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📝 Classify inequalities as identities or contradictions (14 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 14 questions available

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What is Classify inequalities as identities or contradictions?

Definition:
Classifying inequalities as identities or contradictions: An identity inequality is true for all real numbers (e.g., $x + 2 > x + 1$ simplifies to $2 > 1$ , always true). A contradiction inequality has no solution because it simplifies to a false statement (e.g., $x < x - 3$ simplifies to $0 < -3$ , false).

Working:
For $3x + 5 > 3x + 2$ , subtract $3x$ : $5 > 2$ , true for all $x$ , so it is an identity. For $2y - 4 \geq 2y + 1$ , subtract $2y$ : $-4 \geq 1$ , false, so it is a contradiction.

Example:
Classify $4a - 1 \leq 4a + 5$ . Subtract $4a$ : $-1 \leq 5$ , true for all $a$ , so it is an identity.

Reason:
Classification helps understand whether an inequality imposes a restriction (conditional), applies universally (identity), or is impossible (contradiction), which is useful in complex problem-solving.

Easy

Medium

Hard

📝 All Classify inequalities as identities or contradictions MCQs

Q1. After simplifying $3x+7>3x-2$ , how should the inequality be classified?

A.Identity ✅

B.Contradiction

C.Conditional inequality with

x>0

D.Conditional inequality with

x<-3

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Subtracting $3x$ from both sides gives $7>-2$ , which is always true regardless of $x$ . Because every real number satisfies the resulting statement, the original inequality is classified as an identity.

Q2. Which simplified result indicates that an inequality is a contradiction?

0\geq-5

4<9

0>6

✅

x<8

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A contradiction occurs when simplification produces a false numerical statement that cannot be satisfied by any value of $x$ . Since $0>6$ is always false, the original inequality has no solution.

Q3. A student simplifies $5(2x-3)\leq10x+4$ to $-15\leq4$ . What is the best classification of the original inequality?

A.Identity ✅

B.Contradiction

C.Only

x=0

works

D.Only positive

x

values work

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Expanding gives $10x-15\leq10x+4$ . Subtracting $10x$ leaves $-15\leq4$ , which is always true. Therefore every real value of $x$ satisfies the inequality, making it an identity.

Q4. Consider $7x-4>7x+9$ . Without solving for $x$ , what can you conclude after comparing the variable terms?

A.It is an identity because the

x

-terms cancel

B.It is a contradiction because it reduces to

-4>9

✅

C.It has solution

x>13

D.It has solution

x<-13

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The same $7x$ term appears on both sides, so subtracting it eliminates the variable and leaves $-4>9$ . That numerical statement is false, so no real number can satisfy the original inequality.

Q5. A company models two cost estimates as $C_1=4x+120$ and $C_2=4x+90$ , where $x$ is the number of units. The manager asks whether $C_1>C_2$ can ever be false. What classification applies?

A.Identity ✅

B.Contradiction

C.It depends on whether

x

is positive

D.It is true only when

x>30

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Comparing the expressions gives $4x+120>4x+90$ . Subtracting $4x$ leaves $120>90$ , which is always true. Thus the first cost estimate exceeds the second for every allowed value of $x$ , so the inequality is an identity.

Q6. A teacher claims $2(3x+5)\geq6x+8$ is an identity. Which reasoning most effectively verifies the claim?

A.Divide both sides by

x

B.Expand and reduce to

10\geq8

✅

C.Move constants first and obtain

x\geq1

D.Test only

x=0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Expanding the left side produces $6x+10\geq6x+8$ . Subtracting $6x$ from both sides gives $10\geq8$ , which is always true. Therefore the teacher's classification is correct.

Q7. A water tank model compares two expressions: $2(4t-3)+6$ and $8t-1$ . The engineer wants to know whether the first quantity is always at least the second. What should be concluded?

A.It is an identity ✅

B.It is a contradiction

C.It is true only for

t\geq1

D.It is true only for

t\leq1

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Simplifying the first expression gives $8t$ , so the comparison becomes $8t\geq8t-1$ . Subtracting $8t$ leaves $0\geq-1$ , which is always true. Therefore the comparison is an identity, not a contradiction.

Q8. A student solves $4(x-2)<4x-11$ and writes $x<-3$ . What is the student's error?

A.They should divide by

x

before simplifying

B.They failed to cancel identical variable terms, producing a false restriction ✅

C.They reversed the inequality sign unnecessarily

D.They should test only negative values

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Expanding gives $4x-8<4x-11$ . Subtracting $4x$ leaves $-8<-11$ , which is false. Thus there are no solutions. The student incorrectly treated the remaining constants as though a variable were still present.

Q9. A student simplifies $6-2x\leq10-2x$ to $6\leq10$ and calls it an identity. Is the classification correct?

A.Yes, because the variable terms cancel and the statement is always true ✅

B.No, because canceling variable terms is never allowed in inequalities

C.No, because the original inequality requires

x\leq2

D.Yes, but only for

x\geq0

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Adding $2x$ to both sides gives $6\leq10$ , which is always true. The cancellation is valid because the same expression is added to both sides. Therefore every real $x$ satisfies the original inequality, making it an identity.

Q10. On a number line, the solution graph for an inequality shades the entire number line, with no endpoint restriction. Which conclusion is justified?

A.The inequality is a contradiction

B.The inequality is an identity ✅

C.The inequality has exactly one solution

D.The inequality must contain a negative coefficient

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Shading the entire number line means every real number satisfies the inequality. When algebraic simplification produces an always-true statement, the classification is an identity. A contradiction would instead have no shaded points.

Q11. Two students analyze $9x+2\geq9x-5$ . Student A says it is an identity because $2\geq-5$ . Student B says it is a contradiction because the $x$ -terms disappear. Who is correct?

A.Only Student A ✅

B.Only Student B

C.Both students

D.Neither student

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Subtracting $9x$ from both sides produces $2\geq-5$ , a statement that is always true. The disappearance of the variable does not itself indicate a contradiction. Student B confuses variable cancellation with an impossible numerical result.

Q12. Which pair of inequalities has the same classification?

4x+1>4x-6

and

5x-8>5x+3

7x-2\leq7x+4

and

2x+9>2x+15

3x+5\geq3x+1

and

8x-7<8x-10

6x+4<6x+9

and

9x+2>9x-1

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The first inequality in option D reduces to $4<9$ , which is always true, and the second reduces to $2>-1$ , also always true. Therefore both are identities, while the other pairs mix an identity with a contradiction.

Q13. For real $x$ , consider $(k+2)x+5>(k+2)x+9$ . For which value of $k$ does changing the coefficient affect the classification?

k=-2

k=0

k=2

D.No value of

k

changes the classification ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The identical term $(k+2)x$ occurs on both sides for every value of $k$ , so it always cancels. The inequality becomes $5>9$ , which is false regardless of $k$ . Hence the inequality remains a contradiction for all real $k$ .

Q14. A researcher compares $3(2x+1)-x$ with $5x+7$ and claims the comparison is an identity because both sides contain similar variable terms. Which conclusion is mathematically correct for $3(2x+1)-x\geq5x+7$ ?

A.Identity

B.Contradiction ✅

C.True only when

x\geq7

D.True only when

x\leq-7

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Simplifying the left side gives $6x+3-x=5x+3$ . The inequality becomes $5x+3\geq5x+7$ , and subtracting $5x$ gives $3\geq7$ , which is false. Therefore the inequality is a contradiction with no solution.

🔗 Related Topics (MCQs)

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