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Quizzes > Mathematics

Class 7 Algebraic Expressions Quiz

Quick quiz to check your grade 7 algebraic expressions skills. Instant results.

Editorial: Review CompletedCreated By: Carolisa MontgomeryUpdated Aug 28, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art style Class 7 algebraic expressions quiz scene with MCQ sheets pencils shapes on golden yellow background

This Class 7 algebraic expressions quiz helps you practice simplifying, evaluating, and combining like terms while building speed. Get instant feedback on 15 MCQs with clear steps, then sharpen specific skills with the simplifying expressions practice and the evaluating expressions quiz for your next test.

What is the coefficient of y in the term 7y?
y
1
7
0
In algebraic terms, the coefficient is the numerical factor that multiplies the variable. In the term 7y, the coefficient is 7. The variable part is y, and the number in front is always the coefficient. For more details on terms and coefficients see .
Identify the constant term in the expression 5x + 7.
0
12
7
5
A constant term in an expression is a value without any variable attached. In 5x + 7, the term 7 has no variable, so it is the constant. Constants remain the same regardless of variable values. Learn more at .
Evaluate the expression 3a + 2b when a = 2 and b = 3.
13
9
12
8
Substitute a = 2 and b = 3 into 3a + 2b to get 3(2) + 2(3) = 6 + 6 = 12. However, notice a correct evaluation gives 12, not 13. Upon re-evaluation, 6 + 6 = 12, so answer should reflect that. Correction: the calculation yields 12. See steps at .
Which pair of terms are like terms?
2ab and 2a^2b
5y and 5xy
3x and 3y
4x^2 and -5x^2
Like terms have the same variable part raised to the same power. Both 4x^2 and -5x^2 have x squared, so they are like terms. The coefficients differ but that does not affect likeness. More on like terms at .
Simplify the expression 5m + 3m - 2m.
10m
0
4m
6m
Combine like terms: 5m + 3m = 8m, then subtract 2m to get 6m. The variables remain the same and only coefficients change. For more practice see .
What is the value of the expression 2(x + 3) when x = 1?
6
8
7
10
First substitute x = 1: (1 + 3) = 4, then multiply by 2 to get 8. Distributive property also shows 2x + 6 yields 2(1) + 6 = 8. See details at .
Evaluate the sum of -3p and 7p.
4p
10p
-4p
21p
Both terms are like terms. Combine coefficients: -3 + 7 = 4, so the result is 4p. The variable remains p. For combining like terms rules visit .
Simplify the expression 8n - 5n + n.
8n
3n
2n
4n
Combine like terms by adding coefficients: 8n - 5n = 3n, then 3n + n = 4n. Notice the extra n adds one more, so the final result is 4n. For combining steps see .
Simplify the expression 2(x + 3) + 3(x - 1).
2x + 5
x + 9
5x + 1
5x + 3
Apply the distributive property: 2x + 6 + 3x - 3 = (2x + 3x) + (6 - 3) = 5x + 3. Actually the calculation yields 5x + 3. Check answer choices: correct is 5x + 3. For distribution rules see .
Expand and simplify 3(2x - 4) - 5.
6x - 17
6x - 12 - 5
6x - 17
6x - 17
Distribute 3: 6x - 12, then subtract 5 gives 6x - 17. So the fully simplified form is 6x - 17. This uses combining like terms after distribution. More examples at .
Factor the expression 6x^2 + 9x.
x(6x + 9)
3(2x^2 + 3x)
6x( x + 9)
3x(2x + 3)
Both terms share the greatest common factor 3x. Factoring out 3x leaves 2x + 3. Thus the factored form is 3x(2x + 3). For GCF and factoring see .
Simplify the expression 5a - (2a + 3).
5a - 2a + 3
3a - 3
3a + 3
7a - 3
Distribute the negative sign: 5a - 2a - 3 = (5a - 2a) - 3 = 3a - 3. Always change signs inside parentheses when a negative is in front. See details at .
Combine like terms: 4x + 5 - 2x + 7.
6x + 2
2x + 2
4x + 12
2x + 12
Group like terms: (4x - 2x) + (5 + 7) = 2x + 12. Coefficients combine separately from constants. For combining steps see .
Evaluate the expression 1/2 x + 3/4 x when x = 4.
7
5
6
4
Combine like fractions: 1/2x + 3/4x = (2/4 + 3/4)x = 5/4 x. Substitute x = 4 gives 5/4 * 4 = 5. The final answer is 5. More at .
Factor by grouping: x^2 + 3x + 2x + 6.
x^2 + 5x + 6
(x + 1)(x + 6)
x(x + 5) + 2(x + 3)
(x + 2)(x + 3)
Group terms: x^2 + 3x + 2x + 6 = x(x+3) + 2(x+3) = (x+2)(x+3). Factoring by grouping uses common binomial factors. For grouping examples visit .
Factor x^2 - 9 completely.
(x - 3)(x + 3)
(x - 9)(x + 1)
x(x - 9)
(x + 3)^2
x^2 - 9 is a difference of squares: a^2 - b^2 = (a - b)(a + b). Here a = x and b = 3, giving (x - 3)(x + 3). More on difference of squares at .
Expand the perfect square trinomial (x + 5)^2.
x^2 + 5x + 25
x^2 + 10x + 25
x^2 + 25
x^2 + 5x + 5
Use (a+b)^2 = a^2 + 2ab + b^2 with a = x and b = 5: x^2 + 2·x·5 + 25 = x^2 + 10x + 25. Perfect square trinomials follow this pattern. See .
Factor the quadratic x^2 + 5x + 6.
x(x + 5) + 6
(x + 3)^2
(x + 2)(x + 3)
(x + 1)(x + 6)
Find two numbers that multiply to 6 and add to 5: 2 and 3. So the factorization is (x + 2)(x + 3). This is standard for trinomials with leading coefficient 1. Further reading at .
Simplify the expression (3x)(2x + 4).
5x + 4
3x + 2x + 4
6x^2 + 12x
6x + 4x^2
Distribute 3x to both terms: 3x·2x = 6x^2 and 3x·4 = 12x, giving 6x^2 + 12x. Always multiply coefficients and add exponents on like bases. Learn more at .
Which identity represents (a - b)^2?
a^2 - 2ab + b^2
a^2 + 2ab + b^2
a^2 - b^2
a^2 + b^2 - 2ab
The square of a binomial (a - b)^2 expands to a^2 - 2ab + b^2 by applying (a - b)(a - b). The middle term always features twice the product of a and b with a negative sign. More at .
Factor by grouping: x^3 + x^2 - x - 1.
(x^2 - 1)(x + 1)
(x^2 - x) + (-1)(x + 1)
(x^2 - 1)(x + 1)
(x + 1)(x^2 - 1)
Group as (x^3 + x^2) + (-x - 1) = x^2(x + 1) -1(x + 1) = (x + 1)(x^2 - 1). Then note x^2 - 1 is a difference of squares: (x - 1)(x + 1), so fully (x + 1)^2(x - 1). For grouping see .
Simplify the expression 4x^2y · 3xy^2.
4x^4y^3
12x^3y^3
12x^2y^2
7x^3y^3
Multiply coefficients: 4·3=12. Add exponents for like bases: x^2 · x = x^(2+1) = x^3, and y · y^2 = y^(1+2) = y^3. So the product is 12x^3y^3. More at .
Factor completely: 4x^2 - 9y^2.
(2x - 3y)(2x + 3y)
(4x - 9y)(x + y)
(2x - 3y)^2
2x(2x - 3y)
This is a difference of squares: (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y). Always identify perfect squares and apply the formula a^2 - b^2 = (a - b)(a + b). See .
Expand and simplify (x + 2)^3.
x^3 + 3x^2 + 6x + 8
x^3 + 6x + 8
x^3 + 6x^2 + 12x + 8
x^3 + 8
Use the binomial expansion: (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. Here a = x, b = 2 gives x^3 + 3·x^2·2 + 3·x·4 + 8 = x^3 + 6x^2 + 12x + 8. More at .
Factor completely: x^4 - 16.
(x^2 - 4)(x^2 + 4)
(x - 4)(x + 4)
(x^2 - 16)(x^2 + 1)
(x^2 - 2)(x^2 + 8)
First recognize difference of squares: x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4). Then further factor x^2 - 4 = (x - 2)(x + 2), giving (x - 2)(x + 2)(x^2 + 4). For multi-step factorization see .
0
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Study Outcomes

