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

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:

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 ]
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 ]
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 ]

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 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 ]
can now be
written in the algebraic form

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 ] can now be written in the algebraic form

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 ] can now be written in the algebraic form

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

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

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).

Step 2: Substitute this value for

0.06x+0.07y+0.12z |
= | ||

= | |||

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 | |

= | 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

Step 4: Substitute this value of

Step 5: Solve for

Step 6: Substitute this value of

Step 7: Substitute 20,000 for

The solution is: and In terms of the original problems, Alicia plans to invest in a safe investment, in a relatively safe investment, and $60,000 in a risky investment. 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 *x* first.

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).

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).

Step : We now have two equations with two variables.

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.

Step 6: Substitute 60,000 for

Step 7: Substitute 60,000 for

The solution is: and In terms of the original problems, Alicia plans to invest in a safe investment, in a relatively safe investment, and $60,000 in a risky investment.

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

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..

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.

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.

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.

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.

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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