Question

Solve the system of equations:

$$\begin{gathered} 2x-2y+3z=6 \\ 4x-3y+2z=0 \\ -2x+3y-7z=1 \end{gathered}$$

Short answer

There is no solution of given system of linear equations.

Step-by-step solution

Step 1. Given Information

We are given a system of linear equation,

$$\begin{gathered} 2x-2y+3z=6-\left(1\right) \\ 4x-3y+2z=0-\left(2\right) \\ -2x+3y-7z=1-\left(3\right) \end{gathered}$$

Step 2. Solving the equations 

Adding first and third equation, we get

$$\begin{gathered} 2x-2x-2y+3y+3z-7z=6+1 \\ y-4z=7-\left(4\right) \end{gathered}$$

Multiplying by $2$in first equation and subtraction it from second equation,

$$\begin{gathered} 2(2x-2y+3z)=2\left(6\right) \\ 4x-4y+12z=12 \end{gathered}$$

Subtracting, we get

role="math" localid="1644577982732" $$\begin{gathered} 4x-4x-4y+3y+6z-2z=12-0 \\ -y+4z=12-\left(5\right) \end{gathered}$$

Now, adding fourth and fifth equation, we get

$$\begin{gathered} y-y-4z+4z=7+12 \\ 0\ne 19 \end{gathered}$$

This is not true hence the given system has no solution.