LESSON 25: Solving Systems by Combination/Elimination

Algebra Success T551

LESSON 25: Solving Systems by Combination/Elimination

[OBJECTIVE]

The student will fi nd solutions to systems of equations by combination/elimination.

[MATERIALS]

Student pages S217–S229Transparencies T563, T565, T567, T569, T571, T573, T575, T578Systems of Equations Foldable from Lesson 23Optional: graphing calculators

[ESSENTIAL QUESTIONS]

1. Is it possible to combine any two linear equations and have one of the variables cancel?

2. What happens when using the combination method if the lines are parallel or the same line?

[GROUPING]

Cooperative Pairs, Whole Group, Individual

[LEVELS OF TEACHER SUPPORT]

Modeling (M), Guided Practice (GP), Independent Practice (IP)

[MULTIPLE REPRESENTATIONS]

SOLVE, Graph, Algebraic Formula, Verbal Description

[WARM-UP] (5 minutes – IP) S217 (Answers on T562.)

• Have students turn to S217 in their books to begin the Warm-Up. Students will practice fi nding the solution to a system of equations by graphing and by substitution. Monitor students to see if any of them need help during the Warm-Up. Give students 3 minutes to complete the problems and then spend 2 minutes reviewing the answers as a class. {Algebraic Formula}

[HOMEWORK]: (5 minutes)

Take time to go over the homework from the previous night.

[LESSON]:(47–55 minutes – M, GP, IP)

Algebra SuccessT552

SOLVE Problem (2 minutes – GP) T563, S218 (Answers on T564.)

Have students turn to S218 in their books, and place T563 on the overhead. The fi rst problem is a SOLVE problem. You are only going to complete the S step with students at this point. Tell students that during the lesson they will learn how to fi nd the solution of a system of equations by combination or elimination. They will use this knowledge to complete this SOLVE problem at the end of the lesson. {SOLVE}

Solving Systems (5 minutes – M, GP) T563, S218 (Answers on T564.)

Use the following activity to introduce students to problems that may arise when solving systems by graphing or substitution. Make sure your students understand that graphing fractions is not an exact process and could lead to mistakes. Using substitution with fractions also leads to mistakes. {Algebraic Formula, Verbal Description}




MODELING

Problems in Solving Systems by Substitution and Graphing

Step 1: Discuss Problem 1 with students. Remind students that the fi rst step in solving a system of equations by substitution is to identify a variable that can be easily isolated. Explain that since each variable has a coeffi cient, students would have to subtract or add and then divide to isolate any of the variables.

Step 2: Work with students to complete Problems 2 and 3. Remind students that the fi rst step in solving a system of equations by graphing is to write each of the equations in slope-intercept form.

First Equation:

5x + 3x + 3x y = 16y = 16y

– 5x x x – 5x Subtract 5x from both sides.x from both sides.x

3y3

= -5x3

+ 163

Divide all terms by 3.

y = y = y-5x3

+ 163

Second Equation:

4x – 3x – 3x y = 2y = 2y

– 4x – 4x – 4x x Subtract 4x Subtract 4x x from both sides.

-3y-3

= -4x-3

+ 2-3

Divide all terms by -3.

y = y = y 4x3

– 23

Step 3: Discuss Problem 4 with students. Ask students what problems could occur if they tried to graph the lines described by the two equations above. Most of your students should know that it is hard to graph with fractions. Ask students what problems could occur if they tried to substitute one of these equations into the other. Most students do not like to work with fractions, and so they will say there is a good chance of making a mistake in calculations with all of the fractions. Ask, “If there was an easier way,would you like to try it?”

Algebra SuccessT554


Combination/Elimination (8 minutes – M, GP) T565, T567, S219, S220

(Answers on T566, T568.)

Have students turn to S220 in their books, and place T567 on the overhead. Use the following modeling activity to introduce students to the combination/elimination method. {Algebraic Formula, Verbal Description}

MODELING

Introduce the Combination/Elimination Method

Step 1: Discuss Problems 5 and 6 on S220 (T567) with students. Students should know that they are looking for a solution of (x, y) when solving systems of equations. So the two things they are looking for are the x-coordinate and the y-coordinate.

Step 2: For Problem 7, ask students to look at the system of equations on T567. Ask, “Do the coeffi cients of the same variable have the same absolute value?” In this case the answer is yes, because the coeffi cients of y are 3 y are 3 yand -3, and -3 has an absolute value of 3.

Have students turn to S219 (T565) in their books. Explain to students that, since the answer to the question is yes, they should follow the “Yes” arrow to the next box, where they will fi nd the next question.

Step 3: Have students turn back to S220, look at the system of equations, and discuss Problem 8. Ask, “Do the coeffi cients have opposite signs?” The answer to this question is yes. The coeffi cients of y are 3 and y are 3 and y -3, which are opposites.

Step 4: Have students write the statement, “Combine the equations to solve for one variable” in the appropriate box on S219. Have students turn back to S220. Model for students how to combine the two equations and solve for the variable to complete Problem 9.

5x + 3x + 3x y = 16 Start by adding the like terms:y = 16 Start by adding the like terms:y

4x – 3x – 3x y = 2y = 2y 5x + 4x + 4x x = 9x = 9x x 3x 3x y + y + y -3y = 0 16 + 2 = 18y = 0 16 + 2 = 18y

9x = 18 We are left with 9x = 18 We are left with 9x x equal to 18.x equal to 18.x

9x9

= 189

Divide both sides by 9 to isolate the variable x.

x = 2x = 2x


Step 5: Have students complete Problems 10–13. The value of x is 2. Students x is 2. Students xcan plug that value into one of the original equations to fi nd the value of y. This is the same process students used in substitution.

