We shall solve systems of three linear equations algebraically. Learning Objective We shall solve systems of three linear equations algebraically. CFU What are we going to do today? In your own words, what does Solve mean? Algebraically means __________. Activate Prior Knowledge Solve each system of two equations algebraically. CFU How do you solve systems of two linear equations? Connection Students you already know how to solve systems of two linear equations. We will now use the same techniques to solve systems of three linear equations in this lesson. Academic Vocabulary Algebraically Techniques 

Concept Development A system of linear equations (or linear system) is a collection of linear equations involving the same set of variables x, y, & z. The solution set is the where all three lines meet at (x, y, z). Examples: You can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned Algebra-1. Academic Vocabulary  Linear System Elimination Substitution 1 CFU: How are we going to solve system of three linear equations?

/ Use elimination to solve the system of equations. Skill Development/Guided Practice Use elimination to solve the system of equations. Steps to Complete the Skill: 1 4 2 3 Eliminate one variable. Eliminate another variable. Then solve for the remaining variable. Use one of the equations in your 2-by-2 system to solve for x. Substitute for x and y in one of the original equations to solve for z. / Academic Vocabulary Eliminate Substitute CFU What are the four steps to solve a systems of three equations algebraically?

You now have a 2-by-2 system. 5x + 9y = –1 Skill Development/Guided Practice (continued) Use elimination to solve the system of equations. CFU How did I eliminate down to two equations? Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second & third equation is 1. –x + y + 2z = 7 –x + y + 2z = 7 Multiply equation - by –2, and add to equation . 1 2 1 –2(2x + 3y + z = 1) –4x – 6y – 2z = –2 2 –5x – 5y = 5 4 Use equations and to create a second equation in x and y. 1 3 –x + y + 2z = 7 –x + y + 2z = 7 Multiply equation - by –2, and add to equation . 1 3 1 –2(–3x – 4y + z = 4) 6x + 8y – 2z = –8 3 1 5x + 9y = –1 5 –5x – 5y = 5 4 You now have a 2-by-2 system. 5x + 9y = –1 5

Step 2 Eliminate another variable. Skill Development/Guided Practice (continued) Use elimination to solve the system of equations. CFU How do you solve systems to two linear equations? Step 2 Eliminate another variable. Then solve for the remaining variable. –5x – 5y = 5 4 Add equation to equation . 4 5 5x + 9y = –1 5 4y = 4 Solve for y. y = 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. 4 –5x – 5y = 5 –5x – 5(1) = 5 Substitute 1 for y. 1 –5x – 5 = 5 Solve for x. –5x = 10 x = –2 1

Step 4 Substitute for x and y in one of the original Skill Development/Guided Practice (continued) Use elimination to solve the system of equations. CFU How do you solve for Z when you know x=-2 & y=1? x = –2 y = 1 z = ? Step 4 Substitute for x and y in one of the original equations to solve for z. 2x +3y + z = 1 2 1 2(–2) +3(1) + z = 1 Solve for z. –4 + 3 + z = 1 z = 2 1

Step 1 Eliminate one variable. Provide matched teacher/student problems Use elimination to solve the system of equations. CFU How did I eliminate down to two equations? Step 1 Eliminate one variable. Multiply equation by –1 and add to equation . 1 2 5x – y + 2z = 4 5x – y + 2z = 4 1 –1(2x – y + 3z = 7) –2x + y – 3z = –7 3 3x – z = –3 4 Multiply equation by 3 and add to equation . 3 2 3(2x – y + 3z = 7) 6x – 3y + 9z = 21 2 –9x + 3y – 6z = –12 –9x + 3y – 6z = –12 3 –3x + 3z = 9 5

z = 3  Step 1 Eliminate one variable. 3x – z = –3 Provide matched teacher/student problems Use elimination to solve the system of equations. CFU How did I eliminate down to two equations? Step 1 Eliminate one variable. 3x – z = –3 Now you have a 2-by-2 system. 4 –3x + 3z = 9 5 Step 2 Eliminate another variable. Then solve for the remaining variable. 3x – z = –3 4 –3x + 3z = 9 5 2z = 6 z = 3 

z = 3   Step 1 Eliminate one variable. 3x – z = –3 Provide matched teacher/student problems Use elimination to solve the system of equations. CFU How do you solve for z & x? Step 1 Eliminate one variable. 3x – z = –3 Now you have a 2-by-2 system. 4 –3x + 3z = 9 5 Step 2 Eliminate another variable. z = 3  Step 3 Use one of the equations in your 2-by-2 system to solve for x. 3x – z = 3 3x – 3 = 3 3x = 6 x = 2 

z = 3    Step 2 Eliminate another variable. Provide matched teacher/student problems Use elimination to solve the system of equations. CFU How do you solve for z & x? Step 2 Eliminate another variable. z = 3  Step 3 system to solve for x x = 2  Step 4 Substitute for x and z in one of the original equations to solve for y. 

Relevance It is relevant to solve systems of three linear equations algebraically:

Relevance It is relevant to solve systems of three linear equations algebraically:

Relevance It is relevant to solve systems of three linear equations algebraically: CFU Does anyone else have any other reason why it is relevant? Which reason is most relevant to you? Why? You can give me my reason or one of your own.

Solve any systems of three linear equations algebraically. Closure Solve any systems of three linear equations algebraically. Steps to Complete the Skill: 1 4 2 3 Eliminate one variable. Eliminate another variable. Then solve for the remaining variable. Use one of the equations in your 2-by-2 system to solve for x. Substitute for x and y in one of the original equations to solve for z. CFU (Strategic questions: How did I/you eliminate one 3-by-3 to 2-by-2 system?) What did you learn today about _______   Independent Practice/Periodic Reviews