Question

Solve each system of equations. If the system has no solution, say that it is inconsistent.

$$\begin{gathered} 2x-3y-z=0 \\ 3x+2y+2z=2 \\ x+5y+3z=2 \end{gathered}$$

Short answer

The solution set for the given system is$\left\{\right(x,y,z)=(\frac{6-4z}{13},\frac{4-7z}{13},z)/z\in R\}$

Step-by-step solution

Given information

We are given an equation

$$\begin{gathered} 2x-3y-z=0\left(1\right) \\ 3x+2y+2z=2\left(2\right) \\ x+5y+3z=2\left(3\right) \end{gathered}$$

Multiply equation 1 by 2 and equation 2 by 3And then add both the equation

We get,

$$\begin{gathered} 2(2x-3y-z=0) \\ 4x-6y-2z=0\left(4\right) \\ 3(3x+2y+2z=2) \\ 9x+6y+6z=5\left(5\right) \end{gathered}$$

Now we add both the equation

$4x-6y-2z=0+9x+6y+6z=413x+4z=4$

Hence we have

$$\begin{gathered} 13x+4z=4 \\ x=\frac{4-4z}{13} \end{gathered}$$

Multiply equation 1 by 3 and then add it to equation 3

We get,

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

And now we add,

$3x+2y+2z=2-3x-15y-9z=-6-13y-7z=-4$

Hence we get,$$\begin{gathered} -13y-7z=-4 \\ y=\frac{-7z+4}{13} \end{gathered}$$

Conclusion

The solution set for the given equation is

$\left\{\right(x,y,z)=(\frac{6-4z}{13},\frac{4-7z}{13},z)/z\in R\}$