SYSTEMS OF EQUATIONS in THREE VARIABLES

It is often desirable or even necessary to use more than one variable to model a situation in a field such as business, science, psychology, engineering, education, and sociology, to name a few. When this is the case, we write and solve a system of equations in order to answer questions about the situation.

If a system of linear equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. If the system has an infinity number of solutions, it is dependent. Otherwise it is independent.

A linear equation in three variables is an equation equivalent to the equation

$$\begin{eqnarray*}&& \\ Ax+By+Cz+D &=&0 \\ && \end{eqnarray*}$$

where A, B, C, and D are real numbers and A, B, C, and D are not all 0. This is the equation of a plane.

Example 4:

Alicia inherited $120,000 from her uncle. She is trying to decide how to invest it. She decides to divide the money into three parts. She will invest one part of the money in a safe investment, one part in a relatively safe investment, and one part in a risky investment. One option is to invest the safe part at 6%, the relatively safe part at 7%, and the risky part at 12%. Her anticipated annual income is $11,000. The other option is to invest the safe part at 5%, the relatively safe part at 6.5%, and the risky part at 18%. Her anticipated income is $14,100. How much money did she put into each part of her inhertiance.?

There are three unknowns:

$$\begin{eqnarray*}&& \\ 1 &:&\text{ The amount of money invested in the safe in... ... &:&\text{ The amount of money in the risky investment} \\ && \end{eqnarray*}$$

Let's rewrite the paragraph that asks the question we are to answer.

The first sentence can be rewritten as [ The amount of money invested in the safe investment ] + [ The amount of money invested in the relativey safe investment ] + [ The amount of money in the risky investment ] $% =\$120,000.\bigskip$ The second sentence can be rewritten 0.06 times [ The amount of money invested in the safe investment ] + 0.07 times [ The amount of money invested in the relativey safe investment ] + 0.12 times [ The amount of money in the risky investment ] $=\$11,000.\bigskip$ The third sentence can be rewritten 0.05 times [ The amount of money invested in the safe investment ] + 0.065 times [ The amount of money invested in the relativey safe investment ] + 0.18 times [ The amount of money in the risky investment ] $=\$14,100.$

It is going to get boring if we keep repeating the phrases

$$\begin{eqnarray*}&& \\ 1 &:&\text{ The amount of money invested in the safe in... ... &:&\text{ The amount of money in the risky investment} \\ && \end{eqnarray*}$$

Let's create a shortcut by letting symbols represent these phrases. Let

$$\begin{eqnarray*}&& \\ x &=&\text{ The amount of money invested in the safe in... ... &=&\text{ The amount of money in the risky investment} \\ && \end{eqnarray*}$$

in the three sentences, and then rewrite them.

The first sentence [ The amount of money invested in the safe investment ] + [ The amount of money invested in the relativey safe investment ] + [ The amount of money in the risky investment ] $=\$120,000$ can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ x+y+z &=&120,000 \\ && \\ && \end{eqnarray*}$$

The second sentence 0.06 times [ The amount of money invested in the safe investment ] + 0.07 times [ The amount of money invested in the relativey safe investment ] + 0.12 times [ The amount of money in the risky investment ] $=\$11,000$ can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ 0.06x+0.07y+0.12z &=&\$11,000 \\ && \\ && \end{eqnarray*}$$

The third sentence 0.05 times [ The amount of money invested in the safe investment ] + 0.065 times [ The amount of money invested in the relativey safe investment ] + 0.18 times [ The amount of money in the risky investment ] $=\$14,100$ can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ 0.05x+0.065y+0.18z &=&14,100 \\ && \\ && \end{eqnarray*}$$

We have converted the problem from one described by words to one that is described by three equations.

x+y+z = 120,000 (1)

$$\$11,000$$ 0.06x+0.07y+0.12z = (2)

0.05x+0.065y+0.18z = 14,100 (3)

We are going to show you how to solve this system of equations three different ways:

1)        Substitution, 2)        Elimination 3)        Matrices

SUBSTITUTION:

The process of substitution involves several steps:

Step 1:        Solve for one of the variables in one of the equations. It makes no difference which equation and which variable you choose. Let's solve for x in equation (1).

$$\begin{eqnarray*}&& \\ x+y+z &=&120,000 \\ && \\ x &=&120,000-y-z \\ && \\ && \end{eqnarray*}$$

