SYSTEMS OF EQUATIONS in THREE VARIABLES

It is often desirable or even necessary to use more than one variable to model situations 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.

Example 3:

The ABC Shipping Company charges $2.90 for all packages weighing less than or equal to 5 lbs, $5.20 for packages weighting more than 5 lbs and less than 10 lbs, and $8.00 for all packages weighing 10 lbs or more. The number of packages weighing 5 lbs or less is 50% more than the number of packages weighing 10 lbs or more. One day shipping charges for 300 orders was $1,508. Find the number of packages in each category. .

There are three unknowns:

$$\begin{eqnarray*} && \\ 1 &:&\text{ The number of packages weighing 5 lbs or ... ...&\text{ The number of packages weighing 10 lbs or more} \\ && \end{eqnarray*}$$

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

[ The number of packages weighing 5 lbs or less ] $+$ [ The number of packages weighing between 5 lbs and 10 lbs ] $+$ [ The number of packages weighing 10 lbs or more ] $=300.\bigskip$ The $2.90 per package charge for [ The number of packages weighing 5 lbs or less ] $+$ the $5.20 per package charge for [ The number of packages weighing between 5 lbs and 10 lbs ] $+$ the $8.00 per package charge for [ The number of packages weighing 10 lbs or more ] $=\$1,508\bigskip$ [ The number of packages weighing 5 lbs or less ] $=1.50$ times [ The number of packages weighing 10 lbs or more ]

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

$$\begin{eqnarray*} && \\ 1 &:&\text{ The number of packages weighing 5 lbs or ... ...&\text{ The number of packages weighing 10 lbs or more} \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} && \\ x &=&\text{ The number of packages weighing 5 lbs or ... ...t{ The number of packages weighing more 10 lbs or more} \\ && \end{eqnarray*}$$

in the three sentences, and then rewrite them.

The sentence [ The number of packages weighing 5 lbs or less ] $+$ [ The number of packages weighing between 5 lbs and 10 lbs ] $+$ [ The number of packages weighing 10 lbs or more ] $=300.$ can now be written as

$$\begin{eqnarray*} x+y+z &=&300 \\ && \\ && \end{eqnarray*}$$

The $2.90 per package charge for [ The number of packages weighing 5 lbs or less ] $+$ the $5.20 per package charge for [ The number of packages weighing between 5 lbs and 10 lbs ] $+$ the $8.00 per package charge for [ The number of packages weighing 10 lbs or more ] $=\$1,508$ can now be written as

$$\begin{eqnarray*} \$2.90x+\$5.20y+\$8.00z &=&\$1,508 \\ && \\ && \end{eqnarray*}$$

The sentence [ The number of packages weighing 5 lbs or less ] $=1.50$ times [ The number of packages weighing 10 lbs or more ] can now be written as

$$\begin{eqnarray*} && \\ x &=&1.50z \\ &&or \\ x-1.50z &=&0 \end{eqnarray*}$$

$\bigskip$

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

$$x+y+z = 300$$ (1)

$$\$2.90x+\$5.20y+\$8.00z = \$1,508$$ (2)

$$x-1.50z = 0$$ (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) because the equation only has two variables.

$$\begin{eqnarray*} && \\ x-1.50z &=&0 \\ && \\ x &=&1.50z \\ && \\ && \end{eqnarray*}$$

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.

$$x+y+z = 300$$

$$\left( 1.50z\right) +y+z = 300$$

$$y+2.50z = 300$$ (4)

$$\$2.90x+\$5.20y+\$8.00z = \$1,508$$

$$\$2.90\left( 1.50z\right) +\$5.20y+\$8.00z = \$1,508$$

$$\$5.20y+\$12.35z = \$1,508$$ (5)

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

$$\begin{eqnarray*} && \\ y+2.50z &=&300 \\ && \\ y &=&-2.50z+300 \\ && \\ && \\ && \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} && \\ \$5.20y+\$12.35z &=&\$1,508 \\ && \\ \$5.20\left(... ...+\$12.35z &=&\$1,508 \\ && \\ -0.65z &=&-52 \\ && \\ && \end{eqnarray*}$$

Step 5:         Solve for $z$.

$$\begin{eqnarray*} && \\ -0.65z &=&-52 \\ && \\ z &=&80 \\ && \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} && \\ y+2.50z &=&300 \\ && \\ y+2.50\left( 80\right) &=&300 \\ && \\ y &=&100 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute $100$ for $y$ and $80$ for $z$ in equation (1) and solve for $x$.

$$\begin{eqnarray*} && \\ x+y+y &=&300 \\ && \\ x+100+80 &=&300 \\ && \\ x &=&120 \\ && \end{eqnarray*}$$

