Lesson 5.2.4 – System of Equations (Elimination Method)
Teacher Lesson Plan
Lesson:
5.2.4 – Supplement
System of Equations (Elimination Method)
CC Standards
8.EE.C.8
Analyze and solve pairs of simultaneous linear equations.
8.EE.C.8a (Calculator – No)
Understand that solutions to a system of two linear equations in two variables correspond to
points of intersection of their graphs, because points of intersection satisfy both equations
simultaneously.
8.EE.C.8b (Calculator – No)
Solve systems of two linear equations in two variables algebraically, and estimate solutions by
graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6
have no solution because 3x + 2y cannot simultaneously be 5 and 6.
8.EE.C.8c (Calculator – Yes)
Solve real-world and mathematical problems leading to two linear equations in two variables. For
example, given coordinates for two pairs of points, determine whether the line through the first
pair of points intersects the line through the second pair
Calculator
Teacher call, but I would lean towards “yes” on this lesson. Most of the lesson pertains to 8.EE.8c.
Objective
The students will practice solving system of equations using the elimination method.
Mathematical Practices
#1 Make sense of problems and persevere in solving them.
#5 Use appropriate tools strategically.
#6 Attend to precision.
#7 Look for and make sense of structure.
Teacher Input
Bellwork:
Homework:
Introduction:
Lesson:
Review bellwork.
Review previous night’s homework.
Introduce the lesson as directed by PowerPoint.
Teach as directed by the PowerPoint. Students notes coincide with the lesson.
Extra Practice
Classwork
Extra Practice
Page 4
Page 5 (can be used for homework or extra practice)
Closure
Teacher selected
1|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Student Notes
SECTION 1: Warm-up Activity
Look at each pair of terms below. Determine what you can multiply one (or both) terms by to make them
cancel each other out.
Example
Example
SECTION 2:
Term #1
−𝟓𝒙
𝟏𝟐𝒚
−𝟔𝒙
𝟐𝒙
𝒚
𝟐𝒙
Term #2
𝟏𝟎𝒙
𝟒𝒚
𝟑𝟎𝒙
𝟒𝒙
𝒚
−𝟑𝒙
Answer
Multiply Term #1 by: 2
Multiply Term #2 by: -3
Solving Systems using the Elimination Method
STEP 1:
Line up the two equations in Standard Form.
STEP 2:
Eliminate one of the variables.
To do this, look for coefficients that have the same variable but with opposite signs.
STEP 3:
If this does not exist, multiply one or both of the equations by a number that will create this
situation.
STEP 4:
Combine to make one equation…. Then solve that equation.
STEP 5:
Use the resulting answer to find the other variable by plugging it into either one of the original
equations.
Example 1:
Example 2:
You Try
1)
−3𝑥 + 2𝑦 = 8
3𝑥 + 4𝑦 = 16
2|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Guided Practice
1)
9𝑥 + 𝑦 = 13
3𝑥 + 2𝑦 = −4
You Try
2)
6𝑥 + 2𝑦 = −10
6𝑥 − 4𝑦 = −34
SECTION 3:
3)
7𝑥 − 6𝑦 = 4
−𝑥 + 12𝑦 = −34
4)
3𝑥 + 2𝑦 = 11
2𝑥 − 3𝑦 = −10
Real-World Systems of Equations
1) The Sunny Meadows Safe Haven for Pets charge $200 to adopt a dog and $100 to adopt a
cat. On April 30, National Adopt a Pet Day, 20 pets were adopted and $3,200 was collected.
PART A:
Write a system of equations to represent this situation.
PART B:
Solve the system to determine how many of each were adopted.
Dogs adopted: ____
Cats adopted: ____
2) The admission at a fair is $2 for children and $4 for adults. On a certain day 500 people enter the fair and
$1,300 is collected.
PART A:
Write a system of equations to represent this situation.
PART B:
Solve the system.
How many child admissions were there?
How many adult admissions were there?
3|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Classwork
Name___________________________ Date__________
Lesson 5.2.4
Period: ____
System of Equations
Using the Elimination Method
Solve each system of equations by elimination. Write your answer as a coordinate pair (x, y).
1.
x + y = 12
x–y=2
4.
7.
-3x – 2y = -3
6x + 2y = 10
5. 4x – 2y = 7
-2x + 2y = 2
3x + 6y = 9
2x – y = 3
-2x + y = -3
10.
2. 3x + 4y = 9
8. 2x – 3y = 11
-2x + 3y = -2
3. -x + y = 4
-2x + y = 2
6. 5x – 6y = 8
2x + 2y = 18
9. 5x + 3y = -9
2x – 5y = -16
Tatiana and Jill each improved their yards by planting rose bushes and geraniums.