Step 2:        Substitute this value for x in equations (2) and (3). This will change equations (1) and (2) to equations in the two variables y and z. Call the changed equations (4) and (5), respectively..

$$\$11,000$$ 0.06x+0.07y+0.12z =

$$0.06\left( 120,000-y-z\right) +0.07y+0.12z \$11,000$$ =

7200-0.06y-0.06z+0.07y+0.12z = 11,000

0.01y+0.06z = 3,800

y+6z = 380,000 (4)

0.05x+0.065y+0.18z = 14,100

$$0.05\left( 120,000-y-z\right) +0.065y+0.18z$$ = 14,100

6,000-0.05y-0.05z+0.065y+0.18z = 14,100

0.015y+0.13z = 8,100

3y+26z = 1,620,000 (5)

Step 3:        Solve for y in equation (4).

$$\begin{eqnarray*}&& \\ y+6z &=&380,000 \\ && \\ y &=&-6z+380,000 \\ && \\ && \\ && \\ && \end{eqnarray*}$$

Step 4:        Substitute this value of y in equation (5). This will give you an equation in one variable.

$$\begin{eqnarray*}&& \\ 3y+26z &=&1,620,000 \\ && \\ 3\left( -6z+380,000\ri... ...z+1,140,000+26z &=&1,620,000 \\ && \\ 8z &=&480,000 \\ && \end{eqnarray*}$$

Step 5:         Solve for z.

$$\begin{eqnarray*}&& \\ 8z &=&480,000 \\ && \\ z &=&60,000 \\ && \\ && \end{eqnarray*}$$

Step 6:        Substitute this value of z in equation (4) and solve for y.

$$\begin{eqnarray*}&& \\ y+6z &=&380,000 \\ && \\ y+6\left( 60,000\right) &=&380,000 \\ && \\ y &=&20,000 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute 20,000 for y and 60,000 for z in equation (1) and solve for x.

$$\begin{eqnarray*}&& \\ x+y+y &=&300 \\ && \\ x+20,000+60,000 &=&120,000 \\ && \\ x &=&40,000 \\ && \end{eqnarray*}$$

The solution is: $x=\$40,000,\ y=\$20,000,$ and $z=\$60,000.$ In terms of the original problems, Alicia plans to invest $\$40,000$ in a safe investment, $\$20,000$ in a relatively safe investment, and $60,000 in a risky investment. Step 8:        Check the solutions:

$$\begin{eqnarray*}&& \\ 40,000+20,000+60,000 &=&120,000\rightarrow Yes \\ && ... ...ght) &=&\$14,100\rightarrow Yes \\ && \\ && \\ && \\ && \end{eqnarray*}$$

ELIMINATION:

The process of elimination involves several steps: First you reduce three equations to two equations with two variables, and then to one equation with one variable.

Step 1:        Decide which variable you will eliminate. It makes no difference which one you choose. Let us eliminate x first.

$$\begin{eqnarray*}&& \\ \begin{array}{l} (1) \\ \\ (2) \\ \\ (3) \end... ...0.18z & = & 14,100 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

Step 2:        Multiply both sides of equation (1) by -0.06 and then add the transformed equation (1) to equation (2) to form equation (4).

$$\begin{eqnarray*}&& \\ \begin{array}{l} (1) \\ \\ (2) \\ \\ (4) \end... ... 0.06z & = & 3,800 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

Step 3:        Multiply both sides of equation (1) by -0.05 and then add the transformed equation (1) to equation (3) to form equation (5).

$$\begin{eqnarray*}&& \\ \begin{array}{l} (1) \\ \\ (3) \\ \\ (5) \end... ... 0.13z & = & 8,100 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

Step :        We now have two equations with two variables.

$$\begin{eqnarray*}&& \\ (4) &:&0.01y+0.06z=3,800 \\ && \\ (5) &:&0.015y+0.13z=8,100 \\ && \\ && \end{eqnarray*}$$

Step 4:        Multiply both sides of equation (4) by -1.5 and add the transformed equation (4) to equation (5) to create equation (6) with just one variable.

