Questions: Honors Algebra Il: Section 3 IXL: Solve a System of Equation: IXL - Solve a system of equ ixl.com/math/level-m/solve-a-system-of-equations-in-three-variables-using-substitutic My IXL Learning Level M > E. 12 Solve system of equations in three variables using substitution X 8 H Solve the system of equations by substitution. -x + 3y + 2z = -4 x - y + 3z = -10 x = -6 Submit Work it Not feeling ready yet Solve a system of equations using any method Type here to search

Solution Steps

To solve the system of equations using substitution, we can follow these steps:

1. Substitute the value of \( x \) from the third equation into the first and second equations.
2. Solve the resulting two-variable system of equations for \( y \) and \( z \).
3. Use the values of \( y \) and \( z \) to find \( x \).

1. Substitute \( x = -6 \) into the first and second equations.
2. Solve the resulting equations for \( y \) and \( z \).
3. Use the values of \( y \) and \( z \) to find \( x \).

We start with the system of equations: \[ \begin{align_} -x + 3y + 2z &= -4 \quad (1) \\ x - y + 3z &= -10 \quad (2) \\ x &= -6 \quad (3) \end{align_} \] Substituting \( x = -6 \) into equations (1) and (2), we get: \[ \begin{align*} 3y + 2z + 6 &= -4 \quad \Rightarrow \quad 3y + 2z = -10 \quad (4) \\

• y + 3z + 6 &= -10 \quad \Rightarrow \quad -y + 3z = -16 \quad (5) \end{align*} \]

Now we solve the system of equations (4) and (5): \[ \begin{align_} 3y + 2z &= -10 \quad (4) \\ -y + 3z &= -16 \quad (5) \end{align_} \] From equation (5), we can express \( y \) in terms of \( z \): \[ y = 3z + 16 \] Substituting this expression for \( y \) into equation (4): \[ 3(3z + 16) + 2z = -10 \] This simplifies to: \[ 9z + 48 + 2z = -10 \quad \Rightarrow \quad 11z + 48 = -10 \quad \Rightarrow \quad 11z = -58 \quad \Rightarrow \quad z = -\frac{58}{11} \]

Now substituting \( z = -\frac{58}{11} \) back into the expression for \( y \): \[ y = 3\left(-\frac{58}{11}\right) + 16 = -\frac{174}{11} + \frac{176}{11} = \frac{2}{11} \]

Final Answer