Q 4.70 - Intermediate Algebra

Question

Solve the system by equations:

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

Step-by-Step Solution

Answer

There are infinitely many number of solutions of given system of equations.

Step 1. Given Information

We are given a system of linear equation,

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

Step 2. Solving the equations

Adding first and second equation, we get

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

Now, multiplying by $3$ in second equation, we get

$$\begin{gathered} 3(-x+2y-3z)=3(-4) \\ -3x+6y-9z=-12 -\left(5\right) \end{gathered}$$

Now, subtracting third and fifth equation, we get

$$\begin{gathered} -3x+3x+6y-2y-9z-7z=-12+0 \\ 4y-16z=-12 \\ y-4z=-3 -\left(6\right) \end{gathered}$$

Now, subtracting sixth and fourth equation, we get

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

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