L3 · Equations with parentheses

Learning goal

“I can solve equations containing parentheses - by expanding with the distributive property or by dividing both sides first - including cases with a negative or fractional multiplier outside the brackets, and verify every solution by substitution.”

Parentheses — grouping symbols that package several terms into a single unit that must be treated together: in 3(x + 2), the package is (x + 2).

Distributive property — a multiplier outside parentheses multiplies EVERY term inside: 3(x + 2) = 3x + 6, and 3(x - 4) = 3x - 12.

Expand — to rewrite a packaged expression without parentheses, using the distributive property.

Package-first route — the alternative strategy: divide both sides by the outside multiplier first (3(x + 2) = 21 -> x + 2 = 7), then unwrap the package.

Worked example

Solve 3(x + 2) = 21. Route 1 - expand first: the 3 multiplies every term inside: 3x + 6 = 21. Then it is a Grade 7 two-step: 3x = 15, x = 5. Route 2 - package first: divide both sides by 3: x + 2 = 7. Then x = 5. Verify in the original: 3(5 + 2) = 3 x 7 = 21. Say it plainly: parentheses package terms into one unit. You may unwrap the package first (expand) or shrink the whole equation first (divide) - both are preservation of equality, and both must end at the same answer.

Check yourself — explain it

In expanding 3(x + 2), why is the result 3x + 6 and not 3x + 2?

Paired problem — example still visible

Solve 4(x - 1) = 20 -> x = 6 (expand: 4x - 4 = 20, so 4x = 24; or package-first: x - 1 = 5. Check: 4(6 - 1) = 20)

Faded completion — last step is yours

"2(x + 7) = 22: expand to 2x + 14 = 22, then 2x = 8, so x = ___" -> 4

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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