📝 Identity and Inverse Properties of Addition and Multiplication (35 MCQs)

📖 From Digital SAT Algebra • 1. Basics of Algebra • 35 questions available

←Previous Topic:Commutative and Associative Properties in Algebra📝 Identity and Inverse Properties of Addition and MultiplicationZero property of multiplication:Next Topic→

What is Identity and Inverse Properties of Addition and Multiplication?

Definition:
The identity properties state that adding $0$ to a number leaves it unchanged (additive identity: $a+0=a$ ) and multiplying by $1$ leaves it unchanged (multiplicative identity: $a \times 1 = a$ ); the inverse properties state that adding the opposite gives $0$ (additive inverse: $a+(-a)=0$ ) and multiplying by the reciprocal gives $1$ (multiplicative inverse: $a \times \frac{1}{a} = 1$ , for $a \neq 0$ ).

Working:
Use identity properties to simplify expressions by recognizing $+0$ or $\times 1$ as neutral; use inverse properties to cancel terms or factors when solving equations, such as adding/subtracting to isolate variables or multiplying/dividing to remove coefficients.

Example:
Solve $x + 5 = 0$ and $3x = 6$ .
Solution: For first, add inverse of 5: $x = -5$ ; for second, multiply by inverse of 3: $x = 2$ .

Reason:
These properties are essential for solving linear equations, as they allow us to isolate the variable by undoing" operations, and they form the basis for algebraic manipulation in all areas."

Easy

Medium

Hard

📝 All Identity and Inverse Properties of Addition and Multiplication MCQs

Q1. A student simplifies $18+(-18)+7$ by first replacing $18+(-18)$ with $1$ . What should the student have used instead, and what is the correct result?

A.The multiplicative identity;

26

B.The additive inverse;

7

✅

C.The additive identity;

1

D.The multiplicative inverse;

7

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The additive inverse of $18$ is $-18$ , and their sum is the additive identity $0$ , not $1$ . Therefore $18+(-18)+7=0+7=7$ . The mistake comes from confusing additive and multiplicative identities.

Q2. A machine processes an input $x$ using $x+0$ in one mode and $x\times1$ in another mode. Which conclusion best explains why the machine's output remains unchanged in either mode?

A.Both

0

and

1

are additive inverses

0

is the additive identity and

1

is the multiplicative identity ✅

0

and

1

are both multiplicative inverses

D.Both

0

and

1

are additive identities

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Adding $0$ leaves any number unchanged, so $0$ is the additive identity. Multiplying by $1$ also leaves any number unchanged, making $1$ the multiplicative identity. These identities preserve the original input.

Q3. A rectangular garden has side lengths $x$ and $5$ . A designer wants an algebraic expression that leaves the area unchanged after an extra multiplication step. Which expression correctly models this requirement?

5x\times0

5x+1

5x\times1

✅

5x+(-5x)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The original area is $5x$ . Multiplying it by $1$ does not change its value, so $5x\times1=5x$ . Multiplying by $0$ destroys the area value, while adding $1$ changes it.

Q4. A student claims that because $9+(-9)=0$ and $9\times\frac{1}{9}=1$ , both pairs demonstrate the same type of inverse relationship. Which analysis is most accurate?

A.Both are additive inverse pairs because their results are identities

B.Both are multiplicative inverse pairs because their results are identities

C.The first pair uses additive inverses, while the second uses multiplicative inverses ✅

D.The first pair uses identities, while the second uses zero factors

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The pair $9$ and $-9$ are additive inverses because their sum is the additive identity $0$ . The pair $9$ and $\frac{1}{9}$ are multiplicative inverses because their product is the multiplicative identity $1$ .

