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=\left(3\right)+\left(-2\right)+\left(1\right)=2&& \text{True}\end{align}[/latex]

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

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

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

Figure 4

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&& \left(1\right)\\-y+z&=4,000&& \left(2\right) \\ 3x+4y+7z&=67,000&& \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\\ 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 \\ \hfill \\ \hfill \\ y+4\left(7,000\right)&=31,000 \\ y&=3,000 \\ \hfill \\ \hfill \\ 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.

Looking at the coefficients of [latex]x[/latex], we can see that we can eliminate [latex]x[/latex] by adding equation (1) to equation (2).

[latex]\begin{align}x - 3y+z&=4 \\ -x+2y - 5z&=3 \\ \hline -y - 4z&=7\end{align}[/latex] [latex]\begin{align}&&(1) \\ &&(2) \\ &&(4) \end{align}[/latex]

Next, we multiply equation (1) by [latex]-5[/latex] and add it to equation (3).

[latex]\begin{align}−5x+15y−5z&=−20 \\ 5x−13y+13z&=8 \\ \hline 2y+8z&=−12 \end{align}[/latex] [latex]\begin{align} \left(1\right)&\text{ multiplied by }−5 \\ (3)& \\ (5)& \end{align} [/latex]

Then, we multiply equation (4) by 2 and add it to equation (5).

[latex]\begin{align}−2y−8z&=14\\ 2y+8z&=−12\\ \hline 0&=2\end{align}[/latex] [latex]\begin{align}\left(4\right)&\text{ multiplied by }2 \\ (5)& \\ \text{ }& \end{align} [/latex]

The final equation [latex]0=2[/latex] is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

Analysis of the Solution

In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.

First, we can multiply equation (1) by [latex]-2[/latex] and add it to equation (2).

[latex]\begin{align} −4x−2y+6z&=0\\ 4x+2y−6z&=0 \end{align}[/latex] [latex]\begin{align}\left(1\right)&\text{multiplied by }−2 \\ (2)& \end{align} [/latex]

We do not need to proceed any further. The result we get is an identity, [latex]0=0[/latex], which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by [latex]-2[/latex], and adding it to equation (1). We then perform the same steps as above and find the same result, [latex]0=0[/latex].

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

[latex]\begin{align}2x+y−3z&=0 \\ x−y+z&=0 \\ \hline 3x−2z&=0 \end{align}[/latex]

We then solve the resulting equation for [latex]z[/latex].

[latex]\begin{gathered}3x - 2z=0 \\ z=\frac{3}{2}x \end{gathered}[/latex]

We back-substitute the expression for [latex]z[/latex] into one of the equations and solve for [latex]y[/latex].

[latex]\begin{gathered}2x+y - 3\left(\frac{3}{2}x\right)=0 \\ 2x+y-\frac{9}{2}x=0 \\ y=\frac{9}{2}x - 2x \\ y=\frac{5}{2}x\end{gathered}[/latex]

So the general solution is [latex]\left(x,\frac{5}{2}x,\frac{3}{2}x\right)[/latex]. In this solution, [latex]x[/latex] can be any real number. The values of [latex]y[/latex] and [latex]z[/latex] are dependent on the value selected for [latex]x[/latex].

Analysis of the Solution

As shown in Figure 5, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

Figure 5