5x + 3x + 3x y = 16 Write one of the original equations.y = 16 Write one of the original equations.y

5(2) + 3y = 16 Plug in the value of x as 2.x as 2.x

10 + 3y = 16 Simplify by multiplying.y = 16 Simplify by multiplying.y

–10 –10 Subtract 10 from both sides to isolate 3y.

3y3

= 63

Divide both sides by 3 to isolate y.

y = 2y = 2y

If x = 2 and x = 2 and x y = 2, the solution is (2, 2). Have students check the solution y = 2, the solution is (2, 2). Have students check the solution yas they have in the past two lessons:

5x + 3x + 3x y = 16 4y = 16 4y x – 3x – 3x y = 2 Original equationsy = 2 Original equationsy

5(2) + 3(2) = 16 4(2) – 3(2) = 2 Plug in x = 2 and x = 2 and x y = 2.y = 2.y

10 + 6 = 16 8 – 6 = 2 Multiply.

16 = 16 2 = 2 Add or subtract.

Both statements are true, so the solution is correct.

Examples (15 minutes – M, GP) T565, T569, T571, S219, S221, S222

(Answers on T566, T570, T572.)

Have students turn to S221 in their books, and place T569 on the overhead. Use the following modeling activities to model Examples 1–6 on S221 and S222 (T569 and T571) with students. {Algebraic Formula}


Algebra SuccessT556

MODELING

Example 1

Step 1: Ask students to look at the system of equations in Example 1 on S221 (T569). Ask, “Do the coeffi cients of the same variable have the same absolute value?” In this case the answer is yes, because the coeffi cients of y are 5 and y are 5 and y -5, and -5 has an absolute value of 5.

Have students turn to S219 (T565) in their books. Explain to students that, since the answer to the question is yes, they should follow the “Yes” arrow to the next box, where they will fi nd the next question: “Do the coeffi cients have opposite signs?”

Step 2: Have students turn back to S221, look at the system of equations, and answer the question, “Do the coeffi cients have opposite signs?” The answer to this question is yes. The coeffi cients of y are 5 and y are 5 and y -5, which are opposites.

Have students continue to follow the “Yes” arrow on S219 to the last box: “Combine the equations to solve for one variable.”

Step 3: Have students add the two equations and solve for the variable.

3x + 5x + 5x y = 8 Add the like terms in both equations.y = 8 Add the like terms in both equations.y2x – 5x – 5x y = y = y -3 3x + 2x + 2x x = 5x = 5x x 5x 5x y + y + y -5y = 0 8 + y = 0 8 + y -3 = 5

5x5

= 55

Divide both sides by 5 to isolate x.

x = 1x = 1x

Step 4: The value of x is 1. Have students plug that value into one of the original x is 1. Have students plug that value into one of the original xequations to fi nd the value of y.

3x + 5x + 5x y = 8 Write the original equation.y = 8 Write the original equation.y 3(1) + 5y = 8 Plug in the value of 1 for y = 8 Plug in the value of 1 for y x. 3 + 5y = 8 Simplify by multiplying.y = 8 Simplify by multiplying.y –3 –3 Subtract 3 from both sides to isolate the 5y.

5y5

= 55


y = 1y = 1y

The solution is (1, 1), because x = 1 and x = 1 and x y = 1.y = 1.y

Step 5: Have students check the solution by plugging (1, 1) into both equations.

3x + 5x + 5x y = 8 2y = 8 2y x – 5x – 5x y = y = y -3 Original equation 3(1) + 5(1) = 8 2(1) – 5(1) = -3 Plug in x = 1 and x = 1 and x y = 1.y = 1.y 3 + 5 = 8 2 – 5 = -3 Multiply. 8 = 8 -3 = -3 Add or subtract.





MODELING

Example 2

Step 1: Ask students to look at the system of equations in Example 2 on S221 (T569). Ask, “Do the coeffi cients of the same variable have the same absolute value?” In this case the answer is yes, because the coeffi cients of x are 4 and 4.x are 4 and 4.x

Have students turn to S219 (T565) in their books. Explain to students that, since the answer to the question is yes, they should follow the “Yes” arrow to the next box, where they will fi nd the next question: “Do the coeffi cients have opposite signs?”

Step 2: Have students turn back to S221, look at the system of equations, and answer the question, “Do the coeffi cients have opposite signs?” The answer to this question is no. The coeffi cients of x are 4 and 4, which x are 4 and 4, which xhave the same sign.

Have students turn to S219 (T565) in their books. Explain to students that, since the answer to the question is no, they should follow the “No” arrow to the next box. In that box, have students write the statement “Multiply one equation by -1.”

Explain to students that, in order to get the coeffi cients to have opposite signs, they should multiply one equation by -1. Have students multiply the second equation by -1:4x + 2y = 2 y = 2 y → 4x + 2x + 2x y = 2 y = 2 y → 4x + 2x + 2x y = 2y = 2y

4x 4x 4 – 3x – 3x y = y = y -13 → -1(4x1(4x1(4 – 3x – 3x y = y = y -13) → -4x4x4 + 3x + 3x y = 13y = 13y

Step 3: Demonstrate to students that the next step, shown on S219, is to combine the equations and solve for one variable.

Have students add the two equations and solve for y. Then have students plug in the value of y into one of the original equations and solve for y into one of the original equations and solve for y x:

4x + 2y = 2 Add the like terms in both equations. -4x + 3y = 13 4x + x + x -4x = 0x = 0x x x x 2y + 3y + 3y y = 5y = 5y y y y 2 + 13 = 15

5y5

= 155


y = 3y = 3y 4x + 2x + 2x y = 2 Write one of the original equations.y = 2 Write one of the original equations.y 4x + 2(3) = 2 Plug in the value of 3 for x + 2(3) = 2 Plug in the value of 3 for x y. 4x + 6 = 2 Multiply.x + 6 = 2 Multiply.x –6 –6 Subtract 6 from both sides to isolate 4x.