Q5. On a number line, point $P$ represents $a$ , while point $Q$ is located the same distance from $0$ on the opposite side. If $a>0$ , which operation involving $a$ and $Q$ must produce the origin?

a\times Q=1

a+Q=0

✅

a-Q=1

a\times Q=0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Points equally distant from zero on opposite sides represent additive inverses. Thus $Q=-a$ , so $a+Q=a+(-a)=0$ . The graph interpretation identifies the inverse through symmetry about the origin.

Q6. Two methods are proposed to simplify $12(0)+12$ and $12(1)+12$ . Method A says both expressions equal $12$ , while Method B says they equal $12$ and $24$ , respectively. Which method is correct?

A.Method A, because

0

and

1

both preserve multiplication

B.Method A, because both are identities

C.Method B, because

12\times0=0

while

12\times1=12

✅

D.Method B, because

12\times0=12

while

12\times1=1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Method B is correct. Multiplication by $0$ gives $12\times0=0$ , so the first expression is $12$ . Multiplication by $1$ preserves $12$ , making the second expression $12+12=24$ .

Q7. For a nonzero number $x$ , an expression is constructed as $x+\frac{1}{x}$ . A student argues that this expression must equal $1$ because $x$ and $\frac{1}{x}$ are inverses. Which response best evaluates the reasoning?

A.Correct, because every inverse pair produces

1

B.Incorrect, because multiplicative inverses produce

1

when multiplied, not when added ✅

C.Correct, because additive and multiplicative inverses behave identically

D.Incorrect, because inverse numbers can never be combined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The reasoning confuses the two inverse operations. For nonzero $x$ , $x$ and $\frac{1}{x}$ are multiplicative inverses because $x\times\frac{1}{x}=1$ . Their sum generally is not $1$ , so $x+\frac{1}{x}$ cannot be simplified to $1$ solely from the inverse relationship.

Q8. A student simplifies $x+0$ and claims the result must be $0$ because zero is involved in the expression. Which evaluation correctly identifies the mistake?

A.Zero always makes an expression zero

B.Zero is the additive identity, so

x+0=x

✅

C.Zero is an additive inverse, so

x+0=-x

D.The expression cannot be simplified without knowing

x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The student has confused the effect of zero under addition with its effect under multiplication. When zero is added to any number or algebraic expression, the original value remains unchanged, so $x+0=x$ .

Q9. A bank records a daily balance as $B+0$ when no deposit or withdrawal occurs. If $B=2750$ , what does this model tell us about the balance after the transaction?

A.It becomes

0

B.It doubles to

5500

C.It remains

2750

✅

D.It becomes

-2750

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The expression $B+0$ represents adding no change to the existing balance. Therefore, with $B=2750$ , the new balance is $2750+0=2750$ . The zero represents no net change rather than cancellation.

Q10. Two students simplify $3x+0$ . Student A writes $3x$ , while Student B writes $3$ . Which student is correct, and why?

A.Student A, because adding zero leaves the entire expression unchanged ✅

B.Student B, because zero removes the variable

C.Both are correct for different values of

x

D.Neither is correct because

0

cannot be added to an expression

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Student A is correct because the entire quantity $3x$ is unchanged when zero is added. The variable and coefficient remain intact, giving $3x+0=3x$ . Student B incorrectly removes the variable.

Q11. A graph shows a function $y=f(x)+0$ for every plotted $x$ -value. Compared with the graph of $y=f(x)$ , what should a careful observer conclude?

A.The graph shifts upward by

1

B.The graph shifts downward by

1

C.The graph is reflected across the

x

-axis

D.The two graphs coincide at every point ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Adding zero to every output does not change any $y$ -coordinate. Therefore, every point $(x,f(x))$ remains exactly where it was, so $y=f(x)+0$ has the same graph as $y=f(x)$ .

Q12. A temperature sensor records $T$ degrees, then applies the adjustment $+0$ . Another sensor applies $+(-T)$ . If $T=18$ , how do the two results compare?

