Example: Determining Whether an Ordered Triple Is a Solution to a System

Determine whether the ordered triple [latex]\left(3,-2,1\right)[/latex] is a solution to the system.

[latex]\begin{array}{l}\text{ }x+y+z=2\hfill \\ 6x - 4y+5z=31\hfill \\ 5x+2y+2z=13\hfill \end{array}[/latex]

Answer: We will check each equation by substituting in the values of the ordered triple for [latex]x,y[/latex], and [latex]z[/latex].

[latex]\begin{array}{ccccc}\begin{array}{r}\hfill x+y+z=2\\ \hfill \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}6x - 4y+5z=31\\ \hfill 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ \hfill 18+8+5=31\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}5x+2y+2z=13\\ \hfill 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ \hfill \text{}15 - 4+2=13\\ \hfill \text{True}\end{array}\end{array}[/latex]

The ordered triple [latex]\left(3,-2,1\right)[/latex] is indeed a solution to the system.

Example: Solving a System of Three Equations in Three Variables by Elimination

Find a solution to the following system:

[latex]\begin{array}{ll}\text{ }x - 2y+3z=9\hfill & \text{(1)}\hfill \\ \text{ }-x+3y-z=-6\hfill & \text{(2)}\hfill \\ 2x - 5y+5z=17\hfill & \text{(3)}\hfill \end{array}[/latex]

Answer: There will always be several choices as to where to begin, but the most obvious first step here is to eliminate [latex]x[/latex] by adding equations (1) and (2).

[latex]\frac{\begin{array}{ll}\text{ }\text{}x - 2y+3z=9\hfill & \text{(1)}\hfill \\ \text{ }-x+3y-z=-6\hfill & \text{ (2)}\hfill \end{array}}{\begin{array}{ll}\text{ }\text{}\text{}y+2z=3\hfill & \text{ (3)}\hfill \end{array}}[/latex]

The second step is multiplying equation (1) by [latex]-2[/latex] and adding the result to equation (3). These two steps will eliminate the variable [latex]x[/latex].

[latex]\begin{array} \hfill−2x+4y−6z=−18 \hfill& \left(1\right)\text{ multiplied by }−2 \\ 2x−5y+5z=17 \hfill& \left(3\right) \\ \text{_____________________________} \\ \hfill−y−z=−1\left(5\right)\hfill& \end{array}[/latex]

In equations (4) and (5), we have created a new two-by-two system. We can solve for [latex]z[/latex] by adding the two equations.

[latex]\frac{\begin{array}{l}\begin{array}{l}\hfill \\ \text{}y+2z=3\text{ }\left(4\right)\hfill \end{array}\hfill \\ -y-z=-1\text{ }\left(5\right)\hfill \end{array}}{z=2\text{ }\left(6\right)}[/latex]

Choosing one equation from each new system, we obtain the upper triangular form:

[latex]\begin{array}{ll}\text{}\text{}x - 2y+3z=9\text{ }\hfill & \left(1\right)\hfill \\ \text{ }y+2z=3\hfill & \left(4\right)\hfill \\ \text{ }z=2\hfill & \left(6\right)\hfill \end{array}[/latex]

Next, we back-substitute [latex]z=2[/latex] into equation (4) and solve for [latex]y[/latex].

[latex]\begin{array}{l}y+2\left(2\right)=3\hfill \\ \text{ }y+4=3\hfill \\ \text{ }y=-1\hfill \end{array}[/latex]

Finally, we can back-substitute [latex]z=2[/latex] and [latex]y=-1[/latex] into equation (1). This will yield the solution for [latex]x[/latex].

[latex]\begin{array}{r}\hfill x - 2\left(-1\right)+3\left(2\right)=9\\ \hfill \text{ }x+2+6=9\\ \hfill \text{ }x=1\end{array}[/latex]

The solution is the ordered triple [latex]\left(1,-1,2\right)[/latex].