4x4

= -44


x = x = x -1

The solution is (-1, 3), because x = x = x -1 and y = 3.y = 3.y

Algebra SuccessT558

Step 4: Have students check the solution by plugging (-1, 3) into both original equations.

4x + 2x + 2x y = 2 4y = 2 4y x – 3x – 3x y = y = y -13 Original equations.

4(-1) + 2(3) = 2 4(-1) – 3(3) = -13 Plug in x = x = x -1 and y = 3.y = 3.y

-4 + 6 = 2 -4 – 9 = -13 Multiply.

2 = 2 -13 = -13 Add or subtract.


MODELINGExample 3

Step 1: Ask students to look at the system of equations in Example 3 on S221 (T569). Ask, “Do the coeffi cients of the same variable have the same absolute value?” In this case the answer is no, because the coeffi cients of x are 4 and 3, and the coeffi cients of x are 4 and 3, and the coeffi cients of x y are y are y -2 and 4.

Have students turn to S219 (T565) in their books. Explain to students that, since the answer to the question is no, they should follow the “No” arrow to the next box. In that box, have students write the statement “Choose xor y and multiply to create a common coeffi cient.” Have students multiply y and multiply to create a common coeffi cient.” Have students multiply ythe fi rst equation by 2 to create a common coeffi cient:

4x4x4 – 2x – 2x y = y = y -4 → 2(4x 2(4x 2(4 – 2x – 2x y = y = y -4) → 8x 8x 8 – 4x – 4x y = y = y -8 3x + 4x + 4x y = y = y -14 → 3x + 4x + 4x y = y = y -14 → 3x + 4y = y = y -14

Step 2: Have students turn back to S221, look at the system of equations, and answer the question, “Do the coeffi cients have opposite signs?” The answer to this question, now, is yes. The coeffi cients of y are 4 and y are 4 and y -4, which are opposites.

Have students continue to follow the “Yes” arrow on S219 to the last box: “Combine the equations to solve for one variable.”




Step 3: Have students add the two equations and solve for x. Then have students plug in the value of y into one of the original equations and solve for y into one of the original equations and solve for y y:

8x – 4x – 4x y = y = y -8 Add the like terms in both equations.3x + 4y = -14 8x + 3x + 3x x = x = x 11x -4y + 4y + 4y y = 0y = 0y y -8 + -14 = -22

11x11

= -2211


x = x = x -2

4x – 2x – 2x y = y = y -4 Write one of the original equations.

4(-2) – 2y = y = y -4 Plug in the value of -2 for x.-8 – 2y = y = y -4 Simplify by multiplying.

+ 8 + 8 Add 8 to both sides to isolate -2y.

-2y-2

= 4-2

Divide both sides by -2 to isolate y.

y = y = y -2

The solution is (-2, -2), because x = x = x -2 and y = y = y -2.

Step 4: Have students check the solution by plugging (-2, -2) into both original equations.

4x – 2x – 2x y = y = y -4 3x + 4x + 4x y = y = y -14 Original equations.

4(-2) – 2(-2) = -4 3(-2) + 4(-2) = -14 Plug in x = x = x -2 and y = y = y -2.

-8 + 4 = -4 -6 + -8 = -14 Multiply.

-4 = -4 -14 = -14 Add.


Use the completed diagram on S219 (T565) and the steps above to complete Examples 4–6 on S222 (T571) with students. Explain to students that it may be necessary to multiply both equations by different numbers to get common coeffi cients.

In Example 5, students will be left with the incorrect statement 0 = 72. Remind students that when the variables cancel out, and students are left with an incorrect statement, the system of equations has no solution. In Example 6, students will be left with the correct statement 0 = 0. Remind students that when the variables cancel out, and students are left with a correct statement, the system of equations has infi nitely many solutions.

Algebra SuccessT560


More Combination/Elimination (10 minutes – GP, IP) T565, T573, S219, S223 (Answers on T566, T574.)

Have students complete the four problems on S223 in cooperative pairs. Monitor them closely. If you feel like students are struggling, complete these problems as a class. Make sure students are using their graphic organizer on S219. Give students 8 minutes to complete the problems and use 2 minutes to review the answers. {Algebraic Formula}

SOLVE Problem (7 minutes – GP) T575, S224 (Answers on T576.)

Remind students that the SOLVE problem is the same one from the beginning of the lesson. Complete the SOLVE problem with your students. They may have problems writing the two equations which represent the situation. {SOLVE, Algebraic Formula}

Systems Foldable (5 minutes – M, GP, IP)

Have students get out their foldables that they started in Lesson 23 and label the third fl ap. On the inside, complete the section for fi nding the solution to the system of equations by combination /elimination. Create a transparency to model for students each fl ap.

If time permits... (8 minutes – IP) S225 (Answers on T577.)

Have students fi nd the solutions to the systems of equations on S225. Students can work in cooperative pairs or independently. Give students 6 minutes to complete the problems. Use 2 minutes to review the answers. {Algebraic Formula}

Instructions for using the graphing calculator to solve systems of equations with matrices are included on S226 (T578). Please use these if they are appropriate for your students.


[CLOSURE]: (5 minutes)

• To wrap up the lesson, go back to the essential questions and discuss them with students. • Is it possible to combine any two linear equations and have one of the variables

cancel? (Yes, you can combine any two linear equations, but you may have to multiply one or both of the equations to get one of the variables to cancel out.)

• What happens when using the combination method if the lines are parallel or the same line? (In a system with no solution, the variables will cancel out (In a system with no solution, the variables will cancel out (and you will end up with an untrue statement, and in a system with infi nitely many solutions the variables will cancel out and you will end up with a true statement.)

[HOMEWORK]: Assign S227–S229 for homework. (Answers on T579–T581.)

