Chapter 11, Problem 15RE

To determine

Solve each system of equations using the method of substitution or the method of elimination.

Answer to Problem 15RE

  $\boxed{(x=-1,y=2,z=-3)}$

Explanation of Solution

Given information:

Solve each system of equations using the method of substitution or the method of elimination. If the system has no solution, say that it is inconsistent.

  $\{\begin{array}{l}x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16\end{array}$

Calculation:

In order to solve equations containing three unknown variables, we have to eliminate one variable at a time using any two equations until an equation with a single variable remains.

Multiply both sides of the equation $2x-y+3z=-13$ by $2$ so as to make the coefficients of $y$

  $4x-2y+6z=-26$

Add the equations $4x-2y+6z=-26$ and $x+2y-z=6$

  $\begin{array}{l}4x-2y+6z=-26\\ \begin{array}{cccc}x+& 2y& -z=& 6\\ 5x& & +5z=& -20\end{array}\end{array}$

Divide both sides of the equation $5x+5z=-20$ by $5$

  $x+z=-4$

So, we get an equation containing two variables $x$ and $z$ .

We need another equation containing the variable $x$ and $z$ .

Add the equation $x+2y-z=6$ and $3x-2y+3z=-16$ .

  $\begin{array}{l}x+2y-z=6\\ \begin{array}{c}3x-2y+3z=-16\end{array}\\ 4x+2z=-10\\ 2x+z=-5\end{array}$

So, we get another equation containing the variable $x$ and $z$ .

In order to solve for $x$ , subtract the equation $2x+z=-5$ from the equation $x+z=-4$ .

  $\begin{array}{l}2x+z=-5\\ \begin{array}{c}x+z=-4\end{array}\\ \begin{array}{ccccc}& x=-1& & & \end{array}\end{array}$

So, we get $x=-1$ .

For finding $z$ , substitute $-1$ for $x$ in the equation $x+z=-4$ .

  $-1+z=-4$

  $z=-3$

In order to find $y$ , substitute $-1$ for $x$ and $-3$ for $z$ in the equation $x+2y-z=6$ .

  $-1+2y-(-3)=6$

  $2y=4$

  $y=2$

So, we get $y=2$

Hence, the solution, in ordered triplet form, is $(x=-1,y=2,z=-3)$ .

In order to check the solution, substitute $-1$ for $x$ , $2$ for $y$ , and $-3$ for $z$ in the given system of equations.

Check: $\{\begin{array}{l}x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16\end{array}$ $\begin{array}{l}\left(-1\right)+2(2)-(-3)=6\\ 2(-1)-(2)+3(-3)=-13\\ 3(-1)-2(2)+3(-3)=-16\end{array}$

Hence, the solution checks.