A.Both results are

18

B.The first is

0

, while the second is

18

C.The first is

18

, while the second is

0

✅

D.Both results are

0

because zero is involved

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For the first sensor, $T+0=18$ , because zero is the additive identity. For the second, $T+(-T)=18-18=0$ , because $-T$ is the additive inverse of $T$ . These operations have different effects.

Q13. A student rewrites $5x-7+0$ as $5x$ , arguing that zero has no effect. What should the student have noticed before simplifying?

A.The

-7

also disappears when zero is added

B.Only the term containing zero disappears; the other terms remain ✅

C.Adding zero changes the sign of every term

D.Zero can only be used with constants, not variables

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Adding zero changes nothing in the existing expression. Thus $5x-7+0=5x-7$ . The student incorrectly removed the existing constant $-7$ , confusing the zero term with other terms in the expression.

Q14. For a real number $a$ , consider $a+0$ , $0+a$ , and $a+(-a)+a$ . Which statement correctly compares all three expressions?

A.All three equal

0

B.The first two equal

a

, while the third also simplifies to

a

✅

C.The first equals

0

, while the last two equal

a

D.Only

a+(-a)+a

equals

a

when

a>0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Both $a+0$ and $0+a$ equal $a$ because zero is the additive identity. Also, $a+(-a)+a=0+a=a$ , since $a+(-a)=0$ . The result holds for every real $a$ , including negative and zero values.

Q15. A student simplifies $8x\times1$ as $8$ . Which reasoning correctly identifies the error?

A.The factor

1

changes the coefficient

B.Multiplying by

1

leaves the entire quantity

8x

unchanged ✅

C.The variable

x

must be replaced by

1

D.Multiplication by

1

is undefined for algebraic expressions

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The factor $1$ is the multiplicative identity, so multiplying any number or algebraic expression by $1$ leaves it unchanged. Therefore, $8x\times1=8x$ , not $8$ . The student incorrectly removed the variable.

Q16. A store's inventory system calculates the value of $250$ items and then applies a correction factor of $1$ . What should the system report after the correction?

A.1 item

B.0 items

C.250 items ✅

D.251 items

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A correction factor of $1$ does not alter the original quantity. Thus $250\times1=250$ . The factor represents no multiplicative change, so the inventory remains exactly $250$ items.

Q17. Two students simplify $3(x+4)\times1$ . Student A writes $3x+4$ , while Student B writes $3(x+4)$ . Which evaluation is correct?

A.Student A, because multiplying by

1

removes the parentheses

B.Student B, because multiplying the entire expression by

1

leaves it unchanged ✅

C.Both are correct because

1

has no numerical effect

D.Neither is correct because

1

cannot multiply a variable expression

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Student B is correct because the factor $1$ applies to the entire expression $3(x+4)$ . Therefore, $3(x+4)\times1=3(x+4)$ . Student A changes the expression by incorrectly removing the factor $3$ from the second term.

Q18. A graph represents $y=f(x)$ . A second graph is created using $y=1\cdot f(x)$ . What relationship should be observed between the two graphs?

A.The second graph is twice as high

B.The second graph is reflected across the

x

-axis

C.The two graphs coincide at every point ✅

D.The second graph is shifted upward by

1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Multiplying every output $f(x)$ by $1$ leaves each output unchanged. Thus every point $(x,f(x))$ remains at the same location, so the graphs of $y=f(x)$ and $y=1\cdot f(x)$ coincide.

Q19. A programmer calculates a value using $A=12x$ , then writes $A=(12x)\times1$ before storing it. Another programmer writes $A=12x+1$ . Which comparison is correct?

A.Both methods produce the same value

B.The first preserves

12x

, while the second increases it by

1

✅

C.The first increases

12x

1

, while the second preserves it

D.Both methods reduce the value because

1

has no effect

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Multiplying by $1$ preserves the original value, so $(12x)\times1=12x$ . In contrast, $12x+1$ adds one to the original expression. The two operations therefore model different transformations.

