Q 4.68 - Intermediate Algebra

Question

Solve the system of equations:

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

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,

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

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

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

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.