[QUIZ ANSWERS] T582–T584

1. B 2. C 3. B 4. A 5. B 6. A 7. C 8. D 9. A 10. B

The quiz can be used at any time as extra homework or to see how students did on understanding solving systems of equations by combination/elimination.

Algebra SuccessT562


Warm–Up

Directions: Find the solution to each system of equations.

1. Find the solution to the system of equations by graphing.

2x + x + x y = 5y = 5yx – x – x y = 4y = 4y

2x2x2 + x + x y = 5y = 5y– 2x2x2 x x – 2x2x2

y = y = y -2x2x2 + 5x + 5x

x – y = 4x – y = 4x – y– x x x – x

-y-1

=-x-1

+ 4-1

y = x – 4y = x – 4y = x

The solution is (3, -1).

2. Find the solution to the system of equations by substitution.

2x – 3x – 3x y = 1 y = 1 y 2x2x2 – 3x – 3x y = 1 y = 1 y x + 2x + 2x y = 4y = 4y x + 2x + 2x y = 4 y = 4 y 2(-2y + 4) – 3y + 4) – 3y y = 1 y = 1 y x + 2(1) = 4x + 2(1) = 4x -4y + 8 – 3y + 8 – 3y y = 1 y = 1 y x + 2 = 4x + 2 = 4x

x + 2x + 2x y = 4 y = 4 y -7y + 8 = 1 y + 8 = 1 y – 2 – 2 –2y –2y –2y y y y – 8 – 8 x = 2x = 2x

x = x = x -2y + 4 y + 4 y -7y-7

=-7-7

y = 1y = 1y The solution is (2, 1).

Here is the key to S217.


TRANSPARENCY MASTER

Directions: Complete the following SOLVE problem with your teacher. You will only complete the S step.

Brett and Colby went to the state fair. Brett bought 10 tickets for rides, played 5 games, and spent a total of $20.00. Colby bought 15 tickets for rides, played 4 games, and spent a total of $23.00. What is the price for one ride ticket and to play one game?

S Underline the question. This problem is asking me to fi nd ___________________________________.

Directions: Complete the following with your teacher.

Suppose we were going to solve the system of equations below by graphing or substitution. 5x + 3x + 3x y = 16y = 16y 4x – 3x – 3x y = 2y = 2y

1. Is there a variable which can be easily isolated?

2. Solve the fi rst equation, 5x + 3x + 3x y = 16, for y = 16, for y y and write it in slope-intercept form, y and write it in slope-intercept form, yy = mx + b.

3. Solve the second equation, 4x – 3x – 3x y = 2, for y = 2, for y y and write it in slope-intercept form, y and write it in slope-intercept form, yy = mx + b.

4. What problems could we have when trying to fi nd the solution by graphing or substitution?


Algebra SuccessT564



Directions: Complete the following SOLVE problem with your teacher. You will only complete the S step.

Brett and Colby went to the state fair. Brett bought 10 tickets for rides, played 5 games, and spent a total of $20.00. Colby bought 15 tickets for rides, played 4 games, and spent a total of $23.00. What is the price for one ride ticket and to What is the price for one ride ticket and to play one game?play one game?

S Underline the question. This problem is asking me to fi nd the price for one ride ticket and the price to play one game.


Suppose we were going to solve the system of equations below by graphing or substitution. 5x + 3x + 3x y = 16y = 16y 4x – 3x – 3x y = 2y = 2y

1. Is there a variable which can be easily isolated? No, all four variables have coeffi cients so you have to divide to isolate them.

2. Solve the fi rst equation, 5x + 3x + 3x y = 16, for y = 16, for y y and write it in slope-intercept form, y and write it in slope-intercept form, yy = mx + b.5x + 3x + 3x y = 16y = 16y–5x –5x3y3

= y = y -5x3

+ 163

y = y = y-5x3

+ 163

3. Solve the second equation, 4x – 3x – 3x y = 2, for y = 2, for y y and write it in slope-intercept form, y and write it in slope-intercept form, yy = mx + b.4x4x4 – 3x – 3x y = 2y = 2y–4x–4x–4 –4x –4x x –4x –4-3y-3

= -4x4x4-3

+ 2-3

y = y = y 4x4x43

– 23

4. What problems could we have when trying to fi nd the solution by graphing or substitution? Because both of the y-intercepts are fractions, it would be hard to insure you are graphing the correct point. This could affect the graphs of the lines and identifying the solution to the system. The fractions are also very hard to work with when substituting.



TRANSPARENCY MASTER

Algebra SuccessT566





TRANSPARENCY MASTER


5. When two lines intersect at one point, we have a solution of (____, ____).

6. What are the two things we are looking for? ___________________________.

Both equations have two variables, but we must fi nd a way to solve for one variable.

Look at the system of equations.

5x + 3x + 3x y = 16y = 16y

4x – 3x – 3x y = 2 y = 2 y

7. Answer the fi rst question on your graphic organizer on S219. Do the coeffi cients of the same variable have the same absolute value?

8. Follow the arrows to the next question. Do the coeffi cients have opposite signs?

9. Follow the arrows to the next statement. Combine the equations to solve for one variable. Solve for the variable.

10. What is the value of x?

11. If the value of x is ________, plug it into one of the original equations to fi nd x is ________, plug it into one of the original equations to fi nd xthe value of y.

12. The solution to the system is (x, y) or (____, ____).

13. Check your solution by plugging it into the original equations.

Algebra SuccessT568


5. When two lines intersect at one point, we have a solution of (x(x( , x, x y).y).y

6. What are the two things we are looking for? the x-coordinate and the y-coordinate.

Both equations have two variables, but we must fi nd a way to solve for one variable.

Look at the system of equations.