Q20. A student claims $1\times(x-5)=1$ because $1$ is the multiplicative identity. Which response best analyzes the reasoning?

A.Correct, because the identity replaces every expression with

1

B.Correct, but only when

x=6

C.Incorrect, because

1

leaves the complete factor

x-5

unchanged ✅

D.Incorrect, because

1

can multiply numbers but never expressions

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The multiplicative identity preserves whatever it multiplies. Therefore, $1\times(x-5)=x-5$ , regardless of the value of $x$ . The student's mistake is treating the identity as though it replaces the other factor with $1$ .

Q21. For a real number $x$ , compare $x\times1$ , $1\times x$ , and $(x\times0)+x$ . Which conclusion is always true?

A.All three expressions equal

0

B.Only

x\times1

equals

x

C.The first two equal

x

, and the third also simplifies to

x

✅

D.The first two equal

1

, while the third equals

0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Both $x\times1$ and $1\times x$ equal $x$ because $1$ is the multiplicative identity. Also, $x\times0=0$ , so $(x\times0)+x=0+x=x$ . Hence all three expressions always equal $x$ .

Q22. A student wants to choose a number that cancels $17$ when added to it. Which choice guarantees the result is the additive identity?

A.-17 ✅

B.17

C.0

D.-1

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The additive inverse of $17$ is $-17$ because their sum is $17+(-17)=0$ . The number $0$ is the additive identity, so adding an opposite cancels the original number completely.

Q23. A company's account has a balance of $-850$ units. A correction is added to make the balance exactly zero. Which correction should be applied?

A.-850 units

B.0 units

C.850 units ✅

D.1 unit

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The opposite of $-850$ is $850$ . Adding these values gives $-850+850=0$ , so the correction must be $850$ units. This demonstrates that opposite quantities cancel when added.

Q24. A student claims that the additive inverse of $x-6$ is $x+6$ . Which expression correctly represents the additive inverse of the entire quantity?

x+6

-x-6

-x+6

✅

6-x+6

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The additive inverse of an entire expression must make its sum with the original equal to zero. Negating $x-6$ gives $-(x-6)=-x+6$ , and $(x-6)+(-x+6)=0$ .

Q25. On a number line, point $P$ is located at $x=-4$ . Point $Q$ is the same distance from zero on the opposite side. Which operation involving the two coordinates produces $0$ ?

(-4)-4

(-4)+4

✅

(-4)\times4

(-4)+(-4)

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The point opposite $-4$ across zero is $4$ . These coordinates are additive inverses, so $(-4)+4=0$ . Their equal distances from zero and opposite directions identify the inverse relationship.

Q26. A temperature is $12^\circ$ below a reference level. A later adjustment adds the exact opposite of this temperature value. What is the resulting numerical change relative to the reference level?

-24^\circ

-12^\circ

0^\circ

✅

24^\circ

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Twelve degrees below the reference is represented by $-12$ . Its additive inverse is $12$ , and $-12+12=0$ . Therefore, the adjustment exactly cancels the original numerical displacement.

Q27. A student simplifies $7x+(-7x)+5$ as $14x+5$ . Which reasoning best identifies the error and the correct result?

A.The opposite terms should be multiplied, giving

49x^2+5

B.The opposite terms cancel because

7x+(-7x)=0

, leaving

5

✅

C.The negative term changes

7x

into

-7x

, leaving

5-7x

D.The terms combine to

7x+5

because negative signs are ignored

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The quantities $7x$ and $-7x$ are additive inverses, so their sum is zero. Therefore, $7x+(-7x)+5=0+5=5$ . The student's error is treating opposite terms as though their magnitudes should be added.

Q28. For a real number $x$ , an expression is given by $x+(-x)+x+(-x)+12$ . Without knowing $x$ , what can be concluded?

