## Precalculus (6th Edition) Blitzer

The solution set is $\left\{ \left( 6,3,5 \right) \right\}$
We have the system given as: \begin{align} & x+y+2z=19\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & y+2z=13\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\ & z=5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \end{align} Put the value $z=5$ into equation (II). Then, \begin{align} & y+2z=13 \\ & y+2\left( 5 \right)=13 \\ & y+10=13 \\ & y=3 \end{align} Now, put the value of $y=3$ and $z=5$ into equation (I) and then find the value of x: \begin{align} & x+y+2z=19 \\ & x+3+2\left( 5 \right)=19 \\ & x+3+10=19 \\ & x+13=19 \end{align} Finally, \begin{align} & x=19-13 \\ & x=6 \end{align} Hence, the solution set is $\left\{ \left( 6,3,5 \right) \right\}$.