5x + 3x + 3x y = 16y = 16y 4x – 3x – 3x y = 2 y = 2 y

7. Answer the fi rst question on your graphic organizer on S219. Do the coeffi cients of the same variable have the same absolute value? yes

8. Follow the arrows to the next question. Do the coeffi cients have opposite signs?yes

9. Follow the arrows to the next statement. Combine the equations to solve for one variable. Solve for the variable.

5x + 3x + 3x y = 16y = 16y4x4x4 – 3x – 3x y = 2y = 2y

9x =x =x 18

9x9

= 189

x = 2x = 2x

10. What is the value of x? x = 2x = 2x

11. If the value of x is x is x 2, plug it into one of the original equations to fi nd the value of y. 5x + 3x + 3x y = 16 5(2) + 3y = 16 5(2) + 3y y = 16y = 16y

10 + 3y = 16y = 16y –10 –10

3y3

= 63

y = 2y = 2y

12. The solution to the system is (x, y) or (2, 2).

13. Check your solution by plugging it into the original equations.

5x + 3x + 3x y = 16 4y = 16 4y x = 16 4x = 16 4 – 3x – 3x y = 2y = 2y

5(2) + 3(2) = 16 4(2) – 3(2) = 2

10 + 6 = 16 8 – 6 = 2

16 = 16 2 = 2





TRANSPARENCY MASTER

Directions: Follow along with your teacher to complete Examples 1–6.

Example 1: 3x + 5x + 5x y = 8y = 8y 2x – 5x – 5x y = y = y -3

Check:

Example 2: 4x + 2x + 2x y = 2 y = 2 y 4x – 3x – 3x y = y = y -13

Check:

Example 3: 4x – 2x – 2x y = y = y -4 3x + 4x + 4x y = y = y -14

Check:

Algebra SuccessT570



Directions: Follow along with your teacher to complete Examples 1–6.

Example 1: 3x + 5x + 5x y = 8 y = 8 y 3x + x + x 5y = 8 y = 8 y (1, 1) 2x – 5y = -3 3(1) + 5y = 8y = 8y

5x5

= 55

3 + 5y = 8y = 8y

x = 1 x = 1 x –3 –3

5y5

= 55

y = 1y = 1yCheck: 3x + 5x + 5x y = 8 2y = 8 2y x = 8 2x = 8 2 – 5x – 5x y = y = y -3 3(1) + 5(1) = 8 2(1) – 5(1) = -3 3 + 5 = 8 2 – 5 = -3 8 = 8 -3 = -3

Example 2: 4x + 2x + 2x y = 2 y = 2 y → 4x4x4 + 2x + 2x y = 2 y = 2 y → 4x 4x 4 + 2x + 2x y = 2y = 2y 4x 4x 4 – 3x – 3x y = y = y -13 → -1(4x1(4x1(4 – 3x – 3x y = y = y -13) → -4x4x4 + 3x + 3x y = 13y = 13y 4x4x4 + 2(3) = 2x + 2(3) = 2x

5y5

= 155

4x4x4 + 6 = 2x + 6 = 2x

y = 3 y = 3 y –6 –6

(-1, 3) 4x4x44

=-44

Check: 4x 4x 4 + 2x + 2x y = 2 4y = 2 4y x = 2 4x = 2 4 – 3x – 3x y = y = y -13 x = x = x -1 4(-1) + 2(3) = 2 4(-1) – 3(3) = -13

-4 + 6 = 2 -4 – 9 = -13 2 = 2 -13 = -13

Example 3: 4x – 2x – 2x y = y = y -4 → 2(4x 2(4x 2(4 – 2x – 2x y = y = y -4) → 8x 8x 8 – 4x – 4x y = y = y -8 3x + 4x + 4x y = y = y -14 → 3x + 4x + 4x y = y = y -14 → 3x + 4x + 4x y = y = y -14 4(-2)–2y = y = y -4

11x11

= -2211

-8 – 2y = -4

x = x = x -2 +8 +8

-2y-2

= 4-2

(-2, -2) y = -2

Check: 4x4x4 – 2x – 2x y = y = y -4 3x + 4x + 4x y = y = y -14 4(-2) – 2(-2) = -4 3(-2) + 4(-2) = -14

-8 + 4 = -4 -6 + -8= -14 -4 = -4 -14 = -14



TRANSPARENCY MASTER

Example 4: 2x + 5x + 5x y = 26y = 26y

3x – 7x – 7x y = y = y -19

Check:

Example 5: 4x – 2x – 2x y = 10y = 10y

6x – 3x – 3x y = y = y -3

Check:

Example 6: 2x + 3x + 3x y = 6y = 6y

4x + 6x + 6x y = 12y = 12y

Check:

Algebra SuccessT572



Example 4:

2x + 5x + 5x y = 26 y = 26 y → 7(2x7(2x7(2 + 5x + 5x y = 26) y = 26) y → 14x 14x 14 + 35x + 35x y = 182 2y = 182 2y x = 182 2x = 182 2 + 5x + 5x y = 26y = 26y 3x – 7x – 7x y = y = y -19 → 5(3x – 7x – 7x y = y = y -19) → 15x – 35y = y = y -95 2(3) + 5y = 26y = 26y

29x29

= 8729

6 + 5y = 26y = 26y

x = 3 x = 3 x –6 –6

(3, 4) 5y5

= 205

y = 4y = 4y

Check: 2x 2x 2 + 5x + 5x y = 26 3y = 26 3y x – 7x – 7x y = y = y -19

2(3) + 5(4) = 26 3(3) – 7(4) = -19

6 + 20 = 26 9 – 28 = -19

26 = 26 -19 = -19

Example 5:

4x – 2x – 2x y = 10 y = 10 y → 6(4x6(4x6(4 – 2x – 2x y = 10) y = 10) y → -1(24x1(24x1(24 – 12x – 12x y = 60) y = 60) y -24x24x24 + 12x + 12x y = y = y -606x – 3x – 3x y = y = y -3 → 4(6x – 3x – 3x y = y = y -3) → 24x 24x 24 – 12x – 12x y = y = y -12 24x24x24 – 12x – 12x y = y = y -12 0 ≠ -72

No Solution

Check: No check because there is no solution.