A.The expression equals

2x+12

B.The expression equals

12

✅

C.The expression equals

0

for every

x

D.The expression equals

4x+12

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Pairing each number with its additive inverse gives $x+(-x)=0$ . The expression contains two such cancelling pairs, so it becomes $0+0+12=12$ . The result is independent of the value of $x$ .

Q29. A student needs a number that, when multiplied by $8$ , produces $1$ . Which value satisfies the requirement?

A.-8

B.0

\frac{1}{8}

✅

D.-1

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The multiplicative inverse of a nonzero number is its reciprocal. Since $8\times\frac{1}{8}=1$ , the required value is $\frac{1}{8}$ . Multiplying by $0$ would instead produce $0$ .

Q30. A recipe uses $\frac{3}{5}$ of a container for one batch. A calculation needs a factor that reverses multiplication by $\frac{3}{5}$ . Which factor should be used?

\frac{3}{5}

\frac{5}{3}

✅

-\frac{3}{5}

\frac{2}{5}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The reciprocal of $\frac{3}{5}$ is $\frac{5}{3}$ . Multiplying them gives $\frac{3}{5}\times\frac{5}{3}=1$ , so $\frac{5}{3}$ reverses the multiplicative effect of $\frac{3}{5}$ .

Q31. A student claims that the multiplicative inverse of $-7$ is $7$ because the two numbers are opposites. Which response is correct?

A.Correct, because opposites always multiply to

1

B.Correct, because changing the sign creates a reciprocal

C.Incorrect, because the multiplicative inverse of

-7

-\frac{1}{7}

✅

D.Incorrect, because negative numbers cannot have multiplicative inverses

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The additive inverse of $-7$ is $7$ , but the multiplicative inverse is its reciprocal, $-\frac{1}{7}$ . Their product is $(-7)(-\frac{1}{7})=1$ , confirming the reciprocal relationship.

Q32. A graph contains a point representing $x=\frac{2}{3}$ . Another calculation requires a number whose product with $\frac{2}{3}$ is $1$ . Which coordinate represents that required number?

-\frac{2}{3}

\frac{1}{3}

\frac{3}{2}

✅

D.-2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The required number must satisfy $\frac{2}{3}\times y=1$ . Solving gives $y=\frac{3}{2}$ , which is the reciprocal of $\frac{2}{3}$ . The negative value would instead produce $-1$ .

Q33. A measurement is multiplied by $0.25$ , and a second step must restore the original measurement. Which factor should the second step use?

A.-0.25

B.-0.5

C.-2

D.-4 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: To restore a value after multiplication by $0.25=\frac{1}{4}$ , the calculation must multiply by its reciprocal, $4$ . Since $\frac{1}{4}\times4=1$ , the original measurement is recovered.

Q34. A student simplifies $\frac{6}{7}\times\frac{7}{6}\times x$ as $1+x$ . What is the correct simplification, and why?

1+x

, because inverse factors are added

x

, because the reciprocal factors multiply to

1

✅

2x

, because the fractions cancel each other

\frac{42}{42x}

, because

x

is also inverted

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The fractions $\frac{6}{7}$ and $\frac{7}{6}$ are multiplicative inverses, so their product is $1$ . Therefore, $1\times x=x$ . The student's mistake is treating multiplication by an inverse pair as though it required addition.

Q35. For nonzero $x$ , a student argues that $x+\frac{1}{x}=1$ because $x$ and $\frac{1}{x}$ are multiplicative inverses. Which statement most precisely evaluates the claim?

A.Always true, because every inverse relationship produces

1

B.True only when

x=1

C.False, because multiplicative inverses produce

1

when multiplied, not necessarily when added ✅

D.False, because

x

and

\frac{1}{x}

can never be used in the same expression

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Multiplicative inverses satisfy $x\times\frac{1}{x}=1$ for nonzero $x$ . Their sum generally has a different value. For example, when $x=2$ , the sum is $2+\frac{1}{2}=\frac{5}{2}$ , not $1$ .

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