To solve the system of equations, we have:
- \( x + 5y - 3z = -5 \) (Equation 1)
- \( -x + y + 4z = 10 \) (Equation 2)
- \( -x - 4y + 2z = 0 \) (Equation 3)
Let's solve this step by step.
Step 1: Isolate x in Equation 2
From Equation 2, we can express \( x \): \[ -x + y + 4z = 10 \implies x = y + 4z - 10 \tag{Equation 4} \]
Step 2: Substitute Equation 4 into Equations 1 and 3
Substituting \( x \) from Equation 4 into Equations 1 and 3.
For Equation 1: \[ (y + 4z - 10) + 5y - 3z = -5 \] \[ y + 4z - 10 + 5y - 3z = -5 \] \[ 6y + z - 10 = -5 \] \[ 6y + z = 5 \tag{Equation 5} \]
For Equation 3: \[ -(y + 4z - 10) - 4y + 2z = 0 \] \[ -y - 4z + 10 - 4y + 2z = 0 \] \[ -5y - 2z + 10 = 0 \] \[ 5y + 2z = 10 \tag{Equation 6} \]
Step 3: Solve the new system of equations (Equations 5 and 6)
Now we have:
- \( 6y + z = 5 \) (Equation 5)
- \( 5y + 2z = 10 \) (Equation 6)
We can express \( z \) in terms of \( y \) using Equation 5: \[ z = 5 - 6y \tag{Equation 7} \]
Now substitute Equation 7 into Equation 6: \[ 5y + 2(5 - 6y) = 10 \] \[ 5y + 10 - 12y = 10 \] \[ -7y + 10 = 10 \] \[ -7y = 0 \implies y = 0 \]
Step 4: Find z using y = 0
Substituting \( y = 0 \) back into Equation 7: \[ z = 5 - 6(0) = 5 \]
Step 5: Find x using y = 0 and z = 5
Substituting \( y = 0 \) and \( z = 5 \) into Equation 4: \[ x = 0 + 4(5) - 10 = 20 - 10 = 10 \]
Solution
Thus, the solution to the system of equations is: \[ x = 10, \quad y = 0, \quad z = 5 \]
So, the final answer is: \[ x = 10, \quad y = 0, \quad z = 5 \]