Question

Solve the system by equations:

$$\begin{gathered} x+y-z=0 \\ 2x+4y-2z=6 \\ 3x+6y-3z=9 \end{gathered}$$

Short answer

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

Step-by-step solution

Step 1. Given Information

We are given a system of linear equation,

$$\begin{gathered} x+y-z=0-\left(1\right) \\ 2x+4y-2z=6-\left(2\right) \\ 3x+6y-3z=9-\left(3\right) \end{gathered}$$

Step 2. Solving the equations 

Multiplying by $3$in second equation and $2$in third equation, we get

localid="1644580091504" $$\begin{gathered} 3(2x+4y-2z)=3\left(6\right) \\ 6x+12y-6z=18-\left(4\right) \end{gathered}$$

$$\begin{gathered} 2(3x+6y-3z=2(9) \\ 6x+12y-6z=18-(5) \end{gathered}$$

Now, subtracting the fourth and fifth equations, we get

$$\begin{gathered} 6x-6x+12y-12y-6z+6z=18-18 \\ 0=0 \end{gathered}$$

This is true, hence the system has infinitely many solutions.