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
Core coverage follows the detailed SAT skills in Appendix B of the September 2026 framework. The conditional School Day Essay is separate. The framework is maintained rather than issued as an annual subject syllabus.
Original teaching and explained self-check practice are available. Independent subject-specialist review and difficulty calibration remain pending; this is not an exam-board endorsement or a mastery certificate.
Official source specification / guide