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Linear equations and inequalities - Mathematics Year 10

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Introduction

An equation asks where two expressions are equal. An inequality asks which values make one side larger or smaller, and its answer may be an interval rather than one number.

Solve eight equations and six inequalities, including fractions, brackets, special solution sets and contexts. Check values in the original statement and explain when a boundary belongs to the answer.

Solve eight equations and check the original statements

A solution makes the original equation true. Equivalent transformations preserve its solution set: expand correctly, add or subtract the same expression on both sides, and multiply or divide by a nonzero constant. A convenient written rule such as “move it over” must stand for one of those justified operations.

For the source 4(2 x −1)=3 x+16, expand to 8 x −4=3 x+16. Subtract 3 x from both sides to get 5 x −4=16, then add 4 to obtain 5 x=20. Divide by 5 to findx=4. Substitution into the original gives left 4×7=28 and right 12+16=28. Checking only the simplified line 5 x=20 would miss a possible earlier expansion error.

Eight complete equations. The final two rows deliberately have special solution sets, so not every equation should be forced into a single numerical answer. Here x ranges over real numbers unless a context later narrows it.

Task	Original equation	Equivalent intermediate statement	Complete solution
1	4(2 x −1)=3 x+16	8 x −4=3 x+16; 5 x=20	x=4
2	3(x+2)=2 x+11	3 x+6=2 x+11	x=5
3	(x −1)/3+2=5	x −1=9	x=10
4	(2 x+3)/5=(x −1)/2	4 x+6=5 x −5	x=11
5	7−2 x=3 x −8	15=5 x	x=3
6	0.4 x+0.6=2.2	4 x+6=22	x=4
7	2(3 x −4)=6 x −8	6 x −8=6 x −8	Every real x
8	5(x+1)=5 x+9	5=9 after cancelling 5 x	No real solution

Row 2 subtracts 2 x and 6 to givex=5; substitution gives 3×7=21 and 10+11=21. Row 3 subtracts 2 first, then multiplies by 3: x −1=9, so x=10. The whole bracketx −1 is divided by 3; the source solution gives 9/3+2=5.

For row 4, multiply both sides by 10, the common nonzero denominator multiple. This yields 2(2 x+3)=5(x −1), then 4 x+6=5 x −5 and x=11. Original left 25/5=5 and right 10/2=5 agree. Dividing only part of a bracket or forgetting one side’s multiplier would destroy equivalence.

In row 5, add 2 x and 8 to both sides to obtain 15=5 x, so x=3. For row 6, multiply all terms by 10 to remove finite decimals, getting 4 x+6=22 and x=4. Both original sides then equal 2.2; exact decimal scaling avoids an unnecessary approximate answer.

In row 7, the two sides simplify to the same expression, so every real x works. In row 8, the variable terms cancel to the false equality 5=9, so no real x works. These are different outcomes: identity versus inconsistency. Neither cancellation gives permission to announcex=0 without checking the remaining statement.

Solve six inequalities and represent their solution sets

Adding or subtracting the same number preserves inequality order. Multiplying or dividing by a positive number also preserves it. Multiplying or dividing by a negative number reverses it: if 2<5, then −2>−5. Reflection through zero reverses which value lies farther right on a number line.

For the source −2 x+6>10, subtract 6 to obtain −2 x>4. Dividing by −2 reverses the direction and gives $x < - 2.$ Atx=−3, the original left side is 12, which exceeds 10. At the boundaryx=−2, it is 10, which is not strictly greater. Atx=0, it is 6, which fails. Testing one inside value, the boundary and an outside value checks the direction and endpoint.

Six complete inequalities. A strict sign excludes equality; an inclusive sign permits it. The last row has two boundary conditions which must both hold.

Task	Original statement	Equivalent intermediate statement	Complete real solution set
1	−2 x+6>10	−2 x>4	x<−2
2	3 x −5≤10	3 x≤15	x≤5
3	7−x≥2	−x≥−5	x≤5
4	(x+1)/2<4	x+1<8	x<7
5	−3(x −2)≤9	−3 x+6≤9	x≥−1
6	−2≤3 x+1<10	−3≤3 x<9	−1≤x<3

Row 2 divides a non-strict inequality by positive 3, so x≤5 includes 5. Row 3 divides−x≥−5 by −1 and reverses to x≤5. The original left side at 5 is 2, so equality is allowed. Row 4 multiplies by positive 2, then subtracts 1 to givex<7; the boundary 7 fails because the original left side is exactly 4.

Row 5 expands −3(x −2) into −3 x+6, not −3 x −6. Subtract 6 to get −3 x≤3, then divide by −3 and reverse to x≥−1. At −1, the original is 9 and is allowed. At 0, it is 6 and allowed; at −2, it is 12 and fails. Correct expansion and correct order reversal are separate checks.

For row 6, subtract 1 from all three parts, then divide all three by positive 3. The result −1≤x<3 includes −1 but excludes 3. Values 0 and 2 belong, while −2 and 3 do not. The following schematic is not a scale drawing: the filled endpoint means included and the open endpoint means excluded.

       -1             3
<-------●=============○------->

Text alternative: all real values from −1 inclusive to 3 exclusive. For integer-only inputs, the allowed values are −1, 0, 1, 2. A discrete context can narrow the real interval, but must not be silently assumed for a pure algebra question.

