Solving System of Equations Substitution Method

Solving System of Equations Substitution Method

9th Grade

13 Qs

quiz-placeholder

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Solving System of Equations Substitution Method

Solving System of Equations Substitution Method

Assessment

Quiz

Mathematics

9th Grade

Hard

CCSS
8.EE.C.8B, 8.EE.C.8A, HSA.REI.C.7

+2

Standards-aligned

Created by

Anthony Clark

FREE Resource

13 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Example 1

Step 3:

After substituting the first equation into the second one and working it out, I got

x = 1.

What should I do next? Then do it.

Fill in 1 for x in y = 6x

y = 6(1)

y = 6

Fill in 1 for y in y = 6x

1 = 6x

1/6 = x

Fill in 1 for y in 2x + 5y = 32

2x + 5(1) = 32

2x + 5 = 32

2x = 27

x = 27/2

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Example 1

Step 4:

What is the solution to the system:

y = 6x

2x + 5y = 32

Given I got x = 1 from step 2 and y = 6 from step 3.

(1, 6)

(6, 1)

Tags

CCSS.HSA.REI.C.7

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which statement would be correct for step 1 and do it?

I would solve the first equations for y because it has a coefficient of 1.

y = -x

I would solve the first equations for y because it has a coefficient of -1.

-y = -x

I would solve the first equations for x because it has a coefficient of 1.

x = y

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

In step 1, we got x = y. Which shows the correct steps for step 2 (substituting and solving)?

18x + 2y = 0

18(y) + 2y = 0

20y = 0

No Solution

18x + 2y = 0

18(y) + 2y = 0

20y = 0

y = 0

Tags

CCSS.8.EE.C.8B

5.

FILL IN THE BLANK QUESTION

1 min • 1 pt

x - y = 0

18x + 2y = 0

Step 4:

In step 2, we got x = 0 and y = 0, what is the final answer?

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which statement is correct to begin to solve this equation?

The variable is already isolated in the first equation, so we have already solved it.

The variable is already isolated in the first equation, so we have to plug into the second one like this:

-4(4-2x) - 2y = -8

The variable is already isolated in the first equation, so we have to plug into the second one like this:

-4x - 2(4 - 2x) = -8

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Example 3: y = 4 - 2x -4x - 2y = -8 Finish step 2 (which shows correct steps): -4x - 2(4 - 2x) = -8

-4x - 2(4 - 2x) = -8

-4x - 8 + 4x = -8

8 = -8

-4x - 2(4 - 2x) = -8

-4x - 8 - 4x = -8

-8x - 8 = -8

-8x = 0

x = 0

-4x - 2(4 - 2x) = -8

-4x - 8 + 4x = -8

-8 = -8

-4x - 2(4 - 2x) = -8

-4x - 8 - 2x = -8

-6x - 8 = -8

-6x = 0

x = 0

Tags

CCSS.8.EE.C.7B

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