Example 6:

2x + 3x + 3x y = 6 y = 6 y → 2(2x2(2x2(2 + 3x + 3x y = 6) y = 6) y → -1(4x1(4x1(4 + 6x + 6x y = 12) y = 12) y -4x4x4 – 6x – 6x y = y = y -124x + 6x + 6x y = 12 y = 12 y → 4x4x4 + 6x + 6x y = 12 y = 12 y → 4x 4x 4 + 6x + 6x y = 12 y = 12 y 4x 4x 4 + 6x + 6x y = 12y = 12y

0 = 0

Infi nitely Many Solutions

Check: No check because there are infi nitely many solutions.



TRANSPARENCY MASTER

1. 2x – 5x – 5x y = 16 y = 16 y 2. 5x + 2x + 2x y = 3y = 3y

3x + 5x + 5x y = y = y -1 3x + 2x + 2x y = 5y = 5y

Check: Check:

3. 4x + 2x + 2x y = 10 y = 10 y 4. 3x + 2x + 2x y = 7y = 7y

3x – 5x – 5x y = 1 9y = 1 9y x + 6x + 6x y = 21y = 21y

Check: Check:

Algebra SuccessT574



1. 2x – 5x – 5x y = 16 y = 16 y 2. 5x + 2x + 2x y = 3 y = 3 y → 5x + 2x + 2x y = 3y = 3y

3x + 5y = -1 -1(3x + 2x + 2x y = 5y = 5y ) → -3x – 2x – 2x y = y = y -5

5x5

= 155

2x2x22

= -22

x = 3 x = 3 x x = x = x -1

2x 2x 2 – 5x – 5x y = 16 5y = 16 5y x + 2x + 2x y = 3y = 3y2(3) – 5y = 16 5(y = 16 5(y -1) + 2y = 3y = 3y 6 – 5y = 16 y = 16 y -5 + 2y = 3y = 3y –6 –6 +5 +5

-5y-5

= 10-5

2y2

= 82

y = y = y -2 y = 4y = 4y (3, -2) (-1, 4)

Check: Check: 2x 2x 2 – 5x – 5x y = 16 3y = 16 3y x + 5x + 5x y = y = y -1 5x + 2x + 2x y = 3 3y = 3 3y x + 2x + 2x y = 5y = 5y2(3) – 5(-2) = 16 3(3)+5(-2) = -1 5(-1) + 2(4) = 3 3(-1)+2(4) = 5 6 + 10 = 16 9 – 10 = -1 -5 + 8 = 3 -3 + 8 = 5 16 = 16 -1 = -1 3 = 3 5 = 5

3. 5(4x + 2x + 2x y = 10y = 10y ) → 20x + 10x + 10x y = 50y = 50y 4. 3(3x + 2y = 7) → 9x + 6x + 6x y = 21y = 21y2(3x – 5x – 5x y = 1y = 1y ) → 6x – 10x – 10x y = 2y = 2y 9x + 6x + 6x y = 21 y = 21 y → 9x + 6x + 6x y = 21y = 21y

26x26

= 5226

-1(9x + 6x + 6x y = 21)y = 21)y → -9x – 6x – 6x y = y = y -21

x = 2x = 2x 9x + 6x + 6x y = 21y = 21y

4x4x4 + 2x + 2x y = 10 y = 10 y 0 = 04(2) + 2y = 10y = 10y 8 + 2y = 10y = 10y–8 –8

2y2

=y =y 22 Infi nitely Many Solutions

y = 1

(2, 1)

Check: Check:

4x 4x 4 + 2x + 2x y = 10 3y = 10 3y x – 5x – 5x y = 1 y = 1 y No check because there

4(2) + 2(1) = 10 3(2) – 5(1) = 1 are infi nitely many solutions.

8 + 2 = 10 6 – 5 = 1

10 = 10 1 = 1



Brett and Colby went to the state fair. Brett bought 10 tickets for rides, played 5 games, and spent a total of $20.00. Colby bought 15 tickets for rides, played 4 games, and spent a total of $23.00. What is the price for one ride ticket and to play one game?

S Underline the question. This problem is asking me to fi nd ________________________________.

O Identify the facts. Eliminate the unnecessary facts. List the necessary facts.

L Choose an operation or operations. Write in words what your plan of action will be.

V Estimate your answer. Carry out your plan.

E Does your answer make sense? (Compare your answer to the question.) Is your answer reasonable? (Compare your answer to the estimate.) Is your answer accurate? (Check your work.) Write your answer in a complete sentence.

TRANSPARENCY MASTER

Directions: Complete the following SOLVE problem with your teacher.

Algebra SuccessT576



Brett and Colby went to the state fair.Brett and Colby went to the state fair. |Brett bought 10 tickets for rides, |played 5 games,| and spent a total of $20.00.|Colby bought 15 tickets for rides, |played 4 games, |and spent a total of $23.00. |What is the price for one ride ticket and What is the price for one ride ticket and to play one game?to play one game?

S Underline the question. This problem is asking me to fi nd the price for one ride ticket and the price to play one game.O Identify the facts. Eliminate the unnecessary facts. List the necessary facts.

Brett – 10 tickets – 5 games - $20.00 Colby – 15 tickets – 4 games - $23.00 Let x represent the price for one ride ticketx represent the price for one ride ticketx Let y represent the price to play one game.y represent the price to play one game.y

L Choose an operation or operations. Multiplication, Subtraction, Division, Addition

Write in words what your plan of action will be.Create your equations from your assigned variables and use combination to solve for the unknown.

