Q. 8 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 1–10, 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.   

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

Step-by-Step Solution

Answer

The solution of the given system is $(-1,2,-3)$.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16\end{array}\right.$.

We need to solve the system of using the method of substitution or the method of elimination. 

Step 2. Eliminate the variable y

Multiply both sides of the second equation by $2$.

$\begin{array}{rcl}2(2x-y+3z)& =& 2·(-13)\\ 4x-2y+6x& =& -26 ...\left(4\right)\end{array}$

Now add first and the fourth equation

$\begin{array}{rcl}x+2y-z+4x-2y+6z& =& 6+(-26)\\ 5x+5z& =& 6-26\\ 5x+5z& =& -20\\ 5(x+z)& =& 5·(-4)\\ x+z& =& -4 ...\left(5\right)\end{array}$

Now subtract fourth equation from the third equation

$\begin{array}{rcl}3x-2y+3z-(4x-2y+6z)& =& -16-(-26)\\ 3x-2y+3z-4x+2y-6z& =& -16+26\\ -x-3z& =& 10 ...\left(6\right)\end{array}$

Step 3. Solve for z

Add the fifth and sixth equations

$\begin{array}{rcl}x+z+(-x-3z)& =& -4+10\\ x+z-x-3z& =& 6\\ -2z& =& 6\\ z& =& -3\end{array}$

Step 4. Solve for x

Substitute $-3$ for $z$ in the fifth equation

$\begin{array}{rcl}x+z& =& -4\\ x+(-3)& =& -4\\ x-3& =& -4\\ x& =& -4+3\\ x& =& -1\end{array}$

Step 5. Solve for y

Substitute $-1$ for $x$ and $-3$ for $z$ in the second equation and solve for $y$.

$\begin{array}{rcl}2x-y+3z& =& -13\\ 2·(-1)-y+3·(-3)& =& -13\\ -2-y-9& =& -13\\ -y-11& =& -13\\ -y& =& -13+11\\ -y& =& -2\\ y& =& 2\end{array}$

So the solution is given by the ordered triple $(-1,2,-3)$

Step 6. Check the solutions

Substitute $-1$ for $x$, $2$ for $y$, and $-3$ for $z$ in all three equations

$\begin{array}{rcl}x+2y-z& =& 6\\ -1+2·2-(-3)& =& 6\\ -1+4+3& =& 6\\ 3+3& =& 6\\ 6& =& 6\\ & & \\ 2x-y+3z& =& -13\\ 2(-1)-2+3(-3)& =& -13\\ -2-2-9& =& -13\\ -4-9& =& -13\\ -13& =& -13\\ & & \\ 3x-2y+3z& =& -16\\ 3(-1)-2·2+3(-3)& =& -16\\ -3-4-9& =& -16\\ -7-9& =& -16\\ -16& =& -16\end{array}$

So all the three equations are satisfied. Thus the found solution is correct.