[tex]\[ x = 2, \quad y = 3, \quad z = 1 \][/tex]
[tex]\[ \left\{ \begin{align} 6x + y - z &= 14 \\ 4x + 2y - z &= 13 \\ x + 2y + 4z &= 12 \end{align} \right. \][/tex]
Here's how we would solve it:
1. Label the equations for reference:
[tex]\[ \begin{align*} (1) \quad 6x + y - z &= 14 \\ (2) \quad 4x + 2y - z &= 13 \\ (3) \quad x + 2y + 4z &= 12 \end{align*} \][/tex]
2. First, we will eliminate [tex]\( z \)[/tex] from the first two equations:
Subtract equation (2) from equation (1):
[tex]\[ (6x + y - z) - (4x + 2y - z) = 14 - 13 \][/tex]
[tex]\[ 6x + y - z - 4x - 2y + z = 1 \][/tex]
[tex]\[ 2x - y = 1 \quad \Rightarrow \quad (4) \][/tex]
3. Next, we will eliminate [tex]\( y \)[/tex] from equations (3) and (2):
Multiply equation (3) by 2 and subtract equation (2) from it:
[tex]\[ 2(x + 2y + 4z) - (4x + 2y - z) = 2 \cdot 12 - 13 \][/tex]
[tex]\[ 2x + 4y + 8z - 4x - 2y + z = 24 - 13 \][/tex]
[tex]\[ -2x + 2y + 9z = 11 \quad \Rightarrow \quad (5) \][/tex]
4. Now solve equations (4) and (5) together:
Equation (4):
[tex]\[ 2x - y = 1 \][/tex]
Equation (5):
[tex]\[ -2x + 2y + 9z = 11 \][/tex]
Add equations (4) and (5) to eliminate [tex]\( x \)[/tex]:
[tex]\[ (2x - y) + (-2x + 2y + 9z) = 1 + 11 \][/tex]
[tex]\[ y + 9z = 12 \quad \Rightarrow \quad (6) \][/tex]
5. Solve for [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex] using Equation (6):
[tex]\[ y = 12 - 9z \quad \Rightarrow \quad (7) \][/tex]
6. Substitute [tex]\( y \)[/tex] from Equation (7) into Equation (4):
[tex]\[ 2x - (12 - 9z) = 1 \][/tex]
[tex]\[ 2x - 12 + 9z = 1 \][/tex]
[tex]\[ 2x + 9z = 13 \][/tex]
[tex]\[ x = \frac{13 - 9z}{2} \quad \Rightarrow \quad (8) \][/tex]
7. Substitute [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex] into one of the original equations, say Equation (3):
[tex]\[ x + 2y + 4z = 12 \][/tex]
Substitute [tex]\( x \)[/tex] from Equation (8) and [tex]\( y \)[/tex] from Equation (7):
[tex]\[ \left(\frac{13 - 9z}{2}\right) + 2(12 - 9z) + 4z = 12 \][/tex]
[tex]\[ \left(\frac{13 - 9z}{2}\right) + 24 - 18z + 4z = 12 \][/tex]
[tex]\[ \frac{13 - 9z}{2} + 24 - 14z = 12 \][/tex]
Multiply everything by 2 to clear the fraction:
[tex]\[ 13 - 9z + 48 - 28z = 24 \][/tex]
Combine like terms:
[tex]\[ 61 - 37z = 24 \][/tex]
[tex]\[ 37z = 37 \][/tex]
[tex]\[ z = 1 \][/tex]
8. Now, substitute [tex]\( z = 1 \)[/tex] back into Equation (7) to find [tex]\( y \)[/tex]:
[tex]\[ y = 12 - 9(1) = 3 \][/tex]
9. Finally, substitute [tex]\( z = 1 \)[/tex] into Equation (8) to find [tex]\( x \)[/tex]:
[tex]\[ x = \frac{13 - 9(1)}{2} = \frac{13 - 9}{2} = 2 \][/tex]
The solution to the system of equations is:
[tex]\[ (x, y, z) = (2, 3, 1) \][/tex]
So the values are:
[tex]\[ x = 2, \quad y = 3, \quad z = 1 \][/tex]