V EV EV stimate your answer. Both numbers will be greater than 0 but less than 20 Carry out your plan. 10x + 5x + 5x y = 20y = 20y 15x + 4x + 4x y = 23y = 23y

3(10x3(10x3(10 + 5x + 5x y = 20) y = 20) y → -1(30x1(30x1(30 + 15x + 15x y = 60) y = 60) y → -30x30x30 – 15x – 15x y = y = y -60 10x10x10 + 5x + 5x y = 20y = 20y 2(15x 2(15x 2(15 + 4x + 4x y = 23) y = 23) y → 30x 30x 30 + 8x + 8x y = 46 y = 46 y → 30x30x30 + 8x + 8x y = 46y = 46y 10x0x0 + 5(2) = 20x + 5(2) = 20x

-7y-7

= -14-7

10x 10x 10 + 10 = 20x + 10 = 20x

y = 2 -10 -10

10x10

= 1010

x = 1 The price for one ride ticket, x, is $1 and the price to play one x, is $1 and the price to play one x game, y, is $2.y, is $2.y

E Does your answer make sense? (Compare your answer to the question.) Yes.

Is your answer reasonable? (Compare your answer to the estimate.)Yes. Is your answer accurate? (Check your work.)Yes. Write your answer in a complete sentence. The price for one ride ticket,

x, isx, isx $1 and the price to play one game, y, is $2.y, is $2.y

Directions: Complete the following SOLVE problem with your teacher.




1. 2x + 3x + 3x y = y = y -5 2. 4x – 3x – 3x y = 2 y = 2 y → 4x4x4 – 3x – 3x y = 2y = 2y

3x – 3x – 3x y = 15y = 15y -1(4x + 2x + 2x y = y = y -8) → -4x4x4 – 2x – 2x y = 8y = 8y

5x5

= 105

-5y-5

= 10-5

x = 2x = 2x y = y = y -2

2x 2x 2 + 3x + 3x y = y = y -5 4x5 4x5 4 – 3x – 3x y = 2y = 2y

2(2) + 3y = y = y -5 4x5 4x5 4 – 3(x – 3(x -2) = 2

4 + 3y = y = y -5 4x5 4x5 4 + 6 = 2x + 6 = 2x

–4 –4 –6 –6

3y3

=-93

4x4x44

= -44

y = y = y -3 x = x = x -1(2, -3) (-1, -2)

Check: Check:

2x 2x 2 + 3x + 3x y = y = y -5 3x – 3x – 3x y = 15 y = 15 y 4x4x4 – 3x – 3x y = 2 4y = 2 4y x = 2 4x = 2 4 + 2x + 2x y = y = y -8 2(2) + 3(-3) = -5 3(2)–3(-3) = 15 4(-1) – 3(-2) = 2 4(-1)+2(-2) = -8 4 – 9 = -5 6 + 9 = 15 -4 + 6 = 2 -4 + -4 = -8

-5 = -5 15 = 15 2 = 2 -8 = -8

3. 3(2x – 4x – 4x y = y = y -8) → 6x – 12x – 12x y = y = y -24 4. 2(5x + 3x + 3x y = 23y = 23y ) → 10x10x10 + 6x + 6x y = 46y = 46y

2(3x – 6x – 6x y = 6y = 6y ) → 6x – 12x – 12x y = 12y = 12y 3(2x – 2x – 2x y = 6y = 6y ) → 6x – 6y = 18y = 18y-1(6x – 12x – 12x y = y = y -24) →-6x + 12x + 12x y = 24 y = 24 y 16x

16 = 64

16 6x – 12x – 12x y = 12y = 12y x = 4x = 4x 0 ≠ 36 5x + 3x + 3x y = 23y = 23y No Solution 5(4) + 3y = 23y = 23y 20 + 3y = 23 –20 –20

3y3

= 3

y = 1y = 1y (4, 1)

Check: No check because Check: there is no solution. 5x + 3x + 3x y = 23 2y = 23 2y x = 23 2x = 23 2 – 2x – 2x y = 6y = 6y 5(4) + 3(1) = 23 2(4) – 2(1) = 6 20 + 3 = 23 8 – 2 = 6 23 = 23 6 = 6

Algebra SuccessT578


TRANSPARENCY MASTER

Graphing Calculator Instructions

Solving a System of Equations using Matrices

1. Make sure your system of equations is in the correct format.

ax + ax + ax by = by = by c

dx + dx + dx ey = ey = ey f

2. Use the coeffi cients of the variables to make a 2 × 2 matrix and ENTER it in Matrix [A] as

a b

d e

3. Use the answers of the equations to make a 2 × 1 matrix and ENTER it in Matrix [B] as

c

f

4. Come back to the main screen using 2nd MODE (QUIT) after entering the matrices.

5. 2nd x-1 (MATRX)

6. ENTER with 1: [A] highlighted.

7. You will see [A]. Press x-1 so that you see [A]-1.

8. 2nd x-1 (MATRX)

9.

10. ENTER with 2: [B] highlighted

11. You will see [A]-1 [B]. ENTER for your answer x = x = x

y = y = y



Homework

Directions: Find the solution to each system of equations by combination/elimination.

