### Solutions of Linear Systems

A **system of equations** is a set of equations that are used at the same time. A **solution of a system of equations** is an ordered pair (triple, etc.) that is a solution of every equation in the system.

**consistent system**. A consistent system with only one solution is an

**independent system**, and a consistent system with an infinite number of solutions is a

**dependent system**. For some systems, there are no values that make each equation true. A system with no solution is called an

**inconsistent system**.

A system of two equations in two variables can be represented by the graphs of the lines. The solutions of the system are points that are on both lines.

### Solutions of Systems of Linear Equations in Two Variables

Consistent and independent; one solution | Consistent and dependent; infinite number of solutions | Inconsistent; no solution |
---|---|---|

Two lines intersecting at one point | Two lines that coincide with each other | Two parallel lines that do not intersect |

### Solutions of Systems of Linear Equations in Three Variables

Consistent and independent; one solution | Consistent and dependent; infinite number of solutions | Inconsistent; no solution |
---|---|---|

Three planes intersecting at one point | Three planes that intersect at one line | Three planes that do not intersect at a common point |

### Graphing Systems of Linear Equations

Graph each equation.

The first equation is:Graph each equation. Locate any points of intersection.

Both equations are equivalent to:### Substitution

The **substitution method** is a method for solving a system of equations by solving for one variable and then substituting the result back into the other equation or equations.

To solve by substitution:

1. Solve one of the equations for one for the variables.

2. Substitute that expression for the variable in the other equation.

3. Solve for the remaining variable.

4. Write the solution as an ordered pair.

Substitute the expression from Step 1 for the variable in the other equation.

Substitute $6+3y$ for $x$ in the first equation.Substitute the value from Step 3 back into either of the original equations and solve for the other variable.

Substitute –1 for $y$ in the second equation:
Write the values of each variable as an ordered pair.

The solution is $(3, -1)$.

### Elimination

Although the substitution method can be used to solve any system of linear equations, this method sometimes results in equations containing fractions. If it is not possible to solve one of the equations in the system for one of the variables without generating fractions, it may be easier to use another algebraic method to solve the system.

The **elimination method** is a method for solving systems of equations that involves combining multiples of equations to eliminate variables.

To solve by elimination:

1. Multiply both sides of one equation (or both equations, if necessary) by a constant so that one of its terms is the opposite of a term in the other equation.

2. Add the equations to eliminate one of the variables.

3. Solve for the remaining variable.

4. Substitute the value found in Step 3 into either of the original equations and solve for the other variable.

5. Write the solution as an ordered pair.

The elimination method can be expanded and repeated as necessary for a system with two or more equations and variables.

Substitute the value of $x$ found in Step 3 into either of the original equations, then solve for $y$.

Substitute 7 for $x$ in $x-3y=10$.
Write the values of each variable as an ordered pair.

The solution is $(7, -1)$.

Combine two equations to eliminate one of the variables. If necessary, multiply both sides of one equation (or both equations) by a constant so that one of its terms is the opposite of a term in the other equation.

The $x$-terms in equations [1] and [3] are opposites. Add to eliminate the $x$-term. Label the resulting equation as [4].Eliminate the $x$-term from the remaining equation.

Multiply equation [3] by 5 so that its $x$-term is the opposite of the $x$-term in equation [2]. Label the resulting equation as [5].Solve for one of the remaining variables.

Multiply equation [6] by 2 so that its $z$-term is the opposite of the $z$-term in equation [4] from Step 1. Label the resulting equation as [7].Substitute the value found in Step 6 into either of the equations with only two variables, and solve for the other variable.

Substitute –2 for $y$ in equation [4].Substitute the values that have been found into one of the original equations and solve for the remaining variable.

Substitute –2 for $y$ and 4 for $z$ in equation [1].Write the values of each variable as an ordered triple.

The solution is $(6,-2,4)$.

The solution can be checked by substituting the values for $x$, $y$, and $z$ back into the original equations. If all result in true equations, then the solution is correct.