## Precalculus (6th Edition) Blitzer

The values of x, y and z are as follows: $x=\frac{8}{a}$ ; $y=-\frac{3}{b}$ ; $z=-\frac{5}{c}$.
Mark the given system of equations as shown below: $ax-by-2cz=21$ (I) $ax+by+cz=0$ (II) $2ax-by+cz=14$ (III) By eliminating y from equations (I) and (II) to get the equation in variables x and z, we get \begin{align} & \text{ }ax-by-2cz=21 \\ & \underline{\text{ }ax+by+\text{ }cz=\text{ }0} \\ & \text{2}ax\text{ }-\text{ }cz=\text{ 21} \\ \end{align} (IV) By multiplying equation (III) by -1 and eliminating y from equations (I) and (III), we get \begin{align} & \text{ }ax-by-2cz=21 \\ & \underline{-2ax+by-cz=-14} \\ & \text{ }-ax\text{ }-3cz=\text{ }7 \\ \end{align} (V) By multiplying equation (V) by 2 and adding equations (IV) and (V), we get the value of z, \begin{align} & \text{ 2}ax-\text{ }cz=\text{21} \\ & \underline{-ax-6cz=14} \\ & \text{ }-7cz=35 \\ \end{align} (VI) By dividing both sides by -7c, we get, \ \begin{align} & \frac{-7cz}{-7c}=\frac{35}{-7c} \\ & \text{ }z=-\frac{5}{c} \\ \end{align} Substitute the value of z in equation (IV) to get the value of x, \begin{align} & \text{2}ax-c\left( -\frac{5}{c} \right)=\text{21 } \\ & 2ax+5=21 \end{align} Now, subtract 5 from both sides to get: \begin{align} & 2ax+5-5=21-5 \\ & 2ax=16 \end{align} Now, dividing both sides by 2a, we get, \begin{align} & \frac{2ax}{2a}=\frac{16}{2a} \\ & \text{ }x=\frac{8}{a} \\ \end{align} Substitute the values of x and z in equation (I) and we get the value of y, \begin{align} & a\left( \frac{8}{a} \right)-by-2c\left( -\frac{5}{c} \right)=21 \\ & 8-by+10=21 \\ & 18-by=21 \end{align} Now, subtract $18$ from both sides to get: \begin{align} & 18-by-18=21-18 \\ & -by=3 \end{align} By dividing both sides by -b we get, \begin{align} & \frac{by}{b}=-\frac{3}{b} \\ & \text{ }y=-\frac{3}{b} \\ \end{align} Hence, the values are $x=\frac{8}{a}$, $y=-\frac{3}{b}$, and $z=-\frac{5}{c}$.