9.3 Algebraic Manipulation, Equations and Inequalities

Algebra you cannot finish by plugging four numbers

Public MAT Test Content includes algebraic manipulations and the solution of equations and inequalities. The 2015 MAT teachers' booklet is blunt about a school shortcut: items will not be written so that you can substitute each of four easy numeric options into an equation and see which one works. A stem of the form the solution of 5x − 2 = 13 is with options 3, −3, 15, and 2 is exactly the shape they say they avoid, because 3 plugs in immediately.

Independent OpenExamPrep teaching therefore trains equivalent forms, identities, sums and products of roots, completed squares, and systems that ask for a derived expression, not spot x = 3 among four integers. This remains MAT, calculator-free, and unscaffolded multiple-choice. The official 2023 Intake National Report describes 60 items; follow the current session clock if a later sitting changes the composition. It is not QL, and it is not a confidential NBTP paper.

Original parallel to that substitution warning. Consider 2x^2 − 5x − 3 = 0. Factor: (2x + 1)(x − 3) = 2x^2 − 6x + x − 3 = 2x^2 − 5x − 3. The roots are x = 3 and x = −1/2. If four options are 3, −1/2, −3/2, and 5/2, and the stem asks for the product of the roots, then both 3 and −1/2 satisfy the original equation. Substitution cannot eliminate them. The product is 3 × (−1/2) = −3/2, which is c/a. The sum is 3 − 1/2 = 5/2, which is −b/a. MAT can ask for that product; it will not ask you to recognise 3 as the root among four integers.

Factorisation you can finish by hand

Common factor. 6x^2 y − 9x y^2 = 3xy(2x − 3y). Missing 3xy and stopping at 6x − 9y is not factorised.

Difference of squares. a^2 − b^2 = (a − b)(a + b). Worked: 49p^2 − 16 = (7p − 4)(7p + 4). 4 − 9x^2 = (2 − 3x)(2 + 3x), not (4 − 3x)(4 + 3x).

Trinomial. x^2 + 7x + 12 = (x + 3)(x + 4), because 3 + 4 = 7 and 3 × 4 = 12. x^2 − 5x − 14 = (x − 7)(x + 2), because −7 + 2 = −5 and (−7) × 2 = −14.

Worked example: grouping. x^3 + 3x^2 − 4x − 12 = x^2(x + 3) − 4(x + 3) = (x + 3)(x^2 − 4) = (x + 3)(x − 2)(x + 2).

Worked example: identity, not a root hunt. Which expression is identical to x^2 + 5x + 6? Expanding (x + 2)(x + 3) gives x^2 + 5x + 6 for every real x. Checking x = 0 only is not a proof. Expanding is. MAT can ask for the identical factorisation rather than find x.

Worked example: binomial square. (2x − 3)^2 = 4x^2 − 12x + 9, not 4x^2 + 9 and not 4x^2 − 9. The middle term is twice the product. Forgetting it is the algebra version of treating √(a + b) as √a + √b: a sum inside a square is not a square of a sum without the cross term.

Completing the square

To write x^2 + bx as a square, take half of b and square it.

Worked example: rewrite x^2 − 10x + 21. Half of −10 is −5. (−5)^2 = 25.

x^2 − 10x + 21 = (x − 5)^2 − 25 + 21 = (x − 5)^2 − 4.

Expand to check: (x − 5)^2 − 4 = x^2 − 10x + 25 − 4 = x^2 − 10x + 21.

Wrong neighbours: (x − 5)^2 + 4 forgets to subtract the extra 25. (x − 10)^2 − 79 halves incorrectly. (x + 5)^2 − 4 flips the linear sign. Plugging x = 0 into those four expressions can match more than one of them or, after a slip, none of them. Completing the square is the method; substitution of a convenient number is not a MAT plan.

Worked example: solve, then extract a Vieta quantity. x^2 + 8x + 7 = 0.

x^2 + 8x + 7 = (x + 4)^2 − 16 + 7 = (x + 4)^2 − 9 = 0 (x + 4)^2 = 9 x + 4 = ±3 x = −1 or x = −7.

Sum of roots = −1 + (−7) = −8, which is −b/a for a = 1, b = 8. Product = (−1)(−7) = 7 = c/a. A MAT-style stem asks for that sum or product. Substituting four candidate roots is the method they avoid; using Vieta after genuine algebra is the method they leave open.

Linear systems

Two linear equations in two unknowns: substitution or elimination. MAT Proficient-level language includes solving and interpreting systems.

Worked example.

x + 2y = 10 (1) 3x − y = 9 (2)

From (1): x = 10 − 2y. Into (2): 3(10 − 2y) − y = 9 30 − 6y − y = 9 30 − 7y = 9 −7y = −21 y = 3 x = 10 − 6 = 4.

The pair is (4, 3). A stem that asked x equals with options 4, 3, 10, 9 would be the easy-substitution shape. A stem that asks for 2x + y wants 8 + 3 = 11. You must solve first. Option-spotting 10 or 9 from the original constants is not a method. x + y = 7 is another nearby quantity if you stop halfway.

Elimination check. Multiply (2) by 2: 6x − 2y = 18. Add to (1): x + 6x + 2y − 2y = 10 + 18, so 7x = 28, x = 4. Same pair.

Worked example: inconsistent versus unique. 2x + y = 5 and 4x + 2y = 11. Twice the first equation would be 4x + 2y = 10, not 11. Parallel lines, no solution. If the second equation were 4x + 2y = 10, there would be infinitely many solutions. Distinguishing those two is algebraic, not a four-number plug into a single equation.

Inequalities

Adding the same number preserves an inequality. Multiplying or dividing by a negative number reverses the inequality.

Worked example: linear. 2x − 5 < 3 → 2x < 8 → x < 4. If the stem had −2x − 5 < 3, then −2x < 8, so x > −4 after reversing.

Worked example: quadratic. x^2 − 4x − 5 ≤ 0. Factor: (x − 5)(x + 1) ≤ 0. Critical points x = −1 and x = 5. The product of two linear factors is ≤ 0 between the roots, including the ends because of ≤. Solution: −1 ≤ x ≤ 5.

Sign chart:

The negative product lives in (−1, 5). Closed interval because of ≤.

Traps: reporting x ≤ −1 or x ≥ 5, which is the complementary set matching ≥ 0. Reporting −5 ≤ x ≤ 1 from mixing the constants 4, 5, and 1. Using strict inequalities when the stem had ≤.

Language. At least 3 is x ≥ 3. Not more than 5 is x ≤ 5. Those phrases are part of the algebra, not decoration. Twice as much as is multiplication; two more than is addition. Mix those and a correct factorisation still picks a wrong option.

Why four-option substitution fails as a plan

If the correct choice is an identity such as (x − 5)^2 − 4, plugging x = 0 into all four expressions can coincidentally match more than one option. If the correct choice is a sum of roots, the options are not candidate x-values. If the correct choice is 2x + y after a system, the four numbers may all look like coefficients from the stem.

The booklet warning is a teaching point: do the algebra. Cover the options, complete the square or eliminate a variable, then hunt for your result. Unscaffolded MAT will not say factorise. Hence solve. Hence state the sum.

Traps

• Stopping after finding one root of a quadratic
• Forgetting to reverse an inequality after multiplying by a negative
• Treating (x + 2)^2 as x^2 + 4
• Adding equations that do not cancel a variable and calling it elimination
• Plugging four option numbers into one equation and calling that MAT technique
• Using a calculator to check when the sitting does not allow one

Independent OpenExamPrep MAT algebra teaching is identities, completed squares, systems, and inequalities you can finish on paper.