Question: Solve the system. \[ \begin{align*} -x + y + 4z &= 10 \\ x + 3y + 2z &= 4 \\ 2x + 2y - 3z &=…

Solve the system.

\[ \begin{align*} -x + y + 4z &= 10 \\ x + 3y + 2z &= 4 \\ 2x + 2y - 3z &= -7 \end{align*} \]

\[ \begin{align*} x &= -1 \\ y &= 2 \\ z &= 1 \end{align*} \]

Solution

To solve the system of equations: \[ \begin{cases} -x + y + 4z = 10 & \\ x + 3y + 2z = 4 & \\ 2x + 2y - 3z = -7 & \end{cases} \] Let’s solve step by step. First, solve for \(x\) from the second equation: \(x + 3y + 2z = 4\) Rearrange it to express \(x\) in terms of \(y\) and \(z\): \[ x = 4 - 3y - 2z \] Substitute \(x = 4 - 3y - 2z\) into the first and third equations. First equation: \[ -(4 - 3y - 2z) + y + 4z = 10 \] Simplify: \[ -4 + 3y + 2z + y + 4z = 10 \] Combine like terms: \[ 4y + 6z = 14 \] Divide the whole equation by 2: \[ 2y + 3z = 7 \] Third equation: \[ 2(4 - 3y - 2z) + 2y - 3z = -7 \] Distribute: \[ 8 - 6y - 4z + 2y - 3z = -7 \] Combine like terms: \[ -4y - 7z = -15 \] Now solve the simpler system of equations: \[ \begin{cases} 2y + 3z = 7 & \\ 4y + 7z = 15 & \end{cases} \] Multiply the first equation by 2: \[ 4y + 6z = 14 \] Subtract it from the second equation: \[ (4y + 7z) - (4y + 6z) = 15 - 14 \] \[ z = 1 \] Substitute \(z = 1\) into the first equation: \[ 2y + 3(1) = 7 \] \[ 2y + 3 = 7 \] Subtract 3 from both sides: \[ 2y = 4 \] Divide by 2: \[ y = 2 \] Substitute \(y = 2\) and \(z = 1\) into the expression for \(x\): \[ x = 4 - 3(2) - 2(1) \] \[ x = 4 - 6 - 2 \] \[ x = -4 \] The solution is: \( x = -4, \; y = 2, \; z = 1 \)