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Introduction
Equations with Parentheses helps you connect a mathematical rule to a result you can explain and check. Choose between dividing out a common nonzero factor and distributing first. Both valid approaches must agree. This lesson belongs to linear equations and builds toward independent Algebra I problem solving.
Alignment uses 19 TAC §111.39(c). Review lessons sample their listed standards. Readiness and supporting labels describe STAAR emphasis, not whether a topic is required.
Learning objectives
- Identify the quantities and representations used in equations with parentheses.
- Apply the lesson's rule to solve a numerical or contextual problem.
- Explain each mathematical step using the correct vocabulary.
- Verify an answer and correct a likely error.
Prerequisite knowledge
Use distribution, signed arithmetic, and inverse operations.
Previous lesson: Equations with Variables on Both Sides
Key mathematical vocabulary
- variable
- A letter representing an unknown or changing quantity.
- constant
- A value fixed within a problem.
- expression
- A mathematical phrase representing a value.
- equation
- A statement that two expressions are equal.
- solution
- A value making the original conditions true.
Formulas and rules
- a(b+c)=ab+ac
- Equivalent expressions agree for all allowed inputs.
- A solution satisfies every original condition.
Preserve equality while solving
Choose between dividing out a common nonzero factor and distributing first. Both valid approaches must agree.
An equation states that two expressions have the same value. A solution is an input that makes the original statement true. Adding, subtracting, or multiplying both sides by the same number preserves equality; division is permitted only by a nonzero number.
First remove grouping symbols and combine like terms if needed. Gather variable terms on one side and constants on the other, then isolate the variable. An equation can have one solution, no solution, or infinitely many solutions. A false constant statement means no solution; a true identity means all permitted inputs work.
A method you can explain
- Read the question and identify the known information, unknown quantities, and any domain restrictions. Select a suitable representation and tool.
- Apply the relationship explained above. Divide both sides by 12: x+27=30.
- Write each calculation or logical inference on a separate line. Keep units and signs visible. When two methods are possible, compare which one uses the given information most directly.
- Check the result in the ORIGINAL conditions, not merely the last simplified line. Substitute x=3 into the ORIGINAL equation: both sides equal 360.
- Interpret what the answer means. Decide whether its size, sign, and precision make sense; identify a restriction or assumption that would change the conclusion.
Five worked examples
Example 1 · Beginner
Solve 12(x + 27) = 360. Explain each inverse operation.
- Divide both sides by 12: x+27=30.
- Subtract 27: x=3.
- Verification: Substitute x=3 into the ORIGINAL equation: both sides equal 360.
Answer: 3
Verify: Substitute x=3 into the ORIGINAL equation: both sides equal 360.
Avoid this mistake: Changing only one side: apply each operation to both sides.
Example 2 · Basic
Solve 13(x + 27) = 390. Explain each inverse operation.
- Divide both sides by 13: x+27=30.
- Subtract 27: x=3.
- Verification: Substitute x=3 into the ORIGINAL equation: both sides equal 390.
Answer: 3
Verify: Substitute x=3 into the ORIGINAL equation: both sides equal 390.
Avoid this mistake: Dropping a negative coefficient during division.
Example 3 · Intermediate
Solve 14(x + 27) = 420. Explain each inverse operation.
- Divide both sides by 14: x+27=30.
- Subtract 27: x=3.
- Verification: Substitute x=3 into the ORIGINAL equation: both sides equal 420.
Answer: 3
Verify: Substitute x=3 into the ORIGINAL equation: both sides equal 420.
Avoid this mistake: Assuming every equation has one numerical solution: inspect identities and contradictions.
Example 4 · Advanced
A rental costs $2 per hour plus $28. The total is $34. Write an equation and find the hours x.
- Model cost: 2x+28=34.
- Subtract the fixed fee 28: 2x=6.
- Divide by the hourly rate 2: x=3 hours.
Answer: 3
Verify: Substitute x=3 into the ORIGINAL equation: both sides equal 34.
Avoid this mistake: Changing only one side: apply each operation to both sides.
Example 5 · Challenge
Solve 3(x + 28) = 93. Explain each inverse operation.
- Divide both sides by 3: x+28=31.
- Subtract 28: x=3.
- Verification: Substitute x=3 into the ORIGINAL equation: both sides equal 93.
Answer: 3
Verify: Substitute x=3 into the ORIGINAL equation: both sides equal 93.
Avoid this mistake: Dropping a negative coefficient during division.
Common mistakes and corrections
- Changing only one side: apply each operation to both sides.
- Dropping a negative coefficient during division.
- Assuming every equation has one numerical solution: inspect identities and contradictions.
Real-world applications
- Find hours worked from a fixed fee and hourly charge. Define the input and its units, apply the mathematical relationship, and check that the result fits the setting.
- Determine the quantity of supplies affordable at a known total cost. Define the input and its units, apply the mathematical relationship, and check that the result fits the setting.
Topic summary
Choose between dividing out a common nonzero factor and distributing first. Both valid approaches must agree.
An equation states that two expressions have the same value. A solution is an input that makes the original statement true. Adding, subtracting, or multiplying both sides by the same number preserves equality; division is permitted only by a nonzero number.
Frequently asked questions
What should I know before equations with parentheses?
Use distribution, signed arithmetic, and inverse operations.
How can I check my work?
Substitute x=3 into the ORIGINAL equation: both sides equal 360. A numerical check can detect an error; when claiming an identity for all inputs, also explain the algebraic property that makes it true.
Which mistakes should I watch for?
Changing only one side: apply each operation to both sides. Dropping a negative coefficient during division. Assuming every equation has one numerical solution: inspect identities and contradictions.
Can I use another valid method?
Yes. Explain your assumptions, show enough intermediate work to justify the result, and verify it in the original problem. Equivalent expressions and valid alternative methods receive credit.
Mathematical process practice · A.1A–A.1G
- A.1A: Find hours worked from a fixed fee and hourly charge.
- A.1B: Plan, solve, check, and interpret the written response.
- A.1C: Select paper, a calculator, a table, or a graph; explain the choice.
- A.1D: Connect symbols with a verbal explanation and another representation.
- A.1E: Organize the solution with a table, diagram, or labeled equations.
- A.1F: Explain how the representation and the algebraic rule express the same relationship.
- A.1G: Justify each operation and the reasonableness of the conclusion.
Quick revision notes
- Choose between dividing out a common nonzero factor and distributing first. Both valid approaches must agree.
- Changing only one side: apply each operation to both sides.
- Dropping a negative coefficient during division.
- Assuming every equation has one numerical solution: inspect identities and contradictions.
Practice and explain
Try all four multiple-choice questions, then show your reasoning in the written response. Objective scores and AI feedback assess different skills.
Continue learning
Original Cambridge Veritas practice. These are not official STAAR items. Cambridge Veritas is not affiliated with or endorsed by TEA.
Official Algebra I TEKS (§111.39) · TEA assessed curriculum · STAAR blueprint · TEA instructional materials