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Higher - Rearranging and solving

infoWhy this? We are teaching this unit so that pupils can confidently solve a wide range of equations and inequalities, rearrange formulae and interpret the solutions, enabling them to model relationships and tackle complex algebraic problems systematically

scheduleWhy now? We are teaching it now in Year 10 so that pupils can apply these techniques to strengthen their current work on graphs, functions and problem-solving, and to form and solve equations arising from rich, real-life and multi-step mathematical contexts

neurologyYou need to know

  • A linear equation is a statement of equality that can be solved by finding values that make both sides equal; an inequality uses the symbols <, ≤, >, or ≥ to describe ranges of values that satisfy the relation.
  • When solving linear inequalities, multiplying or dividing both sides by a negative number reverses the inequality sign.
  • On a number line, an open circle shows a strict inequality (< or >) and a closed circle shows an inclusive inequality (≤ or ≥); arrows indicate solutions extending to infinity.
  • A two-step linear equation or inequality can be solved by reversing addition/subtraction and multiplication/division in the correct order while maintaining balance on both sides.
  • Equations with the unknown on both sides are solved by collecting like terms, moving all variable terms to one side and constants to the other before isolating the variable.
  • Changing the subject of a formula uses inverse operations applied to both sides while preserving equivalence; if the subject appears on both sides, terms are collected and the subject is factored out before dividing.
  • For any quadratic equation `ax^2+bx+c=0` with `a≠0`, factorising to `(x-r_1)(x-r_2)=0` shows the roots are `x=r_1` and `x=r_2`.
  • The quadratic formula gives the solutions to `ax^2+bx+c=0` as `x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}` provided `a≠0`.
  • The discriminant `\Delta=b^2-4ac` determines the number of real roots of a quadratic: `\Delta>0` gives two distinct real roots, `\Delta=0` gives one repeated real root, and `\Delta<0` gives no real roots.
  • Completing the square rewrites `ax^2+bx+c` (with `a≠0`) in the form `a(x+\frac{b}{2a})^2 - \frac{b^2-4ac}{4a}`; for `a=1`, `x^2+bx` becomes `(x+\tfrac{b}{2})^2-\tfrac{b^2}{4}`.
  • Solving a quadratic by completing the square leads to an equation of the form `(x+p)^2=q`, whose solutions are `x=-p\pm\sqrt{q}` when `q≥0`.
  • Simultaneous linear equations represent two lines; their solution is the ordered pair `(x,y)` at their intersection, which may be a unique point, no solution (parallel lines), or infinitely many solutions (coincident lines).
  • A linear–quadratic simultaneous system represents the intersection(s) of a line and a parabola and can yield 0, 1, or 2 real solutions depending on whether the line cuts, touches, or misses the curve.
  • In equations involving algebraic fractions, any value that makes a denominator zero is excluded from the solution set (domain restriction), and cross-multiplication is valid only when all denominators are non-zero.
  • Clearing denominators in algebraic fraction equations by multiplying through by the lowest common multiple of denominators can create extraneous solutions, which must be checked against domain restrictions.
  • A linear inequality in two variables represents a half-plane; a solid boundary line is used for ≤ or ≥, a dashed boundary line for < or >, and a test point can determine the correct region to shade.
  • A quadratic inequality in one variable can be solved by finding the roots of the related quadratic equation and using a sign diagram (or the shape of the parabola) to determine intervals where the expression is positive or negative.
  • An iterative process forms a sequence `x_{n+1}=f(x_n)` to approach a root; stable iterations show successive terms settling towards a value, which is then taken as an approximate solution to stated accuracy.

rocket_launchYou must be able to

  • Solve a two-step linear equation by undoing operations in reverse order, showing each inverse operation on both sides and verifying the solution by substitution.
  • Solve equations with the unknown on both sides by collecting variable terms on one side, constants on the other, isolating the variable, and checking the solution.
  • Change the subject of a two-step formula by applying inverse operations and, when the subject appears on both sides, factorising the subject before dividing to isolate it.
  • Form and solve two-step equations from a worded context by translating the relationships into algebra, solving, and interpreting the solution with appropriate units.
  • Represent solutions to one-variable inequalities on a number line using open/closed circles and arrows, matching the inequality symbols precisely.
  • Solve linear simultaneous equations by elimination (matching coefficients, adding/subtracting to eliminate one variable) or substitution, and present the solution as an ordered pair.
  • Solve quadratic equations by factorising into linear factors where possible and deducing the roots directly from the zero-product property.
  • Use the quadratic formula accurately, including calculating the discriminant and evaluating surds, and state both roots (or the repeated root) to a specified degree of accuracy.
  • Complete the square to solve a quadratic and to deduce roots, writing the quadratic in vertex form and solving `(x+p)^2=q` when `q≥0`.
  • Solve equations involving algebraic fractions by stating domain restrictions, clearing denominators using the lowest common multiple, solving the resulting equation, and rejecting any values that invalidate a denominator or arise extraneously. Solve linear–quadratic simultaneous equations by substituting the linear expression into the quadratic, solving the resulting quadratic, and pairing each valid x-value with its corresponding y-value. Solve linear inequalities in two variables by drawing the boundary line, choosing solid or dashed styling, testing a point, and shading the correct half-plane. Solve quadratic inequalities in one variable by factorising, identifying critical values (roots), using a sign chart or parabola shape to select solution intervals, and expressing the final solution using inequality or interval notation as required. Find approximate solutions to equations using iteration by applying a given recurrence `x_{n+1}=f(x_n)` with a sensible starting value, iterating to the required accuracy, and stating the answer with correct rounding and error bounds where requested.


Revision Quiz

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