## Precalculus (10th Edition)

Consistent Solution set: $\left\{\left(x,y,z\right)|x=2,y=z-3,z\text{ is any real number}\right\}$
We are given the system of equations: $\begin{cases} x-y+z=5\\ 3x+2y-2z=0 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 1&-1&1&|&5\\3&2&-2&|&0\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: $R_2=-3r_1+r_2$ $\begin{bmatrix} 1&-1&1&|&5\\0&5&-5&|&-15\end{bmatrix}$ $R_2=\dfrac{1}{5}r_2$ $\begin{bmatrix}1&-1&1&|&5\\0&1&-1&|&-3\end{bmatrix}$ $R_1=r_2+r_1$ $\begin{bmatrix}1&0&0&|&2\\0&1&-1&|&-3\end{bmatrix}$ The system has two equations and 3 variables; therefore it has infinitely many solutions. Write the corresponding system of equations: $\begin{cases} x=2\\ y-z=-3 \end{cases}$ Express $y$ in terms of $z$: $x=2$ $y-z=-3\Rightarrow y=z-3$ The system is consistent. The solution set is: $\left\{\left(x,y,z\right)|x=2,y=z-3,z\text{ is any real number}\right\}$