SYSTEMS OF EQUATIONS in THREE VARIABLES

It is often desirable or even necessary to use more than one variable to model a situation in many fields. 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 describes a plane and 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.

Problem 3.1a:

A total of $50,000 is invested in three funds paying 6%, 8%, and 10% simple interest. The yearly interest is $3,700. Twice as much money is invested at 6% as invested at 10%. How much was invested in each of the funds.

Answer: $30,000 is invested at 6%, $5,000 is invested at 8%, and $15,000 is invested at 10%

Solution:

There are three unknowns:

$$\begin{eqnarray*}&& \\ 1 &:&\text{ The amount of money invested at 6\%} \\ 2... ... \\ 3 &:&\text{ The amount of money invested at 10\%} \\ && \end{eqnarray*}$$

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 at 6% ] + [ The amount of money invested at 8% ] + [ The amount of money invested at 10% ] $=\$50,000.\bigskip$ The second sentence can be rewritten 0.060 times [ The amount of money invested at 6% ] + 0.08 times [ The amount of money invested at 8% ] + 0.10 times [ The amount of money invested at 10% ] $=\$3,700.\bigskip$ The third sentence can be rewritten [ The amount of money invested at 6% ] = twice [ The amount of money invested at 10% ].

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

$$\begin{eqnarray*}&& \\ 1 &:&\text{ The amount of money invested at 6\%} \\ 2... ... \\ 3 &:&\text{ The amount of money invested at 10\%} \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*}&& \\ x &=&\text{ The amount of money invested at 6\%} \\ y... ... \\ z &=&\text{ The amount of money invested at 10\%} \\ && \end{eqnarray*}$$

in the three sentences, and then rewrite them.

The first sentence can be rewritten as [ The amount of money invested at 6% ] + [ The amount of money invested at 8% ] + [ The amount of money invested at 10% ] $=\$50,000.$ can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ x+y+z &=&50,000 \\ && \\ && \end{eqnarray*}$$

The second sentence can be rewritten 0.060 times [ The amount of money invested at 6% ] + 0.08 times [ The amount of money invested at 8% ] + 0.10 times [ The amount of money invested at 10% ] $=\$3,700.$ can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ 0.06x+0.08y+0.10z &=&3,700 \\ && \\ && \end{eqnarray*}$$

The third sentence can be rewritten [ The amount of money invested at 6% ] = twice [ The amount of money invested at 10% ] can now be written in the algebraic form

$$\begin{eqnarray*}&& \\ x &=&2z \\ && \\ && \end{eqnarray*}$$

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

x+y+z = 50,000 (1)

0.06x+0.08y+0.10z = 3,700 (2)

x-2z = 0 (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 (3).

$$\begin{eqnarray*}&& \\ x-2z &=&0 \\ && \\ x &=&2z \\ && \\ && \end{eqnarray*}$$

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

x+y+z = 50,000

$$\left( 2z\right) +y+z$$ = 50,000

y+3z = 50,000 (4)

0.06x+0.08y+0.10z = 3,700

$$0.06\left( 2z\right) +0.08y+0.10z$$ = 3,700

0.08y+0.22z = 3,700

8y+22z = 370,000

4y+11z = 185,000 (5)

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

$$\begin{eqnarray*}&& \\ y+3z &=&50,000 \\ && \\ y &=&50,000-3z \\ && \\ && \\ && \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*}&& \\ 4y+11z &=&185,000 \\ && \\ 4\left( 50,000-3z\right)... ...000 \\ && \\ -z &=&-15,000 \\ && \\ z &=&15,000 \\ && \end{eqnarray*}$$

Step 5:        Substitute this value of z in equation (4) and solve for y.

$$\begin{eqnarray*}&& \\ y+3z &=&50,000 \\ && \\ y+3\left( 15,000\right) &=&50,000 \\ && \\ y &=&5,000 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute 5,000 for y and 15,000 for z in equation (1) and solve for x.

$$\begin{eqnarray*}&& \\ x+y+y &=&50,000 \\ && \\ x+\left( 5,000\right) +\left( 15,000\right) &=&50,000 \\ && \\ x &=&30,000 \\ && \end{eqnarray*}$$

