## Precalculus (10th Edition)

Consistent Solution set: $\left\{(x,y,z)|x=\frac{6}{13}-\frac{4}{13}z,y=\frac{4}{13}-\frac{7}{13}z,z\text{ is any real number}\right\}$
We are given the system of equations: $\begin{cases} 2x-3y-z=0\\ 3x+2y+2z=2\\ x+5y+3z=2 \end{cases}$ Write the augmented matrix: $\begin{bmatrix} 2&-3&-1&|&0\\3&2&2&|&2\\1&5&3&|&2\end{bmatrix}$ Perform row operations to bring the matrix to the reduced row echelon form: Interchange $R_1$ and $R_3$ $\begin{bmatrix} 1&5&3&|&2\\3&2&2&|&2\\2&-3&-1&|&0\end{bmatrix}$ $R_2=-3r_1+r_2$ $\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\2&-3&-1&|&0\end{bmatrix}$ $R_3=-2r_1+r_3$ $\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\0&-13&-7&|&-4\end{bmatrix}$ $R_3=-r_2+r_3$ $\begin{bmatrix}1&5&3&|&2\\0&-13&-7&|&-4\\0&0&0&|&0\end{bmatrix}$ $R_2=-\dfrac{1}{13}r_2$ $\begin{bmatrix}1&5&3&|&2\\0&1&\frac{7}{13}&|&\frac{4}{13}\\0&0&0&|&0\end{bmatrix}$ As the last row contains only zeros, the system is consistent, with infinitely many solutions. Write the corresponding system: $\begin{cases} x+5y+3z=2\\ y+\frac{7}{13}z=\frac{4}{13} \end{cases}$ Express $x,y$ in terms of $z$: $y+\frac{7}{13}z=\frac{4}{13}\Rightarrow y=\frac{4}{13}-\frac{7}{13}z$\\ $x+5y+3z=2\Rightarrow x=2-5\left(\frac{4}{13}-\frac{7}{13}z\right)-3z=\frac{4}{13}-\frac{7}{13}z$ The solutions set is: $\left\{(x,y,z)|x=\frac{6}{13}-\frac{4}{13}z,y=\frac{4}{13}-\frac{7}{13}z,z\text{ is any real number}\right\}$