# 7.8: Determining the Type of Linear System

**At Grade**Created by: CK-12

**Practice**Determining the Type of Linear System

### Determining the Type of Linear System

A third option for identifying systems as consistent, inconsistent or dependent is to just solve the system and use the result as a guide.

#### Consistent Systems

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

\begin{align*}10x - 3y &= 3\\ 2x + y &= 9\end{align*}

Let’s solve this system using the substitution method.

Solve the second equation for \begin{align*}y\end{align*}:

\begin{align*}2x + y = 9 \Rightarrow y = -2x + 9\end{align*}

Substitute that expression for \begin{align*}y\end{align*} in the first equation:

\begin{align*}10x - 3y &= 3 \\ 10x - 3(-2x + 9) &= 3\\ 10x + 6x - 27 &= 3\\ 16x &= 30\\ x &= \frac{15}{8}\end{align*}

Substitute the value of \begin{align*}x\end{align*} back into the second equation and solve for \begin{align*}y\end{align*}:

\begin{align*}2x + y = 9 \Rightarrow y = -2x + 9 \Rightarrow y = -2 \cdot \frac{15}{8} + 9 \Rightarrow y = \frac{21}{4}\end{align*}

The solution to the system is \begin{align*}\left ( \frac{15}{8}, \frac{21}{4} \right )\end{align*}. The system is **consistent** since it has only one solution.

#### Inconsistent Systems

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

\begin{align*}3x - 2y & = 4\\ 9x - 6y & = 1\end{align*}

Let’s solve this system by the method of multiplication.

Multiply the first equation by 3:

\begin{align*}3(3x - 2y = 4) \qquad \qquad \qquad \qquad 9x - 6y = 12\!\\ {\;} \qquad \qquad \qquad \qquad \Rightarrow \!\\ 9x - 6y = 1 \qquad \qquad \qquad \qquad \quad \ 9x - 6y = 1\end{align*}

Add the two equations:

\begin{align*}& \qquad 9x - 6y = 4\\ & \qquad \underline{9x - 6y = 1}\\ & \qquad \qquad \ \ 0 = 13 \quad \text{This statement is not true.}\end{align*}

If our solution to a system turns out to be a statement that is not true, then the system doesn’t really have a solution; it is **inconsistent.**

#### Dependent Systems

Solve the following system of equations. Identify the system as consistent, inconsistent or dependent.

\begin{align*}4x + y &= 3\\ 12x + 3y &= 9\end{align*}

Let’s solve this system by substitution.

Solve the first equation for \begin{align*}y\end{align*}:

\begin{align*}4x + y = 3 \Rightarrow y = -4x + 3\end{align*}

Substitute this expression for \begin{align*}y\end{align*} in the second equation:

\begin{align*}12x + 3y &= 9\\ 12x + 3(-4x + 3) &= 9\\ 12x - 12x + 9 &= 9\\ 9 &= 9\end{align*}

This statement is always true.

If our solution to a system turns out to be a statement that is always true, then the system is **dependent.**

A second glance at the system in this example reveals that the second equation is three times the first equation, so the two lines are identical. The system has an infinite number of solutions because they are really the same equation and trace out the same line.

Let’s clarify this statement. An infinite number of solutions does not mean that *any* ordered pair \begin{align*}(x, y)\end{align*} satisfies the system of equations. Only ordered pairs that solve the equation in the system (either one of the equations) are also solutions to the system. There are infinitely many of these solutions to the system because there are infinitely many points on any one line.

For example, (1, -1) is a solution to the system in this example, and so is (-1, 7). Each of them fits both the equations because both equations are really the same equation. But (3, 5) doesn’t fit either equation and is not a solution to the system.

In fact, for every \begin{align*}x-\end{align*}value there is just one \begin{align*}y-\end{align*}value that fits both equations, and for every \begin{align*}y-\end{align*}value there is exactly one \begin{align*}x-\end{align*}value—just as there is for a single line.

### Example

#### Example 1

Identify the system as consistent, inconsistent, or consistent-dependent.

\begin{align*}3x-2y&=4\\ 9x-6y&=1\end{align*}

**Solution:** Because both equations are in standard form, elimination is the best method to solve this system.

Multiply the first equation by 3.

\begin{align*}3(3x-2y=4)&&9x-6y=12\\ & \qquad \qquad \qquad \qquad \Rightarrow & \qquad\\ 9x-6y=1&&9x-6y=1\end{align*}

Subtract the two equations.

\begin{align*}& \ \ 9x-6y=12\\ & \underline{\;\; 9x-6y=1 \;\;}\\ & \qquad \quad \ 0=11 \quad \text{This Statement is not true.}\end{align*}

This is an untrue statement; therefore, you can conclude:

- These lines are parallel.
- The system has no solution.
- The system is inconsistent.

### Review

Find the solution of each system of equations using the method of your choice. State if the system is inconsistent or dependent.

- \begin{align*}3x + 2y = 4\!\\ - 2x + 2y = 24\end{align*}
- \begin{align*}5x - 2y = 3\!\\ 2x - 3y = 10\end{align*}
- \begin{align*}3x - 4y = 13\!\\ y = -3x - 7\end{align*}
- \begin{align*}5x - 4y = 1\!\\ -10x + 8y = -30\end{align*}
- \begin{align*}4x + 5y = 0\!\\ 3x = 6y + 4.5\end{align*}
- \begin{align*}-2y + 4x = 8\!\\ y - 2x = -4\end{align*}
- \begin{align*}x - \frac{1}{2}y = \frac{3}{2}\!\\ 3x + y = 6\end{align*}
- \begin{align*}0.05x + 0.25y = 6\!\\ x + y = 24\end{align*}
- \begin{align*}x + \frac{2}{3}y = 6\!\\ 3x + 2y = 2\end{align*}
- \begin{align*}&3x-4y=13\\ & y=-3x-7\end{align*}
- \begin{align*}&4x+y=3\\ &12x+3y=9\end{align*}
- \begin{align*}&10x-3y=3\\ &2x+y=9\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 7.8.

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Here you'll learn how to solve a system of equations and use the result as a guide in determining the type of system it is.

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