Brackets and the number of solutions
Expand brackets, collect like terms and distinguish one solution, no solution and infinitely many solutions in linear equations.
For learners aged 14+ · free · no account required
Language, curriculum and teaching reference are independent choices. For learners aged 14+: selected linear-equation skills or review. Read the selected programme notes for its level, version and extensions. This is a limited original series, not a full course or an official examination paper.
Brackets and linear equations connect to KS3 algebra; equations and identities connect to KS4. The three solution cases are explicit in Common Core, but are not a verbatim KS3 requirement.
Madrid 3.º ESO: review of brackets, fractions and linear equations (D.4). The no/one/infinitely-many-solution cases are a teaching extension; the regional document does not list all three explicitly. Check the reasoning manually.
France 3e, 2026–27: expanding, simplifying and solving first-degree equations. No/one/infinitely-many-solution classification is selected enrichment, not an explicit complete annual expectation. The new programme reaches 3e in 2028–29.
NRW Gymnasium G9: review of selected year 7/8 skills. Practise brackets, simplifying expressions and solving linear equations; explain the remaining equality. The three solution cases are an original elaboration of the solution-set skill, not three explicitly listed regional requirements. The real-number domain extends the first-stage rational-number scope; the consulted school-plan example introduces real numbers in year 9 (UV9.2).
Japan: review of selected first-year middle-school skills from the 2017 curriculum, fully introduced in 2021. Practise brackets, simplifying expressions and solving linear equations; explain the remaining equality. The no/one/infinitely-many-solution classification is original enrichment, not a first-year curriculum claim. The real-number domain is an extension beyond the rational-number setting of the first year, for this 14+ series.
Korea: review of selected first-year middle-school skills from the 2022 curriculum, introduced for this year in March 2025. It reaches the third year only in March 2027. Practise brackets, simplifying expressions and solving linear equations; explain the remaining equality. The no/one/infinitely-many-solution classification is original enrichment, not a first-year curriculum claim. The real-number domain is an extension beyond the rational-number setting of the first year, for this 14+ series.
Russia: mapping to the selected 2025 FRP edition, reviewing year 7 algebra. Practise brackets, simplifying expressions and solving linear equations; explain the remaining equality. The year 7 programme explicitly includes the number of roots of a linear equation. The real-number domain extends the year 7 scope with a concept introduced in year 8.
CeDeC / INTEF is the August 2014 edition, used as a historical teaching reference. It is not a companion to the 2022 LOMLOE curriculum.
Sésamath 3e, Magnard, June 2021: selected sections of the student workbook (ISBN 9782210115811). This cross-reference does not cover the whole workbook.
QUA-LiS.NRW, 31 January 2020: a suggested school-plan example, especially UV7.6, page 34. It is a teaching reference, not a textbook or every school’s required sequence.
MEXT, July 2017: official mathematics curriculum commentary, selected first-year sections on pages 70–74. It is a teaching reference, not a publisher textbook companion.
Gyeonggi Office of Education, 10 March 2025: assessment guidance, mathematics chapter Ⅱ-1, pages 58–69. The criteria are a teaching reference; this checker does not assign its achievement levels.
FIPI, Moscow 2026: selected guidance on computation, checking answers, written models and units. It is a year 9 teaching reference, not a textbook or a complete OGE preparation course.
Tokyo Shoseki, 2025, 数学002-72: the publisher’s excerpts and teaching articles for pages 94, 96–97, 78, 101 and 104–108 were consulted. They connect equality-preserving operations, signs and coefficient transformations with modelling and checking the original situation. Classification into 0/1/infinitely many solutions and the real-number domain are MomentDesk extensions; this is a limited cross-reference, not the whole textbook.
Visang, Mathematics 1, first edition 1 March 2025, second printing 1 March 2026, ISBN 979-11-7316-034-9: pages 86–87, 89–91 and 92–93 were read in the publisher’s public e-book. They cover justified equation transformations, substitution, coefficients and a model checked against its situation and units. Classification into 0/1/infinitely many solutions and the real-number domain are MomentDesk extensions. Reasoning still needs manual assessment.
Cornelsen, Fundamente der Mathematik 7, NRW G9, first edition 2020, fifth printing 2026, ISBN 978-3-06-040177-2: the publisher’s open preview covers sections 6.2–6.6, pages 160–175. It explains justified transformations, special cases and modelling with units and interpretation. Its solution sets here use rational numbers; MomentDesk’s real-number domain is an extension. This limited reference keeps the original MomentDesk tasks and manual reasoning assessment.
