Question

Solve the system of equations:

$$\begin{gathered} x+2y+6z=5 \\ -x+y-2z=3 \\ x-4y-2z=1 \end{gathered}$$

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,

$$\begin{gathered} x+2y+6z=5-\left(1\right) \\ -x+y-2z=3-\left(2\right) \\ x-4y-2z=1-\left(3\right) \end{gathered}$$

Step 2. Solving the equations

Adding first and second equations, we get

$$\begin{gathered} x-x+2y+y+6z-2z=5+3 \\ 3y+4z=8-\left(4\right) \end{gathered}$$

Now, adding second and third equations, we get

$$\begin{gathered} -x+x+y-4y-2z-2z=3+1 \\ -3y-4z=4-\left(5\right) \end{gathered}$$

Now, Adding fourth and fifth equation, we get

$$\begin{gathered} 3y-3y+4z-4z=8+4 \\ 0\ne 12 \end{gathered}$$

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