Linear equations: meaning, structure and solution counts
Translate one constraint into an equation, solve it, and recognise when no solution or every value is possible.
What you will tackle
- Create and interpret contextual one-variable equations and their terms
- Solve fluently using equivalent transformations and structural shortcuts
- Distinguish zero, one and infinitely many solutions, including parameter conditions
An equation asserts a relationship
In 18 + 7v = 88, v counts visitors, 18 is a fixed fee, and 7 is a cost per visitor. Subtracting 18 and dividing by 7 gives v = 10. Substitution verifies the original relationship; the answer should also satisfy contextual restrictions such as a whole, nonnegative count.
Compare coefficients before dividing
For ax + b = cx + d, collect (a − c)x = d − b. If a ≠ c there is one solution. If a = c and b = d, every real x satisfies the equation. If a = c but b ≠ d, there is no solution. Dividing by a parameter without checking whether it is zero can erase these last two cases.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
Solve 4(2x − 3) = 5x + 9. Then classify 4(x + 2) = 4x + 8 and 4(x + 2) = 4x + 9.
Show a starting hint
Expand, then compare the coefficients of x.
Compare with the explained answer
- Expand the first left side to 8x − 12.
- Subtract 5x and add 12: 3x = 21, so x = 7.
- The other equations reduce to 8 = 8 and 8 = 9.
Answer / response criteria: x = 7; infinitely many real solutions; no solution.
Independent practice
Independent 1
For which k does (k − 2)x + 3 = 11 have one solution? What happens at the excluded k?
Compare with the explained answer
- A unique solution needs the coefficient k − 2 to be nonzero.
- For k ≠ 2, x = 8/(k − 2).
- At k = 2 the equation is 3 = 11, which is false.
Answer / response criteria: One solution for k ≠ 2; no solution at k = 2.
Application practice
Application 1
A taxi charges $6 plus $2.50 per kilometre. A passenger has $31. Form an equation for the distance when the entire budget is spent; interpret every term and check the result.
Compare with the explained answer
- Let d be the distance in kilometres and write 6 + 2.5d = 31.
- Subtract the initial fee: 2.5d = 25; divide to get d = 10.
- The terms are the fixed fee and distance charge; 6 + 2.5(10) = 31.
Answer / response criteria: 10 km, using 6 + 2.5d = 31.
What can you explain now?
A taxi charges $6 plus $2.50 per kilometre. A passenger has $31. Form an equation for the distance when the entire budget is spent; interpret every term and check the result.
Compare with the explanation
10 km, using 6 + 2.5d = 31.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
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Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide