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

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

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

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

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

The ordered triple $\left(3,-2,1\right)$ 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:

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

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

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

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

\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}

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

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

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

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

Next, we back-substitute $z=2$ into equation (4) and solve for $y$.

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

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

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

The solution is the ordered triple $\left(1,-1,2\right)$.

Example: Solving a Real-World Problem Using a System of Three Equations in Three Variables

In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund?

Answer: To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:

$\begin{array}{l}x=\text{amount invested in money-market fund}\hfill \\ y=\text{amount invested in municipal bonds}\hfill \\ z=\text{amount invested in mutual funds}\hfill \end{array}$

The first equation indicates that the sum of the three principal amounts is $12,000.

$x+y+z=12,000$

We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds.

$z=y+4,000$

The third equation shows that the total amount of interest earned from each fund equals $670.

$0.03x+0.04y+0.07z=670$

Then, we write the three equations as a system.

$\begin{array}{l}\text{ }x+y+z=12,000\hfill \\ \text{ }-y+z=4,000\hfill \\ 0.03x+0.04y+0.07z=670\hfill \end{array}$

To make the calculations simpler, we can multiply the third equation by 100. Thus,

$\begin{array}{ll}\text{ }x+\text{ }y+z\text{ }=12,000\hfill & \left(1\right)\hfill \\ \text{ }-y+z\text{ }=4,000\hfill & \left(2\right)\hfill \\ 3x+4y+7z=67,000\hfill & \left(3\right)\hfill \end{array}$

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

$\begin{array}{l}\text{ }x+\text{ }y +\text{ }z=12,000\hfill \\ 3x+4y +7z=67,000\hfill \\ \text{ }-y\text{ }+\text{ }z=4,000\hfill \end{array}$

Step 2. Multiply equation (1) by $-3$ and add to equation (2). Write the result as row 2.

$\begin{array}{l}x+y+z\text{ }=12,000\hfill \\ \text{ }y+4z=31,000\hfill \\ \text{ }-y+z\text{ }=4,000\hfill \end{array}$

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

$\begin{array}{l}x+y+\text{ }z=12,000\hfill \\ \text{ }y+4z=31,000\hfill \\ \text{ }5z\text{ }=35,000\hfill \end{array}$

Step 4. Solve for $z$ in equation (3). Back-substitute that value in equation (2) and solve for $y$. Then, back-substitute the values for $z$ and $y$ into equation (1) and solve for $x$.

$\begin{array}{l}\text{ }5z=35,000\hfill \\ \text{ }z=7,000\hfill \\ \hfill \\ \hfill \\ \text{ }y+4\left(7,000\right)=31,000\hfill \\ \text{ }y=3,000\hfill \\ \hfill \\ \hfill \\ x+3,000+7,000=12,000\hfill \\ \text{ }x=2,000\hfill \end{array}$

John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds.