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    Algebra - Solving equations and inequalities — Edexcel GCSE Mathematics

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    Algebra - Solving equations and inequalities explained

    In a linear equation the unknown appears only to the first power, never squared and never in a denominator.

    Read the full explanation

    Every step keeps the two sides balanced: whatever you do to one side you do to the other. Gather the unknown on one side, gather the numbers on the other, then undo the multiplications and additions in reverse order. To solve 5x − 4 = 3x + 10, take 3x from both sides to get 2x − 4 = 10, add 4 to both sides to get 2x = 14, then divide by 2 so x = 7. Substituting back gives 31 on each side, which is the check worth doing every time. A graphical version asks you to draw y = 5x − 4 and y = 3x + 10 on one grid and read the x coordinate where the lines cross. That reading is approximate, so quote it to the accuracy the grid supports.

    solve quadratic equations algebraically by factorising; find approximate solutions using a graph (Foundation tier)

    A quadratic equation contains a squared term (e.g. x²) and may have two solutions, one repeated solution or none. To solve by factorising, first rearrange so one side equals zero. For x² + bx + c = 0, find two numbers that multiply to make c and add to make b. For example, to solve x² + 2x − 15 = 0, the numbers are +5 and −3, giving (x + 5)(x − 3) = 0. The zero product rule states that if a product is zero, at least one factor must be zero, so x + 5 = 0 or x − 3 = 0, giving x = −5 and x = 3. A repeated root arises when the bracket is squared, e.g. (x − 3)² = 0 gives x = 3 only. Graphically, solutions to f(x) = 0 are the x-coordinates where y = f(x) meets the x-axis; the curve crosses at two distinct roots but only touches at a repeated root.

    solve quadratic equations (including those that require rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph (Higher tier)

    Higher tier only. On top of factorising you need two methods that work on any quadratic, and the habit of tidying up first: x(x + 4) = 12 must become x² + 4x − 12 = 0 before any method applies. Completing the square rewrites x² + 6x + 2 = 0 as (x + 3)² − 7 = 0, so (x + 3)² = 7 and x = −3 ± √7. The formula x = (−b ± √(b² − 4ac)) / (2a) runs off the coefficients alone and is the safe choice when nothing factorises neatly. Graphically, plot y = x² + 6x + 2 and read the x-coordinates where the curve crosses the horizontal axis as approximate roots. Read the wording for the accuracy wanted: an exact or surd answer keeps the root in place, while a rounded answer means working out the two roots separately, once with the plus and once with the minus.

    solve two simultaneous equations in two variables (linear/linear) algebraically; find approximate solutions using a graph (Foundation tier)

    Two linear equations in two letters are solved by the pair of values that fits both at once, which on a grid is the point where the two lines cross. Lines with different gradients cross exactly once, so such a pair has one solution. Elimination is the usual route: match the coefficients of one letter, then add or subtract the equations so that letter disappears. For 2x + 3y = 19 together with x + y = 8, double the second equation to get 2x + 2y = 16, subtract it from the first, and the x terms vanish, leaving y = 3. Put that into x + y = 8 to get x = 5, then test the pair in both original equations. Substitution is the alternative: make one letter the subject and replace it in the other equation. Graphically you draw both lines and read the coordinates of the crossing point, which is why the graph gives an approximate answer.

    solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph (Higher tier)

    Higher tier only. For two linear equations, eliminate one letter: with 2x + y = 7 and x − y = 2, adding removes y, giving 3x = 9, so x = 3 and y = 1. For a linear/quadratic pair, substitution is the method, because adding or subtracting cannot remove a squared term. Make one letter the subject of the linear equation, substitute into the quadratic, and solve the resulting quadratic. From y = 2x and y = x² − 3, x² − 3 = 2x, so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0, giving x = 3 or x = −1, paired with y = 6 and y = −2. Two pairs mean the line cuts the curve twice; one repeated root means it is a tangent. Graphically, the intersection of the two graphs gives the solution; read approximate coordinates from the grid, or draw the line and curve to estimate them.

