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

where A, B, C, and D are real numbers and A, B, C, and D are not all 0.

**Problem 3.1f:**

Solve the following system of equations for x, y and z:

= | -217 | (1) | |

= | 6 | (2) | |

= | -12 | (3) | |

**
**

Since most people cringe at the site of fractions, let's get rid of them.
Multiply each equation by the smallest number that each denominator will
divide into evenly.

We have created three equivalent equivalent equations. Remember equivalent equations have the same solution as the original equation. The equivalent equations are:

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

14x+9y-4z |
= | 63 | |

= | 63 | ||

= | |||

= | 2,205 | ||

-56y+420z-30,380+315y-140z |
= | 2,205 | |

259y+280z |
= | 32,585 | |

37y+40z |
= | 4,655 | (4) |

20x+12y-7z |
= | -48 | |

= | -48 | ||

= | |||

= | -1,680 | ||

-80y+600z-43,400+420y-245z |
= | -1,680 | |

340y+355z |
= | 41,720 | |

68y+71z |
= | 8,344 | (5) |

Step 3: Solve for

Step 4: Substitute this value of

Step 5: Substitute this value of

Step 7: Substitute 35 for

The solution: and

Step 8: Check the solutions in the original equations:

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: Add -2 times equation (1) to 5 equation (2) to form Equation (4). Add -4 times equation (2) to 7 times equation (3) to form equation (5).

Step : We now have two equations with two variables.

Step 4: Add -68 times equation (5) to 37 times equation (5) to create equation (6) with just one variable.

Step 6: Substitute 84 for

Step 7: Substitute 35 for

The solution: and

**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.

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. Multiply Row 1 by
to form a new Row 1.

Step 3: We will now complete with column 1. We want zeros in Cell 21 and Cell 31. We achieve this by adding -14 times Row 1 to Row 2 to form a new Row 2 and, by adding -20 times Row 1 to Row 3 to form a new Row 3.

Step 4: Let's now work with column 2.We want the number 1 in Cell 22. We do this by multiplying row 2 by to form a new row 2.

Step 5: To complete our work with Row 2 we want zeros in cell 12 and cell 32. We achieve this by adding times Row 2 to Row 1 for a new Row 1, and adding times Row 2 to Row 3 for a new Row 3

Step 6: Let's now work the Column 3. We first want the number 1 in Cell 33. We can do this by multiplying Row 3 by to form a new Row 3.

If you would like to go back to the problem page, 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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**Author**: Nancy Marcus

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