Read the explanation
Expand before collecting terms. Once the unknown terms are brought together, the remaining equality tells you whether there is one solution, none, or every allowed value is a solution.
Before you start
Equivalent operations, signed multiplication and combining like terms.
- Multiply every term inside a bracket by the outside coefficient. A minus before the bracket changes each sign.
- Collect unknown terms and constants using equivalent operations on both sides.
- A nonzero coefficient times x equal to a constant has exactly one solution. Divide by that coefficient.
- If the unknown cancels, a true equality has all allowed values as solutions; a false equality has no solution. Do not divide by a zero coefficient.
- In these first two topics, the unknown ranges over real numbers. No variable occurs in a denominator.
Worked examples
Distribute 2, then subtract x and 2 to isolate the unknown.
2(x + 1) = x + 11
2x + 2 = x + 11
x = 9
The same expression appears on both sides after expansion. Every allowed value works.
5(x + 2) = 5x + 10
10 = 10
The unknown cancels but the constants differ. No value can make the equation true.
5(x + 2) = 5x + 12
10 = 12
Common mistakes
- In a negative product, the minus applies to every term inside the bracket.
- A true equality after cancellation does not mean x = 0. Testing one value cannot establish the whole solution set.
Explain your reasoning
Show expansion with signs, combine terms and explain the final equality before naming the solution set.
The checker assesses a number or a selected option. It does not grade your reasoning, written equation or units. Use the rubric with your tutor or for self-checking. Ten tasks are a small practice set, not proof of mastery.
Practise
Enter a number without x = or a unit. Exact equal values are accepted: integers, decimals, fractions and mixed numbers. A decimal point or comma is accepted; ambiguous input such as 1,234 needs a choice of meaning. Keep units and reasoning in your written work.
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Your written work and answers
Tasks and explanations
Solve: 3(x + 4) = 2x + 5. Enter x.
Answer and explanation
−7
Distribute to both terms, then collect x terms and constants.
3x + 12 = 2x + 5
x = −7
3(−7 + 4) = 2(−7) + 5
Find x: 5 − 2(x − 3) = 4x − 1.
Answer and explanation
2
The product −2(x − 3) is −2x + 6.
11 − 2x = 4x − 1
12 = 6x
x = 2
Solve: + = 5.
Answer and explanation
9
Multiply every term by 6 before expanding and combining.
2(x − 1) + (x + 5) = 30
3x + 3 = 30
3x = 27
x = 9
Find x: 0.5(4x − 6) = x + 2.
Answer and explanation
5
Distribute the decimal coefficient to both terms.
2x − 3 = x + 2
x = 5
Solve: −3(2x − 1) + 4 = x − 14.
Answer and explanation
3
Distribute −3 carefully, then move unknown terms using equal operations.
−6x + 7 = x − 14
21 = 7x
x = 3
Classify the solutions of 4(x + 2) = 2x + 14.
- No solution
- Exactly one solution
- Infinitely many solutions
Answer and explanation
Exactly one solution
A nonzero coefficient remains, so there is one solution.
4x + 8 = 2x + 14
2x = 6
x = 3
Classify the solutions of 3(x + 2) = 3x + 8.
- No solution
- Exactly one solution
- Infinitely many solutions
Answer and explanation
No solution
The unknown cancels, leaving an impossible constant equality.
3x + 6 = 3x + 8
6 = 8
Classify the solutions of 7(x − 1) − 2x = 3x + 1.
- No solution
- Exactly one solution
- Infinitely many solutions
Answer and explanation
Exactly one solution
A nonzero coefficient remains, so there is one solution.
5x − 7 = 3x + 1
2x = 8
x = 4
Classify the solutions of 2(x − 5) = 2x − 10.
- No solution
- Exactly one solution
- Infinitely many solutions
Answer and explanation
Infinitely many solutions
Both sides simplify to the same expression. Every real value of x satisfies the equation.
2x − 10 = 2x − 10
0 = 0
Classify the solutions of 6(x − 1) − 2x = 4x + 2.
- No solution
- Exactly one solution
- Infinitely many solutions
Answer and explanation
No solution
The unknown cancels, leaving an impossible constant equality.
4x − 6 = 4x + 2
−6 = 2
Explain your reasoning
Compare 3(x + 1) = 3x + 3 and 3(x + 1) = 3x + 4. Expand each and explain the solution set of each equation.
Answer and explanation
First: 3 = 3, so every allowed value is a solution. Second: 3 = 4, so there is no solution.
A complete response explains cancellation and whether the remaining constant equality is true. Testing one candidate alone is not enough.