    find approximate solutions to equations numerically using iteration (Higher tier only)

    This is Higher tier only. Iteration closes in on a root by improving a guess over and over. The equation is first rearranged so that a single x sits on the left: x² + x − 5 = 0 becomes x = √(5 − x), which gives the iterative formula xₙ₊₁ = √(5 − xₙ). Then you take the starting value the question supplies and feed each answer back in. From x₁ = 2 you get x₂ = 1.7321, x₃ = 1.8077 and x₄ = 1.7867, and the terms settle towards the root near 1.791. On a calculator, work out the first term, then use the answer key inside the formula so that one press produces the next term. Questions normally give you the rearrangement and the starting value, then ask for a stated number of terms or for the root to a given accuracy.

    translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

    Translate worded problems into algebra. This can mean creating an expression (e.g. the cost of 3 pens at x pence each is 3x), a formula from a procedure (e.g. a taxi fare F for m miles at £2 per mile plus a £1.50 fixed charge is F = 1.5 + 2m), or an equation by setting two expressions equal. For an equation, define your unknown (e.g. 'let c be the cost of a coffee in pounds'). Write other quantities in terms of it. If two facts relate two unknowns, derive a pair of simultaneous equations. After solving, always reread the question. It might ask for a derived value, like a total cost, not just your variable. Finally, check your answer makes sense in the context and state units clearly.

    solve linear inequalities in one variable; represent the solution set on a number line (Foundation tier)

    An inequality describes a range of values rather than a single one. Handle it the way you handle an equation, doing the same thing to both sides, with one extra rule: multiplying or dividing both sides by a negative number turns the sign round, so from −2x > 6 you get x < −3. For 3x + 4 ≤ 19, take 4 from both sides to reach 3x ≤ 15, then divide by 3 to reach x ≤ 5. On a number line, mark the boundary with an open circle when that value is not included and a filled circle when it is, then draw an arrow along the values that work. If integers are asked for, list them: the integers satisfying −2 < n ≤ 3 are −1, 0, 1, 2 and 3.

    solve linear inequalities in one or two variable(s), and quadratic inequalities in one variable; represent the solution set on a number line, using set notation and on a graph (Higher tier)

    This is Higher tier only, and it adds three things to the linear work. First, quadratic inequalities: factorise as though it were an equation, find the critical values, then decide which stretch of the curve you want. For x² − x − 6 < 0 the factors are (x − 3)(x + 2), the critical values are −2 and 3, and a sketch of this upward opening curve shows it below the horizontal axis between them, so −2 < x < 3. Second, set notation, which writes that same answer as {x : −2 < x < 3}. Third, inequalities in two letters, where each one gives a boundary line and one side of it: draw the line dashed when the boundary is not included and solid when it is, then shade or label the region where all of the inequalities hold at once.

    Your focus

    1. Describe what makes an equation linear, with the unknown only to the first power and never in a denominator.
    2. Solve a linear equation with the unknown on both sides, giving a fraction or decimal when the answer is not whole.
    3. Justify a solution by substituting it back and showing that both sides come to the same value.
    Show all 38 objectives
    1. Find an approximate solution by drawing both lines on one grid and reading the x coordinate where they cross.
    2. Understand that a quadratic equation may have two solutions, one repeated solution or none.
    3. Rearrange a quadratic equation into the form ax² + bx + c = 0.
    4. Factorise a quadratic expression of the form x² + bx + c.
    5. Apply the zero product rule to find the solution(s) from the factorised form, including a repeated root.
    6. Find approximate solutions to a quadratic equation by identifying where its graph meets the x-axis.
    7. Solve a quadratic by rearranging to zero on one side first, expanding brackets or clearing a fraction as needed.
    8. Calculate both roots from x = (−b ± √(b² − 4ac)) / (2a), keeping the signs of a, b and c.
    9. Complete the square to write a quadratic as (x + p)² + q and read the exact solutions from that form.
    10. Find approximate solutions by plotting the quadratic and reading where the curve crosses the horizontal axis.
    11. Justify the accuracy given, keeping a surd for an exact answer or working both roots out separately when rounding.
    12. Explain why two lines with different gradients give exactly one solution, at the point where they cross.
    13. Solve a linear pair by elimination, matching the coefficients of one letter before adding or subtracting.
    14. Find the second unknown by substituting the first back in, then test both values in the original equations.
    15. Find approximate solutions by drawing both lines and reading the coordinates of the crossing point.
    16. Explain why substitution rather than elimination is the method when one of the two equations is quadratic.
    17. Solve a linear and quadratic pair by making one letter the subject and substituting to reach a quadratic in one letter.
    18. Present the answers as matched pairs of x and y values rather than as four separate numbers.
    19. Interpret a single repeated solution as the line being a tangent to the curve, and two solutions as the line cutting it twice.
    20. Show that rearranging an equation so one x sits on the left produces a given iterative formula.
    21. Work out successive terms from the starting value supplied, using the calculator answer key to feed each result back in.
    22. Find the root to the accuracy stated, carrying enough decimal places in earlier terms to support it.
    23. Show that a root lies between two values by substituting both and stating that the results have opposite signs.
    24. Translate a situation into an algebraic expression or a procedure into a formula, defining all variables used.
    25. Derive an equation, or a pair of simultaneous equations, from the facts given in a problem.
    26. Solve the equation(s) formed and interpret the solution in the original context, giving units.
    27. Check that the solution is reasonable by substituting it back into the original situation.
    28. Solve a linear inequality by treating both sides alike, writing the answer as an inequality and not as an equation.
    29. Explain why the sign turns round when both sides are multiplied or divided by a negative number.
    30. Draw the solution set on a number line with an open or filled circle at the boundary and an arrow the right way.
    31. Write down every integer that satisfies a given inequality, with none missing and none extra.
    32. Find the critical values of a quadratic inequality by factorising or by using the formula.
    33. Explain which stretch of values satisfies the inequality, supported by a sketch of the curve or by testing a value.
    34. Write a solution set in the form asked for, as an inequality, in set notation or on a number line.
    35. Draw each boundary line for inequalities in two letters, dashed or solid to match, and identify the region satisfying all of them.

    Algebra - Solving equations and inequalities exam tips

    Quick Revision Summary (Key Takeaway)

    Solving equations and inequalities in Pearson Edexcel GCSE Mathematics involves finding the value(s) of an unknown that make an equation true or satisfy an inequality. Key skills include using inverse operations to isolate the variable, expanding brackets, solving linear and quadratic equations, and representing solutions on a number line.

    Topic Overview

    Solving equations and inequalities is a fundamental algebraic skill that appears across all areas of GCSE Mathematics. You will learn to manipulate expressions, use inverse operations, and apply systematic methods to find unknown values. This topic underpins many real-world problems, from calculating costs to interpreting data.

    Mastering these techniques is essential for tackling more advanced topics such as simultaneous equations, quadratic equations, and graphical inequalities. It also develops logical reasoning and problem-solving skills that are valuable in everyday life and further study.

    Key Concepts
    • →Equations require finding exact values that make the statement true, while inequalities describe a range of possible values.
    • →Use inverse operations to isolate the variable, performing the same operation on both sides of the equation or inequality.
    • →When multiplying or dividing an inequality by a negative number, the inequality sign must be reversed.
    • →Always expand brackets and simplify expressions before solving.
    • →Represent solutions to inequalities on a number line using open or closed circles and arrows.
    Marking Points
    • a correct first step that does the same thing to both sides, such as subtracting the smaller x term from each side
    • collecting the unknown on one side and the numbers on the other
    • a correct final division even if an earlier arithmetic slip made the number wrong
    • the final value of the unknown, written as a fraction or decimal when it is not a whole number
    • on a graphical question, draw each straight line accurately and read the x coordinate of their intersection
    • Rearranging the equation so that one side is zero before factorising.
    • Correctly factorising the quadratic expression into two brackets.
    • Setting each bracket equal to zero to form separate linear equations.
    • Solving the linear equations to find the solution(s) for x, including a repeated root when a bracket is squared.
    • When using a graph, accurately reading the x-coordinates of the points where the curve meets the x-axis, including a point of tangency for a repeated root.
    • rearranging to a quadratic with zero on one side, including expanding brackets or clearing a fraction first
    • substituting the values of a, b and c into the formula with their signs kept, even if the arithmetic afterwards is wrong
    • the correct value under the root sign before it is simplified
    • a correct completed square form, and a further mark for the solutions that follow from it
    • both roots given to the accuracy the question states, whether that is a surd or a rounded decimal
    • on a graphical question, plotting the curve and reading the x-coordinates of the intersections with the horizontal axis as approximate solutions
    • multiplying one or both equations so that the coefficients of one letter match
    • adding or subtracting correctly to eliminate that letter, even if the arithmetic afterwards is wrong
    • the value of the first unknown
    • substituting back and finding the second unknown, credited even when the first value was wrong
    • on a graphical question, draw each line accurately and read both coordinates of their intersection
    • eliminating one letter correctly when both equations are linear, for example adding 2x + y = 7 and x − y = 2 to get 3x = 9
    • making one letter the subject of the linear equation and substituting it into the quadratic to reach a quadratic in one letter only
    • solving that quadratic by factorising or the formula, with the usual method marks available
    • finding the matching value of the second letter for each root, and presenting the answers as pairs
    • reading approximate solutions from a graph, giving coordinates to the accuracy the grid allows
    • a rearrangement that leads to the given iterative formula when you are asked to derive it
    • a correct first iteration, with the starting value seen substituted into the formula
    • the later iterations, each carried to enough decimal places to support the answer asked for
    • the final root rounded exactly as the question states
    • substituting two values and stating that the results have opposite signs, when asked to show a root lies between them
    • Deriving a correct formula from a described procedure, with all variables clearly defined.
    • Forming a correct expression, equation, or pair of simultaneous equations from the information given.
    • Solving the derived equation(s) correctly, with marks available for method even if the set-up was flawed.
    • Interpreting the solution back in the original context, with correct units and sensible rounding.
    • a correct first step that treats both sides alike, as in an equation
    • the answer written as an inequality rather than as an equation
    • turning the sign round when both sides are divided by a negative number
    • a number line showing the right kind of circle at the boundary and an arrow in the right direction
    • a complete list of the integers in the range, with none missing and none extra
    • the critical values of a quadratic inequality, found by factorising or by using the formula
    • choosing the correct stretch of values, supported by a sketch of the curve or by testing a value
    • the answer in the notation asked for, whether that is an inequality, set notation or a number line
    • each boundary line drawn correctly on a two letter question, dashed or solid to match its inequality