They bought supplies at the same store. Tatiana spent $210 on 9 rose bushes and
12 geraniums. Jill spent $40 on 3 rose bushes and 1 geranium. Find the cost on
one rose bush and the cost of one geranium.
4|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Extra Practice
Name___________________________ Date__________
Lesson 5.2.4
____________________________
Period: ____
System of Equations
____________________________
____________________________
6.
7.
5|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
6|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Student Notes
SECTION 1: Warm-up Activity
Look at each pair of terms below. Determine what you can multiply one (or both) terms by to make them
cancel each other out.
Example
Example
SECTION 2:
Term #1
−𝟓𝒙
𝟏𝟐𝒚
−𝟔𝒙
𝟐𝒙
𝒚
𝟐𝒙
Term #2
𝟏𝟎𝒙
𝟒𝒚
𝟑𝟎𝒙
𝟒𝒙
𝒚
−𝟑𝒙
Answer
Multiply Term #1 by: 2
Multiply Term #2 by: -3
Multiply Term#1 by: 5
Multiply Term#2 by: -2
Multiply either by: -1
Multiply Term#1 by: 3
Multiply Term#2 by: 2
Solving Systems using the Elimination Method
STEP 1:
Line up the two equations in Standard Form.
STEP 2:
Eliminate one of the variables.
To do this, look for coefficients that have the same variable but with opposite signs.
STEP 3:
If this does not exist, multiply one or both of the equations by a number that will create this
situation.
STEP 4:
Combine to make one equation…. Then solve that equation.
STEP 5:
Use the resulting answer to find the other variable by plugging it into either one of the original
equations.
Example 1:
Example 2:
You Try
1)
−3𝑥 + 2𝑦 = 8
3𝑥 + 4𝑦 = 16
(0,4)
7|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Guided Practice
1)
9𝑥 + 𝑦 = 13
3𝑥 + 2𝑦 = −4
(2,-5)
You Try
2)
6𝑥 + 2𝑦 = −10
6𝑥 − 4𝑦 = −34
(-3, 4)
SECTION 3:
3)
7𝑥 − 6𝑦 = 4
−𝑥 + 12𝑦 = −34
4)
(-2, -3)
3𝑥 + 2𝑦 = 11
2𝑥 − 3𝑦 = −10
(1, 4)
Real-World Systems of Equations
1) The Sunny Meadows Safe Haven for Pets charge $200 to adopt a dog and $100 to adopt a
cat. On April 30, National Adopt a Pet Day, 20 pets were adopted and $3,200 was collected.
PART A:
Write a system of equations to represent this situation.
c + d = 20
100c + 200d = 3200
PART B:
Solve the system to determine how many of each were adopted.
Dogs adopted: ____ 12
Cats adopted: ____ 8
2) The admission at a fair is $2 for children and $4 for adults. On a certain day 500 people enter the fair and
$1,300 is collected.
PART A:
Write a system of equations to represent this situation.
c + a = 500
2c + 4a = 1300
PART B:
Solve the system.
How many child admissions were there?
How many adult admissions were there?
350
150
8|Page
Lesson 5.2.4 – System of Equations (Elimination Method)
Classwork
Name_____ Answer Key_____________
Lesson 5.2.4
Date__________
Period: ____
System of Equations
Using the Elimination Method
Solve each system of equations by elimination. Write your answer as a coordinate pair (x, y).
1.
4.
7.
10.
x + y = 12
2. 3x + 4y = 9
3. -x + y = 4
x–y=2
-3x – 2y = -3
-2x + y = 2
Answer: (7, 5)
Answer: (-1, 3)
Answer (2, 6)
6x + 2y = 10
5. 4x – 2y = 7
-2x + 2y = 2
3x + 6y = 9
2x + 2y = 18
Answer: (1, 2)
Answer: (2, ½)
Answer: (5, 4)
2x – y = 3
8. 2x – 3y = 11
6. 5x – 6y = 8
9. 5x + 3y = -9
-2x + y = -3
-2x + 3y = -2
2x – 5y = -16
Answer: infinitely many
Answer: no solution
Answer: (-3, 2)
Tatiana and Jill each improved their yards by planting rose bushes and geraniums.
They bought supplies at the same store. Tatiana spent $210 on 9 rose bushes and
12 geraniums. Jill spent $40 on 3 rose bushes and 1 geranium. Find the cost on
one rose bush and the cost of one geranium.
9r + 12g = $210
3r + g = $40
Answer: r = $10
g = $10
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Lesson 5.2.4 – System of Equations (Elimination Method)
Extra Practice
Name______ Answer Key_____________________Date__________
Lesson 5.2.4
one
none
Period: ____
System of Equations
Infinitely many
one
5.
6.
7.
Solution: (1, 3)
Solution: (4, -1)
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