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

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:

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 ]
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 ]
[ The number of packages weighing 5 lbs or less ] times [ The number
of packages weighing 10 lbs or more ]

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

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

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 ] can now be written as

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 ] can now be written as

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

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

(1) | |||

(2) | |||

(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 in equation (3) because the equation only has two variables.

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

(4) | |||

(5) | |||

Step 3: Solve for in equation (4).

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

Step 5: Solve for .

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

Step 7: Substitute for and for in equation (1) and solve for .

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. Step 8: Check the solutions:

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 first because is
missing from equation (3).

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

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

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

Step 6: Substitute for in equation (3) and solve for .

Step 7: Substitute for and for in equation (1) and solve for .

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

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.

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

to the following matrix.

Because then you can read the matrix as , and .

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

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 to form the new Row 2

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

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

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

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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