1. 3x – 2x – 2x y =y =y -16 2. 3x – x – x y = 9y = 9y

2x + 2x + 2x y = 6y = 6y 2x + x + x y = 1y = 1y

5x5

= -105

5x5

= 105

x = -2 x = 2x = 2x

3x – 2x – 2x y = -16 3x – x – x y = 9y = 9y

3(-2) – 2y = y = y -16 3(2) – y = 9y = 9y

-6 – 2y = y = y -16 6 – y = 9y = 9y

+ 6 + 6 –6 –6

-2y-2

= -10-2

-y-1

= 3-1

y = 5 y = 5 y y = y = y -3

(-2, 5) (2, -3)

3. 3(-2x + 3x + 3x y = 3y = 3y ) → -6x + 9x + 9x y = 9y = 9y 4. 3(2x – 3x – 3x y = 7y = 7y ) → 6x6x6 – 9x – 9x y = 21y = 21y

2(3x – 2x – 2x y = y = y -7) → 6x – 4x – 4x y = y = y -14 -6x + x + x 9y = y = y -21 → -6x6x6 + 9x + 9x y = y = y -21

5y5

=-55

0 = 0

y = y = y -1

Infi nitely Many Solutions

-2x2x2 + 3x + 3x y = 3y = 3y -2x2x2 + 3(x + 3(x -1) = 3 -2x2x2 – 3 = 3x – 3 = 3x +3 +3

-2x2x2-2

= 6-2

x = x = x -3

(-3, -1)


Algebra SuccessT580



Homework


5. 5(2x – 2x – 2x y = 4y = 4y ) → 10x – 10x – 10x y = 2y = 2y 0 6. 5x + 3x + 3x y = y = y -13

2(3x + 5x + 5x y = 30y = 30y ) → 6x + 10x + 10x y = 60y = 60y 2x – 3x – 3x y = y = y -1

16x16

= 8016

7x7

= -147

x = 5 x = 5 x x = x = x -2

2x2x2 – 2x – 2x y = 4 5y = 4 5y x + 3x + 3x y = y = y -13

2(5) – 2y = 4 5(y = 4 5(y -2) + 3y = y = y -13

10 – 2y = 4 y = 4 y -10 + 3y = y = y -13

–10 –10 +10 +10

-2y-2

= -6-2

3y3

= -33

y = 3 y = 3 y y = y = y -1

(5, 3) (-2, -1)

7. 2x + 3x + 3x y = y = y -5 → 2x2x2 + 3x + 3x y = y = y -5 8. -4x + 2x + 2x y = 4 y = 4 y

-1(2x + 3x + 3x y = 3)y = 3)y → -2x2x2 – 3x – 3x y = y = y -3 4x – 3x – 3x y = y = y -4

0 ≠ -8 -y-1

= 0-1

y = 0y = 0y

-4x4x4 + 2x + 2x y = 4y = 4y

-4x4x4 + 2(0) = 4x + 2(0) = 4x

-4x4x4-4

= 4-4

No Solution x = x = x -1 (-1, 0)



Homework


9. 4x – 3x – 3x y = 9 y = 9 y → 4x4x4 – 3x – 3x y = 9y = 9y 10. -2x – 5x – 5x y = 14y = 14y

2(2x + 4x + 4x y = 10)y = 10)y → 4x 4x 4 + 8x + 8x y = 20y = 20y -2x + 5x + 5x y = y = y -26

-1(4x1(4x1(4 – 3x – 3x y = 9) y = 9) y → -4x4x4 + 3x + 3x y = y = y -9 -4x4x4-4

= -12-4

11y11

= 1111

x = 3x = 3x

y = 1y = 1y

4x 4x 4 – 3x – 3x y = 9 y = 9 y -2x2x2 – 5x – 5x y = 14y = 14y

4x 4x 4 – 3(1) = 9 -2(3) – 5y = 14y = 14y

4x 4x 4 – 3 = 9 x – 3 = 9 x -6 – 5y = 14y = 14y

+3 +3 +6 +6

4x4x44

= 124

-5y-5

= 20-5

x = 3 x = 3 x y = y = y -4

(3, 1) (3, -4)


Algebra SuccessT582


Quiz

1. If the following system is solved, in which quadrant would the solution lie?

3x + x + x y = y = y -2

2x – x – x y = y = y -8

A. Quadrant I

B. Quadrant II

C. Quadrant III

D. Quadrant IV__________________________________________________________________

2. Which ordered pair is the solution to the system of equations below?

3x + 2x + 2x y = 8y = 8y-3x + 3x + 3x y = y = y -3

A. (-1, -2)

B. (1, 2)

C. (2, 1)

D. (2, -1)__________________________________________________________________

3. Solve the system of equations below.4x + 2x + 2x y = y = y -10

2x + 4x + 4x y = y = y -2

What is the value of y?

A. 3

B. 1

C. -3

D. -4__________________________________________________________________

4. What is the solution of the following system of equations? 2x – 3x – 3x y = 8y = 8y

4x + 2x + 2x y = y = y -16

A. (-2, -4)

B. (-4, -2)

C. No Solution

D. Infi nitely Many Solutions

Name ___________________________________ Date ______________


5. Solve the system of equations below.

2x + 3y = 16y = 16y

2x – 3x – 3x y = y = y -8

What is the value of x?

A. -2

B. 2

C. 4

D. 8__________________________________________________________________

6. Solve the system of equations below.

5x + 2x + 2x y = 16y = 16y

3x – 4x – 4x y = 20

What is the value of y?

A. -2

B. 2

C. 3

D. 4

__________________________________________________________________

7. What is the solution of the following system of equations?

4x + 4x + 4x y = 7y = 7y

2x + 2x + 2x y = y = y -4

A. (-2, -4)

B. (-4, -2)

C. No Solution

D. Infi nitely Many Solutions


Algebra SuccessT584



6x – 4y = y = y -24

3x – 2x – 2x y = y = y -12

A. (-2, 3)

B. (0, 5)

C. No Solution

D. Infi nitely Many Solutions__________________________________________________________________


2x – 2x – 2x y = y = y -2

5x + 3x + 3x y = 19y = 19y

A. (2, 3)

B. (3, 2)

C. Infi nitely Many Solutions

D. No Solution

___________________________________________________________


4x – x – x y = 19y = 19y

2x – y = 11y = 11y

A. (-3, 4)

B. (4, -3)

C. Infi nitely Many Solutions

D. No Solution

LESSON 25: Solving Systems by Combination/Elimination

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Transcript of LESSON 25: Solving Systems by Combination/Elimination