Worksheet
Curriculum source
Japan · first middle-school year review · 2017
中1 · A(2) / A(3) · 70–73 · 2026-10-12
Curriculum sourceKorea · first middle-school year review · 2022
[9수02-02] / [9수02-04] · 2026-10-12
Curriculum sourceTextbook / teaching reference: OpenStax · Elementary Algebra 2e
Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis · Edition: 2020-04-22
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: CeDeC / INTEF · Matemáticas 3° E.S.O.
CeDeC / INTEF · Edition: 2014-08
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: Sésamath 3e · Cahier élève
Association Sésamath · Magnard · Edition: 2021-06
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: QUA-LiS.NRW · Beispiel für einen schulinternen Lehrplan
QUA-LiS.NRW · Edition: 2020-01-31
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: 文部科学省 · 中学校学習指導要領解説 数学編
文部科学省 · Edition: 2017-07
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: 경기도교육청 · 성취평가, 이렇게 실천해요! [중학교편]
경기도교육청 · Edition: 2025-03-10
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: ФИПИ · Методические рекомендации · Математика
И. В. Ященко · И. Р. Высоцкий · П. И. Самсонов · Edition: 2026
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: 東京書籍 · 新編 新しい数学 1
清水美憲・真島秀行ほか108名 · Edition: 2025 · 令和7年度
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: Cornelsen · Fundamente der Mathematik 7 · NRW G9
Andreas Pallack (Hrsg.) · Autorenteam · Edition: 2020
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Textbook / teaching reference: 비상교육 · 중학교 수학 1
이진호・김선희・이소민・황윤희・이윤경 · Edition: 2025-03-01
The linked sections were consulted as a teaching reference. These explanations and tasks are original MomentDesk work. This is a topic cross-reference, not a complete textbook companion.
Next topic: Build an equation from a situation
Worksheet: Brackets and the number of solutions
Show expansion with signs, combine terms and explain the final equality before naming the solution set.
Solve: 3(x + 4) = 2x + 5. Enter x.
Equation, working, units and checkAnswer:
Find x: 5 − 2(x − 3) = 4x − 1.
Equation, working, units and checkAnswer:
Solve: + = 5.
Equation, working, units and checkAnswer:
Find x: 0.5(4x − 6) = x + 2.
Equation, working, units and checkAnswer:
Solve: −3(2x − 1) + 4 = x − 14.
Equation, working, units and checkAnswer:
Classify the solutions of 4(x + 2) = 2x + 14.
- No solution
- Exactly one solution
- Infinitely many solutions
Equation, working, units and checkAnswer:
Classify the solutions of 3(x + 2) = 3x + 8.
- No solution
- Exactly one solution
- Infinitely many solutions
Equation, working, units and checkAnswer:
Classify the solutions of 7(x − 1) − 2x = 3x + 1.
- No solution
- Exactly one solution
- Infinitely many solutions
Equation, working, units and checkAnswer:
Classify the solutions of 2(x − 5) = 2x − 10.
- No solution
- Exactly one solution
- Infinitely many solutions
Equation, working, units and checkAnswer:
Classify the solutions of 6(x − 1) − 2x = 4x + 2.
- No solution
- Exactly one solution
- Infinitely many solutions
Equation, working, units and checkAnswer:
Answer key: Brackets and the number of solutions
−7
Distribute to both terms, then collect x terms and constants.
3x + 12 = 2x + 5
x = −7
3(−7 + 4) = 2(−7) + 5
2
The product −2(x − 3) is −2x + 6.
11 − 2x = 4x − 1
12 = 6x
x = 2
9
Multiply every term by 6 before expanding and combining.
2(x − 1) + (x + 5) = 30
3x + 3 = 30
3x = 27
x = 9
5
Distribute the decimal coefficient to both terms.
2x − 3 = x + 2
x = 5
3
Distribute −3 carefully, then move unknown terms using equal operations.
−6x + 7 = x − 14
21 = 7x
x = 3
Exactly one solution
A nonzero coefficient remains, so there is one solution.
4x + 8 = 2x + 14
2x = 6
x = 3
No solution
The unknown cancels, leaving an impossible constant equality.
3x + 6 = 3x + 8
6 = 8
Exactly one solution
A nonzero coefficient remains, so there is one solution.
5x − 7 = 3x + 1
2x = 8
x = 4
Infinitely many solutions
Both sides simplify to the same expression. Every real value of x satisfies the equation.
2x − 10 = 2x − 10
0 = 0
No solution
The unknown cancels, leaving an impossible constant equality.
4x − 6 = 4x + 2
−6 = 2