    • identifying the region that satisfies every inequality at the same time
    Examiner Tips
    • 💡Substitute your answer into the original equation before moving on; if the two sides disagree you still have time to find the slip.
    • 💡When the equation contains a fraction, multiply every term by the denominator first so you are working with whole numbers.
    • 💡On a graph question read to the nearest small square and say that the answer is approximate, because that is what is being asked for.
    • 💡After factorising, quickly expand the brackets in the margin to check they match the original quadratic expression.
    • 💡Give all real solutions the equation has; do not force two if the factorised form shows a repeated root, and check whether the context rules out a value.
    • 💡For graphical solutions, be as precise as possible when reading from the axes, using the scale provided. A small ruler can help align your eye with the grid lines.
    • 💡Write down a, b and c with their signs before substituting, and put brackets round any negative value.
    • 💡A negative number under the root sign may be a sign slip, but it can also be a valid no-real-roots result, so check the coefficients and then interpret the outcome.
    • 💡Completing the square also hands you the turning point of the curve, which a later part of the same question often asks for.
    • 💡Number your equations and write what you did to them, such as doubling the second, so the elimination can be followed.
    • 💡Check your pair in the equation you did not use for the substitution; both sides should balance.
    • 💡When neither letter matches easily, multiply each equation by the other equation's coefficient of the same letter.
    • 💡Substitute for the letter already on its own, usually y, as that saves rearranging.
    • 💡Set out the final answer as two pairs, for example x = 3 with y = 6 and x = −1 with y = −2, so the pairing is unmistakable.
    • 💡Test one pair in both original equations; it costs one line of working and catches sign slips.
    • 💡For a graphical solution, draw both graphs accurately and state the coordinates of each intersection, checking they satisfy both equations approximately.
    • 💡Keep the full calculator display for every term and round only at the very end.
    • 💡Write each term on its own line with its label, so the examiner can see which iteration is which.
    • 💡If you are asked to show that a root lies between two numbers, substitute both into the equation and say that the two answers have opposite signs.
    • 💡When creating a formula, identify what changes (the variable) and what stays the same (the constant or fixed charge).
    • 💡Before writing an equation, state what your letter stands for and its units (e.g., 'let t be the time in hours').
    • 💡After solving, reread the last line of the question to ensure you are answering what was actually asked.
    • 💡Say your answer aloud, as in x is less than or equal to five, to check the sign is pointing the right way.
    • 💡To avoid dividing by a negative number, add the negative term to both sides instead and work from there.
    • 💡Label the numbers on your number line; an unlabelled line cannot show which value the circle sits on.
    • 💡Sketch the curve, mark the critical values on it and shade the part you want; the sketch settles the direction of the signs for you.
    • 💡If you are unsure which stretch to take, test one value from each stretch in the original inequality.
    • 💡Set notation needs the letter as well as the condition, as in {x : x ≥ 4}, so do not write the condition on its own.
    • 💡Show all steps of your working, especially when expanding brackets or rearranging. Method marks are awarded even if the final answer is wrong.
    • 💡For inequalities, always represent the solution on a number line as requested, using the correct type of circle (open for < or >, closed for ≤ or ≥).
    • 💡Check your solution by substituting it back into the original equation or inequality to ensure it satisfies the condition.
    Common Mistakes
    • moving a term across the equals sign without changing its sign, turning x + 6 = 10 into x = 10 + 6
    • dividing only one side by the coefficient, so 2x = 14 is written as x = 14
    • expanding a bracket onto the first term only, so 3(x + 4) becomes 3x + 4
    • reading the y coordinate of the crossing point when the equation was solved for x
    • Incorrectly stating the solutions from the brackets, e.g. for (x + 5)(x − 3) = 0, writing x = 5 and x = −3. The signs must be reversed, so the correct solutions are x = −5 and x = 3.
    • Forgetting to rearrange the equation to equal zero first, e.g. trying to factorise x² + 2x = 15 directly.
    • Stopping after factorising and giving the brackets as the final answer, instead of solving for x.
    • When solving an equation like x² − 4x = 0 by factorising to x(x − 4) = 0, forgetting the x = 0 solution from the single 'x' factor.
    • Assuming every quadratic has two distinct solutions. Correction: a repeated root such as (x − 3)² = 0 gives one solution, and some quadratics have no real roots.
    • forgetting that the formula begins with −b, so a positive coefficient is substituted without its sign changing
    • taking a to be the coefficient of x rather than the coefficient of x²
    • halving the coefficient of x when completing the square but not subtracting the square of that half, so x² + 6x is written as (x + 3)² instead of (x + 3)² − 9
    • dividing only the first part of the formula by 2a and leaving the root undivided
    • assuming a negative value under the root sign must be a slip, when it can legitimately mean the quadratic has no real roots
    • subtracting when the matched terms have opposite signs, so 3y and −3y should have been added
    • eliminating the letter correctly but forgetting to combine the numbers on the right as well
    • finding one letter and stopping, when the question needs both values
    • substituting the value back into an equation you created rather than one you were given, which repeats any earlier slip
    • expanding a squared bracket wrongly, writing (x + 4)² as x² + 16 instead of x² + 8x + 16
    • finding the two x values and stopping, without the y value that goes with each one
    • pairing the answers the wrong way round, so one solution's x is quoted with the other solution's y
    • trying to eliminate a letter by subtracting the two equations when one contains a squared term, which cannot remove it
    • rounding each term before feeding it back in, so the sequence settles on a value that is not the root
    • typing the starting value in again for every term instead of using the previous answer
    • labelling the terms from the wrong place, so the value called x₃ is really the second iteration
    • giving the root to a different accuracy from the one the question asked for
    • Confusing the fixed and variable parts when creating a formula, for example swapping a standing charge with a per-unit cost.
    • Mixing units within an equation, such as pounds on one side and pence on the other, without conversion.
    • Solving correctly but giving the value of the variable (e.g. x) when the question asked for a related quantity (e.g. 2x + 5).
    • writing the answer as x = 5 when the solution is a whole range of values
    • leaving the sign unchanged after dividing by a negative number, so −2x > 6 is given as x > −3
    • filling in the circle for a strict inequality, or leaving it open when the boundary value is included
    • including the boundary in a list of integers when the inequality is strict, so −2 is listed for −2 < n
    • offering the critical values as the answer, so −2 and 3 are written down instead of the range between them
    • taking the values outside the critical values when an upward opening quadratic is below zero, or the values between them when it is above zero
    • writing a double inequality the wrong way round, such as 3 < x < −2, which no value can satisfy
    • drawing a solid boundary line for a strict inequality on a region question
    • Forgetting to reverse the inequality sign when multiplying or dividing by a negative number. Always check by substituting a test value.
    • Incorrectly expanding brackets, especially when a negative sign is outside. Multiply every term inside the bracket by the factor outside.
    • Thinking that an equation and an inequality are solved in exactly the same way. Inequalities have a range of solutions and require sign reversal when multiplying/dividing by negatives.
    Revision Plan
    1. 1Day 1-2: Revise basic equation solving (one-step and two-step) and check understanding with practice questions.
    2. 2Day 3-4: Learn to solve equations with brackets and unknowns on both sides. Practice expanding and simplifying.
    3. 3Day 5-6: Study inequalities: solving linear inequalities, reversing the sign, and representing solutions on number lines.
    4. 4Day 7-8: Mixed practice: solve a variety of equations and inequalities, including word problems.
    5. 5Day 9-10: Complete past paper questions on this topic, mark your work, and review any mistakes.
    Exam Question Types
    • 📋Solve a linear equation with brackets and unknowns on both sides. Advice: Expand carefully, collect terms, and check your answer.
    • 📋Solve a linear inequality and show the solution on a number line. Advice: Remember to reverse the inequality sign if multiplying/dividing by a negative.
    • 📋Form and solve an equation from a word problem. Advice: Define the variable, translate the problem into an equation, solve, and interpret the answer in context.
    • 📋Solve a quadratic equation by factorising. Advice: Rearrange to =0, factorise, and set each factor to zero to find both solutions.
    Command Word Expectations (PEARSON EDEXCEL)
    Solve

