7.3 Advanced Equations, Inequalities & Simultaneous Equations

Papers G-H expect confident equation solving with brackets and fractions, while Papers I-J push into advanced algebraic equations and transformations, inequalities and simultaneous equations. These questions reward methodical written working: personal calculators are banned, so the numbers are chosen to stay manageable if you keep your steps clean.

Solving Advanced Linear Equations

The goal never changes: get the variable alone by doing the same thing to both sides. The difficulty comes from extra layers — brackets, fractions, and variables on both sides.

Example 1 (brackets and variables on both sides). Solve 5(x - 2) = 3x + 4.

1. Expand the brackets: 5x - 10 = 3x + 4.
2. Subtract 3x from both sides: 2x - 10 = 4.
3. Add 10 to both sides: 2x = 14.
4. Divide by 2: x = 7.

Check by substitution: 5(7 - 2) = 25 and 3(7) + 4 = 25. ✓

Example 2 (fractions). Solve (2x + 1)/3 = 5. Multiply both sides by 3 first: 2x + 1 = 15, so 2x = 14 and x = 7. Clearing the denominator is almost always the best first move.

Example 3 (fractions on both sides). Solve x/2 + x/3 = 10. Multiply every term by 6, the lowest common multiple of 2 and 3: 3x + 2x = 60, so 5x = 60 and x = 12.

Linear Inequalities and Number Lines

An inequality uses <, >, ≤ or ≥ instead of =. Solve it exactly like an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the inequality sign.

Example 4. Solve -2x < 8. Divide both sides by -2 and flip the sign: x > -4. Check with x = 0: -2(0) = 0, and 0 < 8 ✓, whereas x = -5 gives 10 < 8, which fails.

On a number line, an open circle at -4 with shading to the right shows x > -4; a closed (filled) circle is used for ≤ or ≥, where the endpoint is included. ICAS often shows four number lines and asks which one matches a solution — check both the circle type and the shading direction.

Simultaneous Equations

Two equations with two unknowns, such as x + y = 10 and x - y = 4, are called simultaneous equations; the answer is the single pair of values that makes both true at once.

Elimination Method

Add or subtract the equations to remove one variable.

Example 5. Solve x + y = 10 and x - y = 4.

• Add the equations: (x + y) + (x - y) = 10 + 4, so 2x = 14 and x = 7.
• Substitute back: 7 + y = 10, so y = 3.
• Answer: x = 7, y = 3. Check in the second equation: 7 - 3 = 4 ✓.

If nothing cancels directly, multiply one equation first. For 2x + 3y = 12 and 2x - y = 4, subtracting removes 2x immediately: 4y = 8, so y = 2, and then 2x + 6 = 12 gives x = 3.

Substitution Method

Rearrange one equation to make a variable the subject, then substitute into the other.

Example 6. Solve y = 2x + 1 and x + y = 7.

• Replace y in the second equation: x + (2x + 1) = 7, so 3x = 6 and x = 2.
• Then y = 2(2) + 1 = 5. Answer: x = 2, y = 5.

Choose elimination when the coefficients of a variable match (or are easy to match); choose substitution when one equation already says 'y = ...'.

Reading Solutions from Graphs

Each linear equation draws as a straight line, and the intersection point of the two lines is the simultaneous solution. If the graphs of y = 2x + 1 and y = 7 - x cross at (2, 5), then x = 2, y = 5 solves both equations. Parallel lines never meet, so their equations have no solution — they share the same gradient but have different intercepts.

Quadratic Equations (Papers H-J)

Papers H onwards add quadratic equations — equations containing an x² term. The standard form is ax² + bx + c = 0, and ICAS keeps the numbers friendly so that factorising is always enough.

Method: factorise, then apply the null factor law — if two things multiply to zero, at least one of them must be zero.

Example 7. Solve x² + 7x + 12 = 0. Find the pair that multiplies to 12 and adds to 7, namely 3 and 4: (x + 3)(x + 4) = 0. So x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4. A quadratic normally has two solutions, and an option offering only one value is often the trap.

Example 8. Solve x² − 5x = 0. Take out the common factor: x(x − 5) = 0, so x = 0 or x = 5. Never divide both sides by x — that silently destroys the solution x = 0, which is precisely the error the distractors are built from.

Example 9 (difference of two squares). Solve x² − 49 = 0. Factorise as (x − 7)(x + 7) = 0, giving x = 7 or x = −7. Recognising a² − b² = (a − b)(a + b) turns this into a one-line question.

Always rearrange to '= 0' before factorising: x² + 2x = 15 must become x² + 2x − 15 = 0, which factorises to (x + 5)(x − 3) = 0 and gives x = −5 or x = 3.

Recognising Graph Shapes (Papers H-J)

The senior papers ask you to match an equation to its curve rather than plot it point by point, so recognition is enough:

Three checks settle most of these questions without any plotting:

1. Substitute x = 0 to find where the curve cuts the y-axis.
2. Check the sign of the leading coefficient — it decides which way a parabola opens.
3. Count the x-axis crossings: a parabola cuts the x-axis twice, once or not at all, matching the number of solutions of its quadratic equation.

For a parabola, the turning point lies on the vertical line of symmetry, midway between the two x-intercepts. So y = x² − 6x + 8 factorises to (x − 2)(x − 4), crosses the x-axis at x = 2 and x = 4, and therefore turns at x = 3, where y = 9 − 18 + 8 = −1. The vertex is (3, −1).

Transformations at Recognition Level

Papers I-J may show a graph and ask what changed. You only need recognition-level facts:

For straight lines this reduces to what you already know: shifting y = 2x + 1 up by 3 gives y = 2x + 4 — only the intercept changes, and the gradient stays the same. The same shifts apply to the curves above: y = x² + 3 is the parabola y = x² lifted 3 units, and y = (x + 1)² is that parabola moved 1 unit to the left.

Which Papers Need Which Skills

Whatever your paper, the habit that protects marks is the same: substitute your answer back into the original equation as a check.