Figure 1. (credit: "Elembis," Wikimedia Commons)
John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund? Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics.
In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution [latex]\left(x,y,z\right),\text<>[/latex] which we call an ordered triple. A system in upper triangular form looks like the following:
[latex]\beginFigure 2. (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions.
Figure 3
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\begin\hfill x+y+z=2\\ \hfill \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \hfill \text\end& & \begin\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\end& & \begin\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\end\end[/latex]
The ordered triple [latex]\left(3,-2,1\right)[/latex] is indeed a solution to the system.
[latex]\begin
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).
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 \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[/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.
Choosing one equation from each new system, we obtain the upper triangular form:[latex]\begin
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
Figure 4
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?
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]\beginWe 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
[latex]\begin
Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.
[latex]\begin
Step 2. Multiply equation (1) by [latex]-3[/latex] and add to equation (2). Write the result as row 2.
[latex]\begin
[latex]\begin
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
Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as [latex]3=7[/latex] or some other contradiction.
[latex]\begin
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]\frac\text< >x - 3y+z=4\text< >\left(1\right)\hfill \\ -x+2y - 5z=3\text< >\left(2\right)\hfill \end>-y - 4z=7\text< >\left(4\right)>[/latex]\begin
Next, we multiply equation (1) by [latex]-5[/latex] and add it to equation (3).[latex]\begin−5x+15y−5z=−20 \hfill& \left(1\right)\text< multiplied by >−5 \\ 5x−13y+13z=8 \hfill& \left(3\right) \\ \text \\ 2y+8z=−12\hfill&\left(5\right) \end[/latex]
Then, we multiply equation (4) by 2 and add it to equation (5).
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.
In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.
We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line.
[latex]\begin
[latex]\begin −4x−2y+6z=0 \hfill& \text\left(1\right)\text−2 \\ 4x+2y−6z=0\hfill&\left(2\right) \end[/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
We then solve the resulting equation for [latex]z[/latex].
[latex]\beginWe back-substitute the expression for [latex]z[/latex] into one of the equations and solve for [latex]y[/latex].
[latex]\begin
So the general solution is [latex]\left(x,\frac<5>x,\fracx\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].5>
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.
Two overlapping planes intersecting a third. The first overlapping plane" width="487" height="288" />
Figure 5
No, you can write the generic solution in terms of any of the variables, but it is common to write it in terms of x and if needed [latex]x[/latex] and [latex]y[/latex].
solution set the set of all ordered pairs or triples that satisfy all equations in a system of equations
1. Can a linear system of three equations have exactly two solutions? Explain why or why not 2. If a given ordered triple solves the system of equations, is that solution unique? If so, explain why. If not, give an example where it is not unique. 3. If a given ordered triple does not solve the system of equations, is there no solution? If so, explain why. If not, give an example. 4. Using the method of addition, is there only one way to solve the system? 5. Can you explain whether there can be only one method to solve a linear system of equations? If yes, give an example of such a system of equations. If not, explain why not. For the following exercises, determine whether the ordered triple given is the solution to the system of equations. 6. [latex]\begin