The solutions: There were 120 packages that weighed 5 lbs or less, 100 packages that weighed between 5 lbs and 10 lbs, and 80 packages that weighed 10 lbs or more. $\bigskip\bigskip\bigskip$ Step 8:        Check the solutions:

$$\begin{eqnarray*} && \\ 120+100+80 &=&300\rightarrow Yes \\ && \\ \$2.90\... ...( 80\right) &=&0\rightarrow Yes \\ && \\ && \\ && \\ && \end{eqnarray*}$$

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 missing from equation (3).

$$\begin{eqnarray*} && \\ \begin{array}{l} (1) \\ \\ (2) \\ \\ (3) \e... ... - & 1.50z & = & 0 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

Step 2:        Multiply both sides of equation (1) by $-\$5.20$ and then add the transformed equation (1) to equation (2) to form equation (4).

$$\begin{eqnarray*} && \\ \begin{array}{l} (1) \\ \\ (2) \\ \\ (4) \e... ...$2.80z & = & -\$52 \end{array} \end{array} & \\ && \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} (3) &:&x-1.50z=0 \\ && \\ (4) &:&-\$2.30x+\$2.80z=-\$52 \\ && \\ && \end{eqnarray*}$$

Step 4:        Multiply both sides of equation (3) by $\$2.30$ and add to equation (4) to create equation (5) with just one variable.

$\begin{array}{r} (3) \\ \\ (4) \\ \\ (5) \end{array} \ \begin{arr... ...& \$2.80z & = & -\$52 \\ & & & & \\ & & -\$0.65z & = & -\$52 \end{array}$

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

$$\begin{eqnarray*} && \\ -\$0.65z &=&-\$52 \\ && \\ z &=&80 \\ && \\ && \end{eqnarray*}$$

Step 6:        Substitute $80$ for $z$ in equation (3) and solve for $x$.

$$\begin{eqnarray*} && \\ x-1.50z &=&0 \\ && \\ x-1.50\left( 80\right) &=&0 \\ && \\ x &=&120 \\ && \\ && \end{eqnarray*}$$

Step 7:        Substitute $120$ for $x$ and $80$ for $z$ in equation (1) and solve for $y$.

$$\begin{eqnarray*} && \\ x+y+z &=&300 \\ && \\ 120+y+80 &=&300 \\ && \\ y &=&100 \\ && \\ && \end{eqnarray*}$$

In terms of the original problems, $120$ packages weighed 5 lbs or less, $%% 100$ packages weighed more than 5 lbs and less thatn 10 lbs, and $80$ packages weighed 10 lbs or more.Check your answers as before.

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.

$$\begin{eqnarray*} && \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3)... ... \\ 1 & & & & -1.50 & \vert & 0 \end{array} \right] \\ && \end{eqnarray*}$$

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

$$\begin{eqnarray*} && \\ && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3)... ... \\ 1 & & & & -1.50 & \vert & 0 \end{array} \right] \\ && \end{eqnarray*}$$

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 $-2.90$ 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..

$$\begin{eqnarray*} && \\ -2.90\left[ Row\ 1\right] +\left[ Row\ 2\right] &=&\l... ...ght] +\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*} && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{a... ... & & -2.50 & \vert & -300 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 3:        We will now work with column 2. We want 1 in Cell 22, and we achieve this by first interchanging Row 2 and Row 3. Then Multiply the interchanged Row 2 by $-1$ to form the new Row 2

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

$$\begin{eqnarray*} && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{a... ...30 & & 5.10 & \vert & 638 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

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 $-2.30$ times Row 2 to Row 3 5o form a new Row 3.

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

$$\begin{eqnarray*} && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{a... ...0 & & -0.65 & \vert & -52 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

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 $-\frac{1}{0.65}.$

$$\begin{eqnarray*} && \\ -\frac{1}{0.65}\left[ Row\ 3\right] &=&\left[ New\ Row\ 3\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*} && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{a... ... & & 0 & & 1 & \vert & 80 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

Step 6:        Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding $1.50$ times Row 3 to Row 1 for a new Row 1 and adding $-2.50$ times Row 3 to Row 3 for a new Row 3.

$$\begin{eqnarray*} && \\ 1.50\left[ Row\ 3\right] +\left[ Row\ 1\right] &=&\le... ...ht] +\left[ Row\ 2\right] &=&\left[ New\ Row\ 2\right] \\ && \end{eqnarray*}$$

$$\begin{eqnarray*} && \begin{array}{r} (1) \\ \\ (2) \\ \\ (3) \end{a... ... & & 0 & & 1 & \vert & 80 \end{array} \right] \\ && \\ && \end{eqnarray*}$$

You can now read the answers off the matrix: $x=120$, $y=100$, and $z=80$. In terms of the original problems, $120$ packages weighed 5 lbs or less, $%% 100$ packages weighed more than 5 lbs and less thatn 10 lbs, and $80$ packages weighed 10 lbs or more. Check your answers by the method described above.

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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