Hi Everyone!

On this page you will find some material about Lesson 3. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 3: 3-D Systems of Equations **

Table of Contents

### Resources

*In this section you will find some important information about the specific resources related to this lesson: *

*the learning outcomes,**the section in the textbook,**the WeBWorK homework sets,**a link to the pdf of the lesson notes,**a link to a video lesson.*

**Learning Outcomes.**

- Solve a system of three linear equations in three variables.
- Give a geometric interpretation to solving a system of three linear equations in three variables.
- Identify when the system is consistent or inconsistent.
- If the system is consistent, identify when the solution is unique or there are infinitely many solutions.

**Topic**. This lesson covers

Section 3.6: Systems of Linear Equations in Three Variables.

**WeBWorK**. There is one WeBWorK assignment on today’s material:

$3\times 3$-Systems

**Lesson Notes.**

**Video Lesson. **

Video Lesson 3 (based on Lesson 3 Notes)

### Warmup Questions

*These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.*

#### Warmup Question 1

Solve $$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

#### Show Answer Using the Substitution Method

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

$\bigstar$ Substitution Method

From the first equation, we obtain $y=1-2x$. We substitute $y=1-2x$ into the second equation and get

\begin{align*}3x+2(1-2x)&=3\\3x+2-4x&=3\\-x&=1\\ x&=-1\end{align*}

So $y=1-2(-1)=3$. The solution set is $\{(-1,3)\}$.

#### Show Answer Using the Addition Method

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

We multiply both sides of the first equation by $-2$. The system becomes

$$\begin{cases}-4x-2y=-2\\ 3x+2y=3\end{cases}$$

By adding up the two equations, we obtain

$$-x=1,\text{ that is, } x=-1.$$

We substitute $x=-1$ into $2x+y=1$ to obtain $y$:

\begin{align*}2(-1)+y&=1\\-2+y&=1\\y&=3\end{align*}

The solution set is $\{(-1,3)\}$.

#### Show Answer Using the Graphing Method

$$\begin{cases}2x+y=1\\ 3x+2y=3\end{cases}$$

On the graph below the red line represents $2x+y=1$ and the blue line represents $3x+2y=3$. The lines intersect at $(-1,3)$. The solution set is $\{(-1,3)\}$.

### Review

*If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.*

Need a review? Check Lesson 2.

### Quick Intro

*This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.*

**A Quick Intro** **to Systems of Linear Equations**

** in Three Variables**

**Key Words.** System, linear equations, solution to a system, consistent, inconsistent, the Addition Method.

In the warmup question we solved a system of 2 linear equations and 2 variables using: the **Substitution Method**, the **Addition Method** and the **Graphing Method**.

In this lesson we will see how to solve a system consisting of 3 linear equations and 3 variables. **A solution must satisfy the 3 equations simultaneously.**

Each equation can be thought of a plane or a line (if it has 2 unknowns). So we are looking for the place(s) where the 3 planes/lines intersect each other. The intersection point is a point that satisfies all three equations. The possibilities are:

$\bigstar$ there is **only one solution**

$\bigstar$ there are **infinitely many solutions**

$\bigstar$ there is **no solution**

When there is no solution, the system is called **inconsistent**. If there is a solution (no matter how many), the system is said to be **consistent**.

To solve a system in 3 variables, we use the **Addition Method** in order to eliminate one variable and obtain a system of 2 equations in 2 variables.

### Video Lesson

*Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!*

**Video Lesson**

**A description of the video**

In the video you will see how to solve

$$\begin{cases}x+2y-z=-3\\ -x+y-2z=-6\\2x-y-z=1\end{cases}$$

### Try Questions

*Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.*

#### Try Question 1

Solve $$\begin{cases}5x-y+3z=6\\ x+y+2z=3\\-x-2y+z=8\end{cases}$$

#### Show Answer 1

$$\begin{cases}5x-y+3z=6 \qquad\text{(A)}\\ x+y+2z=3\qquad\text{(B)}\\-x-2y+z=8\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}x+y+2z=3\qquad\text{(B)}\\-x-2y+z=8\qquad\text{(C)}\end{cases}$$

Adding (B) and (C) eliminates $x$:

$$-y+3z=11\qquad\text{(D)}$$

Now consider (A) and (B):

$$\begin{cases}5x-y+3z=6\qquad\text{(A)}\\ x+y+2z=3\qquad\text{(B)}\end{cases}$$

Multiply (B) by $-5$ to eliminate $x$:

$$\begin{cases}5x-y+3z=6\qquad\text{(A)}\\ -5x-5y-10z=-15\qquad\text{(-5B)}\end{cases}$$

Adding the two equations gives:

$$-6y-7z=-9\qquad\text{(E)}$$

Consider the system consisting of (D) and (E)

$$\begin{cases}-y+3z=11\qquad\text{(D)}\\-6y-7z=-9\qquad\text{(E)}\end{cases}$$

and solve for $y$ and $z$. Multiply (D) by $-6$:

$$\begin{cases}6y-18z=-66\qquad\text{(-6D)}\\-6y-7z=-9\qquad\text{(E)}\end{cases}$$

Adding the two equations gives $-25z=-75$, so $z=3$. By (D), we have

\begin{align*} -y+3\cdot 3 &=11\\-y+9&=11\\-y&=2\\y&=-2\end{align*}

Substituting $y=-2$ and $z=3$ into (B) gives:

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

We have $x=-1$, $y=-2$ and $z=3$.

