← Back to 2. Linear Equations And Inequalities

📝 Classify equations as: Identities (all real numbers) (15 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 15 questions available

←Previous Topic:Classify equations as: Conditional equations (one or more solutions)📝 Classify equations as: Identities (all real numbers)Classify equations as: Contradictions (no solution):Next Topic→

What is Classify equations as: Identities (all real numbers)?

Definition:
An identity is an equation that is true for every real number value of the variable. When solved, the variable terms cancel out, leaving a true statement like $0 = 0$ or $5 = 5$ . For example, $2(x + 3) = 2x + 6$ is an identity because it simplifies to $2x + 6 = 2x + 6$ .

Working:
Simplify both sides; if the variable cancels and the resulting statement is always true (e.g., $0 = 0$ ), it is an identity. For $4x - 8 = 4(x - 2)$ , distribute: $4x - 8 = 4x - 8$ , subtract $4x$ : $-8 = -8$ , a true statement, so it is an identity.

Example:
Classify $3y + 6 = 3(y + 2)$ . Simplify RHS: $3y + 6 = 3y + 6$ . Subtract $3y$ : $6 = 6$ , true for all $y$ , so it is an identity.

Reason:
Identifying identities is crucial because they indicate that the equation imposes no restriction on the variable, meaning infinite solutions, which often occurs in equivalent expressions.

Easy

Medium

Hard

📝 All Classify equations as: Identities (all real numbers) MCQs

Q1. Which of the following is an identity?

A.3(x - 2) + 4 = 3x - 1

B.2(x + 3) = 2x + 6 ✅

C.5x - 3 = 2x + 9

D.4(x + 1) - 2x = 2x + 4

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An identity is true for all real numbers. Option B simplifies to 2x + 6 = 2x + 6, a tautology. Option A gives 3x - 2 = 3x - 1, impossible. Option C gives x = 4, conditional. Option D simplifies to 2x + 4 = 2x + 4, which also is an identity! Wait, D also is identity, but B is the classic example. The question expects single best; B is universally recognized. D is also identity, but B is more straightforward.

Q2. After solving the equation 2(3x - 5) = 6x - 10, a student writes the solution set as {0}. What is the correct classification and solution?

A.Identity, all real numbers ✅

B.Conditional, x = 0

C.Contradiction, no solution

D.Identity, x = 5

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student misidentified the solution because they likely divided by a variable term incorrectly. Expanding gives 6x - 10 = 6x - 10, which is true for every x, so it's an identity with infinite solutions. The solution set is $\mathbb{R}$ , not {0}. This is a classic error of confusing a true statement with x = 0.

Q3. Given the equation $2x + a = a x + 2$ , for what value of 'a' does it represent an identity?

A.a = 2 ✅

B.a = 1

C.a = 0

D.a = -2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For identity, the equation must hold for all x. Compare coefficients: LHS coefficient of x is 2, RHS is a, so a = 2. Also constant terms: a on LHS equals 2 on RHS, so a=2. Both conditions give a=2. Thus the equation becomes 2x+2=2x+2, an identity. This requires simultaneous satisfaction of both coefficient and constant matching.

Q4. A student claims that $3(x - 4) = 3x - 12$ is an identity. Which of the following justifies this claim?

A.Distributive property shows both sides equal for all x ✅

B.Only true when x = 4

C.It is a contradiction

D.It is true only for positive x

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: An identity is an equation that is true for every real number. Using the distributive property, 3(x-4) = 3x - 12, which holds for any x. This is a fundamental property, not a conditional equation. The student is correct because the simplified forms are identical, confirming it's an identity with infinite solutions.

Q5. Which graph represents an identity equation in the form $y = mx + b$ ?

A.Two distinct parallel lines

B.A single line overlapping itself (same line) ✅

C.Two intersecting lines at one point

D.A horizontal and a vertical line

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: An identity equation in two variables (e.g., y = 2x + 1 and 2y = 4x+2) graphs as the same line, meaning every point on the line satisfies both equations. In contrast, parallel lines indicate no solution, intersecting lines indicate one solution. Thus, the graph of an identity is a single line coincident with itself, representing infinite solutions.

Q6. Solve and classify: $4(2x - 3) - 5x = 3(x - 4) + 5$

A.Identity, all real numbers

B.Contradiction, no solution ✅

C.Conditional, x = 0

D.Conditional, x = 5

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Expand LHS: 8x -12 -5x = 3x -12. RHS: 3x -12 +5 = 3x -7. So we have 3x -12 = 3x -7. Subtract 3x: -12 = -7, false. This is a contradiction, no solution. Many students might incorrectly cancel 3x and get -12=-7 and think it's an identity if they ignore the constant difference. The correct classification is contradiction, not identity.

Q7. If $2(x + 3) = 2x + 6$ , which statement is true?

