It is often desirable or even necessary to use more than one variable to model situations in many fields. 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 7:**

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

= | 20 | (1) | |

3x-5y+2z |
= | 11 | (2) |

= | 7 | (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: Let's get rid of the fractions. We can do this by multiplying
both sides of equation (1) by 12, and both sides of equation (3) by 2.

Step 2: Solve for one of the variables in one of the equations. It makes no difference which equation and which variable to choose. Let's solve for

Step 3: Substitute this value for

= | 11 | ||

= | 11 | ||

= | 11 | ||

-13y-28z |
= | -229 | |

13y+28z |
= | 229 | (4) |

= | 14 | ||

= | 14 | ||

= | 42 | ||

1,200-40y-150z+18y-90z |
= | 42 | |

-22y-240z |
= | -1,158 | |

11y+120z |
= | 579 | (5) |

Step 3: We now have two equations in two unknowns.

Step 4: Solve for

Step 5: Substitute this value of

Step 6: Substitute this value of

Step 7: Substitute 4 for

The solutions: Step 8: Let's check this solution.

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 *z* first.

Step 2: Add equations (1) and (3) to form equation (4).

Step 3: Add 15 times equation (2) and equation (3) to form equation (5).

Step 4: We now have two equations with two variables.

Step 5: Add 25 times equation (4) to -2 times equation (5) to form equation (6).

Step 6. Substitute this value for y in equation (4) and solve for x.

Substitute the value for x and y in equation (1) and solve for z.

Check the solution as we did 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.

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. We would like the number 1 in
cell 11(Row1-Col 1). We can achieve this by multiplying Row 1 by
.

Step 3: Our next objective is get zeros in Cell 12 and Cell 13. We can do this by adding -3 times Row 1 to Row 2 to form a new Row 2, and adding -5 times Row 1 to Row 3 to form a new Row 3.

Step 4: Our new objective is to get a 1 in Cell 22. To achieve this, multiply row 2 by

Step 5: 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 adding times Row 2 to Row 3 to form a new Row 3..

Step 6: We now want a 1 in Cell 33. Do this by multiplying Row 3 by

Step 7: Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 33. We do this by adding times Row 3 to Row 2 to form a new Row 2 and adding times Row 3 to Row 1 to form a new Row 1.

Now you can read the solution from the matrix.

If you would like to test yourself by working some problem similar to this example, click on
**Problem**.

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**Author**: Nancy Marcus

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