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

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

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]\begin{align}x - 2y+3z&=9\\ -x+3y-z&=-6 \\ \hline y+2z&=3 \end{align}[/latex][latex]\hspace{5mm}\begin{gathered}\text{(1})\\ \text{(2)}\\ \text{(4)}\end{gathered}[/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{align}−2x+4y−6z&=−18 \\ 2x−5y+5z&=17 \\ \hline −y−z&=−1\end{align}[/latex][latex]\hspace{5mm}\begin{align}&\text{(2) multiplied by }−2\\&\left(3\right)\\&(5)\end{align}[/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]\begin{align}y+2z&=3 \\ -y-z&=-1 \\ \hline z&=2 \end{align}[/latex][latex]\hspace{5mm}\begin{align}(4)\\(5)\\(6)\end{align}[/latex]

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

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

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

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

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

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:

[latex]\begin{align}&x=\text{amount invested in money-market fund} \\ &y=\text{amount invested in municipal bonds} \\ z&=\text{amount invested in mutual funds} \end{align}[/latex]

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

[latex]x+y+z=12{,}000[/latex]

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

[latex]z=y+4{,}000[/latex]

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

[latex]0.03x+0.04y+0.07z=670[/latex]

Then, we write the three equations as a system.

[latex]\begin{align}x+y+z=12{,}000 \\ -y+z=4{,}000 \\ 0.03x+0.04y+0.07z=670 \end{align}[/latex]

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

[latex]\begin{align}x+y+z=12{,}000 \hspace{5mm} \left(1\right) \\ -y+z=4{,}000 \hspace{5mm} \left(2\right) \\ 3x+4y+7z=67{,}000 \hspace{5mm} \left(3\right) \end{align}[/latex]

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

[latex]\begin{align}x+y+z=12{,}000\hfill \\ 3x+4y +7z=67{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

Step 2. Multiply equation (1) by [latex]-3[/latex] and add to equation (2). Write the result as row 2.

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

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

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ 5z=35{,}000 \end{align}[/latex]

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

[latex]\begin{align}&5z=35{,}000 \\ &z=7{,}000 \\ \\ &y+4\left(7{,}000\right)=31{,}000 \\ &y=3{,}000 \\ \\ &x+3{,}000+7{,}000=12{,}000 \\ &x=2{,}000 \end{align}[/latex]

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