Q 4.67 - Intermediate Algebra

Question

Solve the system of equations:

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

Step-by-Step Solution

Answer

There is no solution of given system of linear equations.

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.