## Precalculus (6th Edition) Blitzer

The required values of x, y and z are: $x=-\frac{9}{a}$ ; $y=\frac{5}{b}$ ; $z=\frac{5}{c}$.
Mark the given system of equations as shown below: $ax-by+2cz=-4$ (I) $ax+3by-cz=1$ (II) $2ax+by+3cz=2$ (III) Add equations (I) and (II) in order to eliminate a, \begin{align} & \text{ }ax-\text{ }by+2cz=-4 \\ & \underline{-ax-3by+\text{ }cz=-\text{ }1} \\ & \text{ }-4by+3cz=-5\text{ }\left( \text{IV} \right) \\ \end{align} By multiplying equation (II) by -2 and adding to eliminate a from equations (I) and (III), we get \begin{align} & \text{ 2}ax+\text{ }by+3cz=2 \\ & \underline{-2ax-6by+2cz=2} \\ & \text{ }-5by\text{ +5}cz=0\text{ }\left( \text{V} \right) \\ \end{align} By multiplying equation (IV) by 5, equation (V) by 4, and adding equations (IV) and (V), we get the value of z, \begin{align} & -20by+15cz=-\text{25} \\ & \underline{-20by-20cz=\text{ }0} \\ & \text{ }-5cz=-25 \\ \end{align} Divide both sides by -5c, \begin{align} & \frac{-5cz}{-5c}=\frac{-25}{-5c} \\ & \text{ }z=\frac{5}{c} \\ \end{align} Substitute the value of z in equation (IV) to get the value of y, \begin{align} & -4by+3c\left( \frac{5}{c} \right)=-5 \\ & -4by+15=-5 \end{align} Now, subtract 15 from both sides to get, \begin{align} & -4by+15-15=-5-15 \\ & -4by=-20 \end{align} By dividing both sides by -4b to get, \begin{align} & \frac{-4by}{-4b}=\frac{-20}{-4b} \\ & \text{ }y=\frac{5}{b} \\ \end{align} Substitute the values of y and z in equation (I) to get the value of x, \begin{align} & ax-b\left( \frac{5}{b} \right)+2c\left( \frac{5}{c} \right)=-4 \\ & ax-5+10=-4 \\ & ax+5=-4 \end{align} Now subtract 5 from both sides, we get, \begin{align} & ax+5-5=-4-5 \\ & ax=-9 \end{align} Now, dividing both sides by a to get, \begin{align} & \frac{ax}{a}=-\frac{9}{a} \\ & \text{ }x=-\frac{9}{a} \\ \end{align} Hence, the values are $x=-\frac{9}{a}$, $y=\frac{5}{b}$, and $z=\frac{5}{c}$.