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Lesson 5.4: Solve Systems using Elimination

Lesson 5.4: Solve Systems using Elimination

Assessment

Presentation

Mathematics

8th Grade

Practice Problem

Hard

CCSS
8.EE.C.8B, HSA.CED.A.3, HSA.REI.C.6

Standards-aligned

Created by

Reed Carbone

Used 19+ times

FREE Resource

8 Slides • 4 Questions

1

Lesson 5.4: Solve Systems using Elimination

By Reed Carbone

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Definition:

Elimination: To solve a system of equations using elimination, we will "eliminate" or cancel out one of our variables by adding or subtracting our two equations.​

Subject | Subject

Some text here about the topic of discussion

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3x - y = 7

2x + y = 8​

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Steps to solve using elimination:​

Step 1: Line up your two equations one on top of the other so your "x's" are above your "x's" and your "y's" are above your "y's". Your constants should also be above your constants.

Step 2: Add (or subtract) your like terms to cancel out one of your variables. (Note: if you cannot cancel one of your variables, you will need to multiply one of your equations by a constant so that you can cancel.)

Step 3: Solve the resulting equation for your remaining variable.

Step 4: Plug that solution back into one of your equations to solve for the other variable.​

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Example:​

2x - 2y = -4

2x + y = 11​

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Example:

2y - 5x = -2

3y + 2x = 35​

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Try it out:​

x + y = 56

x + 2y = 94​

Step 1: Order your equations so x's and y's line up.

Step 2: Subtract your equations.

Step 3: Solve for y.

Step 4: Substitute y into the 1st equation then solve for x.​

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Multiple Choice

Solve the system:

2x - 7y = -13

8x - 7y = 11

1

(3,4)

2

(4,3)

3

(2,6)

4

(6,2)

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Multiple Choice

Solve:

2x + 5y = -23

5x + 13y = -60

1

(3,2)

2

(2,3)

3

(-5,1)

4

(1,-5)

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Multiple Choice

Yesterday, a movie theater sold 279 bags of popcorn. You can buy either small or large bags. A large bag costs $4 and a small bag costs $1. In all, the theater made $567 in popcorn sales. Write a system that can model their sales. (Use s for small and L for large.)

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s + L = 567

4L + 1s = 279

2

s = 567 + L

L = 279 + s

3

s + L = 279

4L + 1s = 567

4

s = 279

L = 567

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Multiple Choice

Solve the previous system of equations:

s + L = 279

4L + 1s = 567

1

L = 183

s = 96

2

s = 183

L = 96

3

s = 204

L = 75

4

s = 75

L = 204

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​Conclusion: Write two complete sentences about what you learned.

Lesson 5.4: Solve Systems using Elimination

By Reed Carbone

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