  1. Identify Algebraic Components -

    Develop the ability to distinguish terms, coefficients, and variables in algebraic expressions class 7 practice questions.

  2. Simplify Expressions -

    Combine like terms and apply basic operations to simplify algebraic expressions with confidence in class 7 algebra questions.

  3. Evaluate Variables -

    Calculate the value of algebraic expressions for given variable assignments, mirroring the format of algebraic expressions class 7 MCQ items.

  4. Apply Problem-Solving Strategies -

    Use efficient methods such as order of operations and term grouping to tackle algebraic expressions quiz problems systematically.

  5. Analyze MCQ Patterns -

    Recognize common question formats and pitfalls in our algebraic expressions quiz to improve accuracy and speed during practice algebra problems class 7.

  6. Leverage Performance Feedback -

    Interpret detailed step-by-step feedback to identify strengths and target areas for improvement in your algebraic expressions class 7 practice questions.

Cheat Sheet

  1. Understanding Variables and Constants -

    In algebraic expressions class 7 practice questions, distinguishing variables (letters like x or y) from constants (fixed numbers like 5 or 10) builds a solid foundation. This clarity boosts your confidence when tackling class 7 algebra questions and algebraic expressions class 7 MCQ sets.

  2. Mastering Coefficients and Like Terms -

    Coefficients are the numerical multipliers in terms such as 4x or −2y, and like terms share the same variable part. For example, adding 4x + 7x = 11x streamlines simplification in practice algebra problems class 7.

  3. Using the Distributive Property -

    The distributive property a(b + c) = ab + ac helps you expand or factor expressions efficiently. Applying it - like turning 3(x + 2) into 3x + 6 - is key to acing any algebraic expressions quiz or algebraic expressions class 7 MCQ.

  4. Forming Expressions from Word Problems -

    Spot keywords such as "sum of," "product of," or "difference" to translate phrases into symbols. For instance, "7 more than twice n" becomes 2n + 7, a trick that shines in algebraic expressions class 7 practice questions.

  5. Evaluating Expressions by Substitution -

    After simplifying, plug in values to verify your work: if x = 2 in 4x - 2, then 4×2 - 2 = 6. Regularly practicing this step cements concepts for class 7 algebra questions and boosts your score on any algebraic expressions quiz.

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