## Intermediate Algebra for College Students (7th Edition)

The solution set is$\underline{\left( 4,0,-5 \right)}$.
The given equation set is\begin{align} & 3x-2y+z=7 \\ & 2x+3y-z=13 \\ & x-y+2z=-6 \end{align}. Calculation: The provided equations are, $3x-2y+z=7$ …(1) $2x+3y-z=13$…(2) $x-y+2z=-6$…(3) Simplify as follows: Step 1: Let’s eliminate$z$and make two equations in two variables, use the addition method in equation (1) and (2) to eliminate the variable$z$, $3x-2y+z+\left( 2x+3y-z \right)=7+13$ $5x+y=20$ …(4) And multiply by $2$ in equation (2), use the addition method in equation (2) and (3) to eliminate the variable$z$, $4x+6y-2z+\left( x-y+2z \right)=26+\left( -6 \right)$ $5x+5y=20$ …(5) Step 2: Subtracting equation (5) from (4) to eliminate the variable$x$, \begin{align} & 5x+y-\left( 5x+5y \right)=20-20 \\ & 5x+y-5x-5y=0 \\ & -4y=0 \\ & y=0 \end{align} Step 3: Put the value of $y$ in equation (4) and simplify as follows, \begin{align} & 5x+\left( 0 \right)=20 \\ & 5x=20 \\ & x=4 \end{align} Step 4: Put the value of $x$ and $y$ in equation (1) and find out the value of$z$, simplify as follows: \begin{align} & 3x-2y+z=7 \\ & 3\left( 4 \right)-2\left( 0 \right)+z=7 \\ & 12+z=7 \\ & z=-5 \end{align} Therefore, these three equations have a solution of$x=4,\,\,y=0\text{ and }z=-5$.