AnswersTN-Algebra I A SC 26-27 Elective FocusSolving Linear Equations: Distributive Property

Solving Linear Equations: Distributive Property Answers

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Use the drops-downs to justify how Heidi arrived at each step.

Answers:
Step 1::distributive property
Step 2::combine like terms
Step 3::subtraction property of equality
Step 4::addition property of equality
Step 5::division property of equality
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⇒ 1

Answer not available
72386

Which of these equations have no solution? Check all that apply.2(x + 2) + 2 = 2(x + 3) + 12x + 3(x + 5) = 5(x – 3)4(x + 3) = x + 124 – (2x + 5) = (–4x – 2)5(x + 4) – x = 4(x + 5) – 1

Question illustration
A
2(x + 2) + 2 = 2(x + 3) + 1
B
2x + 3(x + 5) = 5(x – 3)
C
4(x + 3) = x + 12
D
4 – (2x + 5) = (–4x – 2)
Option D
E
5(x + 4) – x = 4(x + 5) – 1
72387

Which equation is an identity?

A
3(x – 1) = x + 2(x + 1) + 1
Option A
B
x – 4(x + 1) = –3(x + 1) + 1
Option B
C
2x + 3 = (4x + 2) + 2
Option C
D
(6x – 3) = 3(x + 1) – x – 2
Option D
72389

Which best describes Nosaira’s work and answer?

A
Her work is correct, but there is one solution rather than no solution.
B
Her work is correct and her interpretation of the answer is correct.
C
Her work is incorrect. She distributed incorrectly.
D
Her work is incorrect. She moved terms across the equals sign incorrectly.
72393

Use the drops-downs to justify how Heidi arrived at each step.

Answers:
Step 1::distributive property
Step 2::combine like terms
Step 3::subtraction property of equality
Step 4::addition property of equality
Step 5::division property of equality
72395

Which of these equations have no solution? Check all that apply.2(x + 2) + 2 = 2(x + 3) + 12x + 3(x + 5) = 5(x – 3)4(x + 3) = x + 124 – (2x + 5) = (–4x – 2)5(x + 4) – x = 4(x + 5) – 1

Question illustration
A
2(x + 2) + 2 = 2(x + 3) + 1
B
2x + 3(x + 5) = 5(x – 3)
C
4(x + 3) = x + 12
D
4 – (2x + 5) = (–4x – 2)
Option D
E
5(x + 4) – x = 4(x + 5) – 1
72396

Which equation is an identity?

A
3(x – 1) = x + 2(x + 1) + 1
Option A
B
x – 4(x + 1) = –3(x + 1) + 1
Option B
C
2x + 3 = (4x + 2) + 2
Option C
D
(6x – 3) = 3(x + 1) – x – 2
Option D

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