    Find the value(s) of the unknown that make the equation or inequality true. Show all steps of your working.

    Show that

    Prove a given result by logical steps. You must show every stage of your working to gain full marks.

    Represent

    Draw a diagram (e.g., number line) to illustrate the solution set. Use correct notation (open/closed circles, arrows).

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Forgetting to flip the inequality sign when multiplying or dividing by a negative number, or failing to represent the solution set correctly on a number line.
    ❌ Weak Answer (Loses Marks):Solve -3x < 12. Student writes x < -4.
    Example improved answer:Solve -3x < 12. Divide both sides by -3, remembering to reverse the inequality sign: x > -4. On a number line, draw an open circle at -4 and an arrow pointing to the right.
    Examiner Tip: Always check the direction of the inequality after multiplying or dividing by a negative. Use a quick substitution test (e.g., x = 0) to verify the solution satisfies the original inequality.
    Pitfall: Incorrectly expanding brackets, especially when a negative sign precedes the bracket, leading to sign errors and incorrect solutions.
    ❌ Weak Answer (Loses Marks):Solve 2(x - 3) = 10. Student writes 2x - 3 = 10, then x = 6.5.
    Example improved answer:Solve 2(x - 3) = 10. Expand: 2x - 6 = 10. Add 6 to both sides: 2x = 16. Divide by 2: x = 8.
    Examiner Tip: Always multiply every term inside the bracket by the factor outside. Write out the expansion clearly before solving. Check your answer by substituting back into the original equation.
    Step-by-Step Worked Solutions

    Question: Solve the inequality 5 - 2x ≥ 11 and represent the solution on a number line.

    1. 1.Step 1: Identify the inequality: 5 - 2x ≥ 11.
    2. 2.Step 2: Subtract 5 from both sides: -2x ≥ 6.
    3. 3.Step 3: Divide both sides by -2 and reverse the inequality sign: x ≤ -3.
    4. 4.Step 4: Represent on a number line: closed circle at -3, arrow pointing left.
    Final Answer: x ≤ -3, shown with a closed circle at -3 and an arrow to the left.

    Question: Solve the equation 3(2x - 4) = 2(x + 5).

    1. 1.Step 1: Expand both sides: 6x - 12 = 2x + 10.
    2. 2.Step 2: Subtract 2x from both sides: 4x - 12 = 10.
    3. 3.Step 3: Add 12 to both sides: 4x = 22.
    4. 4.Step 4: Divide by 4: x = 5.5.
    Final Answer: x = 5.5
    Active Recall Memory Test
    What must you do to the inequality sign when you multiply or divide both sides by a negative number?
    Key Fact: Reverse the direction of the inequality sign.
    How do you represent x ≤ 3 on a number line?
    Key Fact: Draw a closed circle at 3 and an arrow pointing to the left.
    What is the first step when solving an equation with brackets?
    Key Fact: Expand the brackets by multiplying each term inside by the factor outside.
    How can you check if your solution to an equation is correct?
    Key Fact: Substitute the value back into the original equation and verify that both sides are equal.
    Frequently Asked Questions
    How do I solve an equation with unknowns on both sides?
    First, expand any brackets. Then collect all terms with the variable on one side by adding or subtracting the same term from both sides. Next, collect constant terms on the other side. Finally, divide by the coefficient of the variable to find its value. Always check your answer by substituting back into the original equation.
    When do I flip the inequality sign?
    You flip the inequality sign when you multiply or divide both sides of the inequality by a negative number. For example, if you have -2x > 6, dividing both sides by -2 gives x < -3. This is because multiplying or dividing by a negative reverses the order of the numbers.
    What is the difference between an equation and an inequality?
    An equation states that two expressions are equal and typically has a specific solution (or set of solutions). An inequality states that one expression is greater than, less than, greater than or equal to, or less than or equal to another, and its solution is a range of values. Inequalities are represented on a number line, while equations are not.
    How do I solve a quadratic equation by factorising?
    First, rearrange the equation so that one side is zero. Then factorise the quadratic expression into two brackets. Set each bracket equal to zero and solve for the variable. This gives the two possible solutions. Always check your solutions by substituting them back into the original equation.
    What common mistakes should I avoid in the exam?
    Common mistakes include: not reversing the inequality sign when multiplying/dividing by a negative, incorrectly expanding brackets (especially with negative signs), forgetting to multiply all terms, and not showing enough working. Always show every step, check your answer, and read the question carefully to ensure you answer what is asked.
    How can I practise solving equations and inequalities effectively?
    Start with basic one-step and two-step equations, then progress to equations with brackets and unknowns on both sides. For inequalities, practise solving and representing solutions on number lines. Use past paper questions and mark schemes to familiarise yourself with exam-style wording and expectations. Regular practice and reviewing mistakes will improve your accuracy and speed.