Problem 11 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{aligned} 2 x-4 y+3 z &=17 \\ x+2 y-z &=0 \\ 4 x-y-z &=6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

After performing these steps, you will find the solution to the system of equations: \(x = 1\), \(y = 1\), and \(z = -1\). This satisfies all three original equations, solving the system.

Step 1: Eliminate Some Variables

In the first step, we try to eliminate a variable from two of the equations. This can be achieved by adding or subtracting equations. Let's subtract the second equation from the third to eliminate \(y\). We get: \(4x - x - y + 2y - z + z = 6 - 0 \rightarrow 3x + y = 6\)

Step 2: Substitute to Simplify

Now, we substitute \(y\) from our new equation to the first equation: \(2x - 4(3x - 6) + 3z = 17 \rightarrow 2x - 12x + 24 + 3z = 17 \rightarrow -10x + 3z = -7\). So, the system of equations now stands as: \[ \left\{ \begin{aligned} -10x + 3z = -7 \ x + 2y -z = 0 \ 3x + y = 6 \end{aligned} \right\]

Step 3: Solve The System

From here, the rest becomes easier. Solve for \(x\) in terms of \(z\) from the first equation in the system and substitute into the other equations. Then, solve for \(z\) which can relatively easily be done in the last equation of the system. After finding \(z\), substitute back into the other equations to find the remaining variables \(x\) and \(y\).