**Check:**

$$\begin{cases}5(-1)-(-2)+3\cdot 3\stackrel{?}{=}6 \qquad\checkmark\\ (-1)+(-2)+2\cdot 3\stackrel{?}{=}3\qquad\checkmark\\-(-1)-2(-2)+3\stackrel{?}{=}8\qquad\checkmark\end{cases}$$

So the solution set is $\{(-1,-2,3)\}$.

#### Try Question 2

Solve $$\begin{cases}3x-y+z=7\\ x+y+z=-3\\4x+2z=4\end{cases}$$

#### Show Answer 2

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\\4x+2z=4\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\end{cases}$$

Adding (A) and (B) eliminates $y$:

$$4x+2z=4$$

or

$$2x+z=2\qquad\text{(D)}$$

Now consider the system consisting of (C) and (D):

$$\begin{cases}4x+2z=4\qquad\text{(C)}\\ 2x+z=2\qquad\text{(D)}\end{cases}$$

and solve for $x$ and $z$. Divide (C) by $2$:

$$\begin{cases}2x+z=2\\ 2x+z=2\end{cases}$$

Subtracting the two equations gives $0=0$. So the system has infinitely many solutions.

#### Try Question 3

Solve $$\begin{cases}3x-y+z=7\\ x+y+z=-3\\4x+2z=5\end{cases}$$

#### Show Answer 3

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\\4x+2z=5\qquad\text{(C)}\end{cases}$$

Consider:

$$\begin{cases}3x-y+z=7 \qquad\text{(A)}\\ x+y+z=-3 \qquad\text{(B)}\end{cases}$$

Adding (A) and (B) eliminates $y$:

$$4x+2z=4$$

or

$$2x+z=2\qquad\text{(D)}$$

Now consider the system consisting of (C) and (D):

$$\begin{cases}4x+2z=5\qquad\text{(C)}\\ 2x+z=2\qquad\text{(D)}\end{cases}$$

and solve for $x$ and $z$. Multiply (D) by $2$:

$$\begin{cases}4x+2z=5\\ 4x+2z=4\end{cases}$$

Subtract the two equations to get $0=1$, which is not possible. So there is no solution.

### WeBWorK

*You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.*

**WeBWork**

It is time to do the homework on WeBWork:

$3\times 3$-Systems

When you are done, come back to this page for the Exit Questions.

### Exit Questions

*After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!*

#### Exit Questions

- What does it mean to solve a system of equations?
- What is the purpose of the Method of Addition?
- What is the graphical description of the system of linear equations?

$\bigstar$ Solve the system of equations $$\begin{cases}2x+y-z=4\\x+2y+2z=1\\3x-y+2z=0\end{cases}$$

#### Show Answer

\[\begin{cases}2x+y-z=4\qquad\text{(A)}\\x+2y+2z=1 \qquad\text{(B)}\\ 3x-y+2z=0 \qquad\text{(C)}\end{cases}\]

We start by eliminating $y$ from (A) and (B). Multiply (A) by $-2$ and keep (B).

$$\begin{cases}-4x-2y+2z=-8 \qquad\text{(-2A)}\\ x+2y+2z=1 \qquad\text{(B)}\end{cases}$$

Adding these two equations gives

$$ -3x+4z=-7\qquad\text{(D)}.$$

Now consider (A) and (C).

$$\begin{cases}2x+y-z=4 \qquad\text{(A)}\\ 3x-y+2z=0\qquad\text{(C)}\end{cases}$$

To eliminate $y$, we add these two equations and obtain

$$5x+z=4\qquad\text{(E)}.$$

Consider the system with equations (D) and (E):

$$\begin{cases}-3x+4z=-7\qquad\text{(D)} \\5x+z=4 \qquad\text{(E)}\end{cases}$$

and solve for $x$ and $z$. Multiply (E) by $-4$:

$$\begin{cases} -3x+4z=-7 \qquad\text{(D)}\\ -20x-4z=-16\qquad\text{(-4E)}\end{cases}$$

Addding these equations gives

\begin{align*}-23x&=-23\\x&=1\end{align*}

Substitute $x=1$ into (E):

\begin{align*}5x+z&=4\\5\cdot 1+z&=4\\5+z& = 4\\z&=-1\end{align*}

Now substitute $x=1$ and $z=-1$ into (A):

\begin{align*}2(1)+y-(-1)&=4\\2+y+1&=4\\3+y&=4 \\y&=1\end{align*}

So $x=1$, $y=1$, $z=-1$.

**Check:**

$$\begin{cases}2x+y-z=4\\x+2y+2z=1\\3x-y+2z=0\end{cases}$$

$$\begin{cases}2(1)+1-(-1)\stackrel{?}{=}4 \quad\checkmark\\1+2(1)+2(-1)\stackrel{?}{=}1\quad\checkmark\\3(1)-1+2(-1)\stackrel{?}{=}0\quad\checkmark \end{cases}$$

The solution set is $\{(1,1,-1)\}$.

### Need more help?

*Don’t wait too long to do the following.*

**Need more help?**

Don’t wait too long to do the following.

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.