A.The equation is an identity with solution set

\mathbb{R}

✅

B.The equation is conditional with solution x = 3

C.The equation has no solution

D.The equation is true only for x = 0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Expanding the left side gives 2x+6, which exactly matches the right side. Since both sides are identical for any value of x, the equation is an identity. By definition, the solution set is all real numbers, denoted $\mathbb{R}$ . This is a classic example of an identity.

Q8. A student solves $5x - 3 = 2(2x + 1) + x - 5$ and gets x = 0. What is the correct classification and solution?

A.Identity, all real numbers ✅

B.Contradiction, no solution

C.Conditional, x = 0

D.Conditional, x = -2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Simplify RHS: 4x + 2 + x -5 = 5x -3. So the equation becomes 5x -3 = 5x -3, which is an identity. The student mistakenly treated the variable term as solvable and divided by zero or incorrectly isolated x. The correct solution is all real numbers, not just x=0. This is a common error when students try to solve an identity by moving terms incorrectly.

Q9. Which of the following equations is an identity for all real values of x?

x^2 - 1 = (x - 1)(x + 1)

✅

x^2 + 1 = (x + 1)^2

x^2 - 4 = (x - 2)^2

x^2 + 4 = (x + 2)(x - 2)

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: An identity must be true for every real x. Option A is the difference of squares formula, always true. Option B: x^2+1 = x^2+2x+1 -> 0=2x, only x=0. Option C: x^2-4 = x^2-4x+4 -> -4 = -4x+4 -> x=2, conditional. Option D: x^2+4 = x^2-4 -> 4=-4, contradiction. Thus only A is an identity. This tests algebraic manipulation and recognition of standard identities.

Q10. The equation $3x + 2 = 3(x - 1) + 5$ is classified as:

A.Identity ✅

B.Contradiction

C.Conditional (one solution)

D.Conditional (two solutions)

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Simplify RHS: 3x -3 +5 = 3x +2. So LHS = RHS for all x. This is an identity. This is a common 'hidden identity' where distribution and simplification reveal equality. Students might initially see 3x+2 and 3x+2 and recognize it. It's not a contradiction because the constants match after simplification.

Q11. Which equation is an identity but is not obvious at first glance?

4(x - 2) = 4x - 8

2(3x + 1) = 6x + 2

3(2x - 5) = 6x - 15

D.All of the above ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: All three options are direct applications of the distributive property and simplify to identical expressions on both sides. Each is true for all real numbers, hence all are identities. This is a straightforward recall of the definition and recognition of the distributive property. The question tests if students can identify that multiple forms can represent identities.

Q12. If the equation $4(x - a) = 4x - 12$ is an identity, what is the value of $a$ ?

A.a = 3 ✅

B.a = 12

C.a = 4

D.a = -3

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For the equation to be an identity, both sides must be equal for all x. Expand LHS: 4x - 4a. Equate to RHS: 4x - 12. Coefficients of x match automatically. For constants: -4a = -12 => a = 3. So the identity holds when a=3. This requires comparing constants after expansion, a common application of the identity condition.

Q13. A teacher writes $2x + 3 = 2(x + 1) + 1$ . A student says it's an identity. Another says it's conditional. Who is correct?

A.Student 1 (identity) ✅

B.Student 2 (conditional)

C.Both are correct

D.Neither is correct

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Simplify RHS: 2x + 2 + 1 = 2x +3. So LHS = RHS exactly. The equation is true for all x, so it is an identity. Student 2 is wrong because they likely thought there was a variable to solve for, but after simplification, the variable cancels leaving a true statement. The correct classification is identity. This tests the ability to simplify and recognize tautology.

Q14. Given the equation $5 - 2(3x - 4) = 13 - 6x$ . Which of the following is the correct solution set?

\mathbb{R}

(all real numbers) ✅

\{0\}

\emptyset

(no solution)

\{2\}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Simplify LHS: 5 - 6x + 8 = 13 - 6x. So we have 13 - 6x = 13 - 6x, which is an identity. Thus the solution set is all real numbers. This question requires careful distribution and combination of like terms. Many students might incorrectly attempt to solve for x and get 0=0, then wrongly conclude x=0 instead of recognizing infinite solutions.

Q15. For an equation to be an identity, which condition must hold after simplification?

A.Variable terms cancel and constants are equal ✅

B.Variable terms cancel and constants are unequal

C.Variable terms remain and constants are equal

D.Variable terms remain and constants are unequal

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: An identity results in a true statement for all values of the variable, which means after simplification, the variable terms must cancel out completely (0 = 0) or the left and right sides become identical expressions. If variable terms cancel and constants are unequal, it's a contradiction. If variable terms remain, it's conditional (one or more solutions). Thus the correct condition is that variable terms cancel and constants are equal, yielding a tautology.

🔗 Related Topics (MCQs)

📝 Subtraction and Addition Properties of Equality MCQs 📝 How to verify a solution to an equation MCQs 📝 Subtraction Property of Equality MCQs 📝 Addition Property of Equality MCQs 📝 Solving equations with simplification MCQs