## Precalculus (6th Edition) Blitzer

The solution to the system of equations is $x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t$.
Consider the following system of equations: \begin{align} & x-2y+2z=-2 \\ & 2x+3y-z=1 \end{align} Express the above system in the form of a matrix as below: $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 2 & 3 & -1 & 1 \\ \end{matrix} \right]$ Using the elementary row operations we will obtain the echelon form of a matrix: ${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 0 & 7 & -5 & 5 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{7}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & -2 & 2 & -2 \\ 0 & 1 & -\frac{5}{7} & \frac{5}{7} \\ \end{matrix} \right]$ Rewrite the system corresponding to echelon form of the matrix: $x-2y+2z=-2$ (I) $y-\frac{5}{7}z=\frac{5}{7}$ (II) Since, the system has two equations and three variables, one variable can be chosen arbitrarily. Let $z=t$, then equation (II) gives, $y-\frac{5}{7}\left( t \right)=\frac{5}{7}$ Simplify, $y=\frac{5}{7}+\frac{5}{7}t$ Substitute the values of y and z in equation (I) to obtain: $x-2\left( \frac{5}{7}+\frac{5}{7}t \right)+2\left( t \right)=-2$ It can be further simplified as: $x=-\frac{4}{7}-\frac{4}{7}t$ Hence, the solution to the system is $x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t$.