$\begin{array}{r} (4) \\ \\ (4) \\ \\ (5) \end{array} \ \begin{arr... ... & + & 0.13z & = & 8,100 \\ & & & & \\ & & 0.04z & = & 2,400 \end{array}$

Step 5:        Solve for z in equation (5).

$$\begin{eqnarray*}&& \\ 0.04z &=&2,400 \\ && \\ z &=&60,000 \\ && \\ && \end{eqnarray*}$$

Step 6:        Substitute 60,000 for z in equation (5) and solve for y.

$$\begin{eqnarray*}&& \\ 0.015y+0.13z &=&8,100 \\ && \\ 0.015y+0.13\left( 60... ... && \\ 0.015y &=&300 \\ && \\ y &=&20,000 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute 60,000 for z and 20,000 for y in equation (1) and solve for x.

$$\begin{eqnarray*}&& \\ x+y+z &=&120,000 \\ && \\ x+20,000+60,000 &=&120,000 \\ && \\ x &=&40,000 \\ && \\ && \end{eqnarray*}$$

The solution is: $x=\$40,000,\ y=\$20,000,$ and $z=\$60,000.$ In terms of the original problems, Alicia plans to invest $\$40,000$ in a safe investment, $\$20,000$ in a relatively safe investment, and $60,000 in a risky investment.

MATRICES:

The process of using matrices is essentially a shortcut of the process of elimination. Each row of the matrix represents an equation and each column represents coefficients of one of the variables.

Step 1: Create a three-row by four-column matrix using coefficients and the constant of each equation.

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ...& 0.065 & & 0.18 & \vert & 14,100 \end{array} \right] \\ && \end{eqnarray*}$$

The vertical lines in the matrix stands for the equal signs between both sides of each equation. The first column contains the coefficients of x, the second column contains the coefficients of y, the third column contains the coefficients of z, and the last column contains the constants.

We want to convert the original matrix

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ...& 0.065 & & 0.18 & \vert & 14,100 \end{array} \right] \\ && \end{eqnarray*}$$

to the following matrix.

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ... & \\ 0 & & 0 & & 1 & \vert & c \end{array} \right] \\ && \end{eqnarray*}$$

Because then you can read the matrix as x=a, y=b, and z=c..

Step 2:        We work with column 1 first. The number 1 is already in cell 11(Row1-Col 1). Add -0.06 times Row 1 to Row 2 to form a new Row 2, and add -0.05 times Row 1 to Row 3 to form a new Row 3..

$$\begin{eqnarray*}&& \\ -0.06\left[ Row\ 1\right] +\left[ Row\ 2\right] &=&\lef... ...ht] +\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{arr... ... & & 0.13 & \vert & 8,100 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 3:        We will now work with column 1. We want 1 in Cell 22, and we achieve this by multiplying Row 2 by 100 for a new Row 2.

$$\begin{eqnarray*}&& \\ 100\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{arr... ... & & 0.13 & \vert & 8,100 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 4:        Let's now manipulate the matrix so that there are zeros in Cell 12 and Cell 32. We do this by adding -1 times Row 2 to Row 1 to form a new Row 1, and add -0.015 times Row 2 to Row 3 5o form a new Row 3.

$$\begin{eqnarray*}&& \\ -1\left[ Row\ 2\right] +\left[ Row\ 1\right] &=&\left[ ... ...ht] +\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{arr... ... & & 0.04 & \vert & 2,400 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 5:        Let's now manipulate the matrix so that there is a 1 in Cell 33. We do this by multiplying Row 3 by 25.

$$\begin{eqnarray*}&& \\ 25\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{arr... ... 0 & & 1 & \vert & 60,000 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 6:        Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding 5 times Row 3 to Row 1 for a new Row 1 and adding -6 times Row 3 to Row 3 for a new Row 3.

$$\begin{eqnarray*}&& \\ 5\left[ Row\ 3\right] +\left[ Row\ 1\right] &=&\left[ N... ...ght] +\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{arr... ... 0 & & 1 & \vert & 60,000 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

You can now read the answers off the matrix: $x=\$40,000$, $y=\$20,000$, and $z=\$60,000$. The solutions is: $x=\$40,000,\ y=\$20,000,$ and $z=\$60,000.$ In terms of the original problems, Alicia plans to invest $\$40,000$ in a safe investment, $\$20,000$ in a relatively safe investment, and $60,000 in a risky investment.

If you would like to work a similar example, click on Example.

If you would like to test yourself by working some problem similar to this example, click on Problem.

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