Solving Systems of Equations by Elimination

The video tutorial explores methods for solving systems of equations, focusing on the elimination method. It begins with a theoretical explanation and example, demonstrating how to eliminate variables by adding or subtracting equations. The tutorial then applies this method to a practical word problem involving the cost of candy bars and Fruit Roll-Ups, showing how to set up equations and solve for unknowns. The video emphasizes understanding the process of elimination and verifying solutions.

What is the first step in solving systems of equations using the elimination method?

• Multiply the equations

• Graph the equations

• Substitute one variable

• Add or subtract the equations

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

What principle allows us to add or subtract equations in the elimination method?

• The principle of substitution

• The principle of equivalent equations

• The principle of graphical intersection

• The principle of variable isolation

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

What does the elimination method primarily focus on?

• Substituting variables

• Eliminating variables

• Isolating variables

• Graphing equations

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

What happens to the variable y when we add the two given equations?

• It becomes 8y

• It remains unchanged

• It doubles in value

• It is eliminated

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

After eliminating y, what is the value of x?

• 4

• 28/8

• 7/2

• 3.5

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

How do we find the value of y after determining x?

• Substitute x into any of the original equations

• Graph the equations

• Multiply the equations

• Use the elimination method again

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

How do we verify that our solution satisfies both equations in the system?

• By creating a new equation

• By eliminating another variable

• By graphing the equations

• By substituting the values back into the original equations

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CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

What is the cost of a Fruit Roll-Up in the real-world problem?

• $0.35

• $1.79

• $1.05

• $0.48

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CCSS.7.EE.B.4A

What is the cost of a candy bar in the real-world problem?

• $1.79

• $0.35

• $1.05

• $0.48

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CCSS.7.EE.B.4A

Why do we divide both sides of the equation by 3 to solve for y in the real-world problem?

• To subtract y from the equation

• To double the value of y

• To eliminate y

• To isolate y

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CCSS.6.EE.B.7