The solution is: $x=30,000,\ y=5,000,$ and z=15,000. In terms of the original problems, $\$30,000$ was invested at $6\%$, $\$5,000$ was invested at $8\%$, and $\$15,000$ was invested at $10\%$. Step 8:        Check the solutions:

$$\begin{eqnarray*}&& \\ 30,000+5,000+15,000 &=&50,000\rightarrow Yes \\ && \\... ...( 15,000\right) \rightarrow Yes \\ && \\ && \\ && \\ && \end{eqnarray*}$$

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 y first because y is already missing from equation (3).

x+y+z = 50,000 (1)

0.06x+0.08y+0.10z = 3,700 (2)

x-2z = 0 (3)

Step 2:        Multiply both sides of equation (1) by -0.08 and then add the transformed equation (1) to equation (2) to form equation (4).

$$-0.08\ \left[ Equation\ (1)\right] +\left[ Equation\ (2)\right] =\left[ Equation\ (4)\right]$$

$$\begin{eqnarray*}&& \\ \begin{array}{l} (1) \\ \\ (2) \\ \\ (4) \end... ... & 0.02 & = & -300 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

Step 3:        We now have two equations with two variables.

$$\begin{eqnarray*}&& \\ (4) &:&-0.02x+0.02z=-300 \\ && \\ (3) &:&x-2z=0 \\ && \\ && \end{eqnarray*}$$

Step 4:        Multiply both sides of equation (4) by 50 and add the transformed equation (4) to equation (3) to create equation (5) with just one variable.

$\begin{array}{r} (4) \\ \\ (3) \\ \\ (5) \end{array} \ \begin{arr... ... \\ x & - & 2z & = & 0 \\ & & & & \\ & - & z & = & -15,000 \end{array}$

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

$$\begin{eqnarray*}&& \\ z &=&15,000 \\ && \\ && \end{eqnarray*}$$

Step 6:        Substitute 15,000 for z in equation (3) and solve for x.

$$\begin{eqnarray*}&& \\ x-2z &=&0 \\ && \\ x-2\left( 15,000\right) &=&0 \\ && \\ x &=&30,000 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute 30,000 for x and 15,000 for z in equation (1) and solve for y.

$$\begin{eqnarray*}&& \\ x+y+z &=&50,000 \\ && \\ 30,000+y+15,000 &=&50,000 \\ && \\ y &=&5,000 \\ && \\ && \end{eqnarray*}$$

The solution is: $x=30,000,\ y=5,000,$ and z=15,000. In terms of the original problems, $\$30,000$ was invested at $6\%$, $\$5,000$ was invested at $8\%$, and $\$15,000$ was invested at $10\%$.

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.

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

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

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

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 -1 times Row 1 to Row 3 to form a new Row 3.

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

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ...0 & & -1 & & -3 & \vert & -50,000 \end{array} \right] \\ && \end{eqnarray*}$$

Step 3:        We will now work with column 1. We want 1 in Cell 22, and we achieve this by multiplying Row 2 by 50 for a new Row 2.

$$\begin{eqnarray*}&& \\ 50\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ...0 & & -1 & & -3 & \vert & -50,000 \end{array} \right] \\ && \end{eqnarray*}$$

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 Row 2 to Row 3 5o form a new Row 3.

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

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ... 0 & & 0 & & -1 & \vert & -15,000 \end{array} \right] \\ && \end{eqnarray*}$$

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 -1.

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

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ... 0 & & 0 & & 1 & \vert & 15,000 \end{array} \right] \\ && \end{eqnarray*}$$

Step 6:        Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding Row 3 to Row 1 for a new Row 1 and adding -2 times Row 3 to Row 2 for a new Row 2.

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

$$\begin{eqnarray*}&& \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) ... ... 0 & & 0 & & 1 & \vert & 15,000 \end{array} \right] \\ && \end{eqnarray*}$$

The solution is: $x=30,000,\ y=5,000,$ and z=15,000. In terms of the original problems, $\$30,000$ was invested at $6\%$, $\$5,000$ was invested at $8\%$, and $\$15,000$ was invested at $10\%$.

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If you would like to review the solution to the next problem, click on Problem

If you would like to return to the beginning of the three by three system of equations, click on Example.

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