The actual original Corbettmaths Inequalities preview distinguishes greater-than, greater-than-or-equal, less-than and less-than-or-equal signs. Retrieve the meanings, then justify the sign reversal and boundary inclusion in our six solved statements; recognising a sign alone does not complete a solution. Original: corbettmaths, Inequalities - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Model contexts, test boundaries and repair errors

A context determines the domain and the meaning of a boundary. An invented trip budget has a fixed£8 charge plus£3 per ticket and must not exceed£35. Letn be the nonnegative whole number of tickets. The model is $8 + 3 n \leq 35$ , so 3 n≤27 and n≤9. The largest feasible whole count is 9. Check£8+£27=£35 exactly; ten tickets would cost£38 and fail.

If the wording instead required spending strictly less than£35, the solution would ben<9 and the maximum whole count would be 8. A small wording change affects whether the boundary is permitted. Reading “at most” as a strict inequality would exclude an affordable boundary solution incorrectly.

For unfamiliar equation transfer, 5(2 x+1)=3 x+26 expands to 10 x+5=3 x+26. Subtract 3 x and 5 to get 7 x=21, so x=3. Original left 5×7=35 and right 9+26=35 agree. This checks a different coefficient pattern rather than repeating the original 4-bracket problem.

For unfamiliar inequality transfer, −4 x+3≥15 leads to −4 x≥12 and x≤−3. At −3, the original is 15 and allowed; at −4, it is 19 and allowed; at −2, it is 11 and excluded. The test values confirm the direction and the included equality independently of the rearranged line.

First-error repair. A learner goes from −2 x>4 to x>−2. The division is by a negative number, so the sign must reverse. Their claimed answer includes 0, but 0 makes the original 6>10 false. An explicit counterexample diagnoses the first invalid transformation. Another learner expands 4(2 x −1) as 8 x −1; the first mistake is not distributing 4 to the constant −1.

An equation such as 2 x+3=2 x+3 is true for every real x. An equation 2 x+3=2 x+5 is false for every real x. A pure inequality may also simplify to an always-true or always-false numerical statement. Interpret what remains after cancellation; do not divide by a vanished coefficient or treat every simplified row as a one-answer problem.

Retain eight equation solutions, six full inequality sets, the original-statement substitutions and at least one context with a stated domain. For a boundary question, include an inside value, the boundary and an outside value. Revisit a different task later and record the actual working, including any uncertainty; seeing a solved example is not itself independent performance.

Interactive Tasks

Quiz: Test Your Knowledge

Source 4 times bracket 2 x −1 equals 3 x+16. Solution? (x 4) (!x 3) (!x 5) (!x 0)

Whole x −1 divided by 3, plus 2, equals 5. Solution? (x 10) (!x 8) (!x 6) (!x 16)

Whole 2 x+3 divided by 5 equals whole x −1 divided by 2. Solution? (x 11) (!x 1) (!x −11) (!x 5)

0.4 x+0.6=2.2. Solution? (x 4) (!x 7) (!x 0.4) (!x 1.6)

2 times bracket 3 x −4 equals 6 x −8. Solution set? (Every real x) (!No real x) (!x 0 only) (!x 1 only)

5 times bracketx+1 equals 5 x+9. Solution set? (No real solution) (!Every real x) (!x 0) (!x 4)

Source −2 x+6 greater than 10. Solution? (x less than −2) (!x greater than −2) (!x at least −2) (!x equals −2)

−3 times bracketx −2 is at most 9. Solution? (x at least −1) (!x at most −1) (!x greater than 1) (!x at most 3)

Which integers satisfy −1≤x<3? (-1, 0, 1, 2) (!0, 1, 2, 3) (!-1, 0, 1, 2, 3) (!0, 1, 2)

Budget 8+3 n at most 35. Maximum whole tickets? (9) (!8) (!10) (!11)

Memory Game

Solution	Input making the original statement true
Equivalent	Preserving the complete solution set
Strict	Excluding equality at a boundary
Inclusive	Allowing equality at a boundary
Identity	Equality for all permitted values
Inconsistent	Having no permitted solution

Drag and Drop

x=4	Source bracket-equation solution
Every real x	Solution of the identity equation in row 7
No real solution	Outcome of the inconsistent equation in row 8
x<−2	Strict source inequality solution
−1≤x<3	Two-boundary interval in row 6

Crossword Puzzle

EQUATION	Statement asserting equal values
INEQUALITY	Statement comparing unequal or possibly equal values
BOUNDARY	Endpoint controlling membership in a solution set
STRICT	Excluding equality
INCLUSIVE	Including equality
REVERSAL	Change of inequality direction under a negative multiplier

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Check the source solution: Substitute 4 into both original equation sides.
Interpret strictness: Explain why −2 is excluded from the source inequality.
Read an interval: Describe the sketch without relying on colour.
Recognise special outcomes: Distinguish every real input from no solution.

Standard

Solve eight equations: Show equivalent transformations and original checks.
Solve six inequalities: Retain direction, endpoint inclusion and full sets.
Represent the compound interval: Give real interval and separate integer restriction.
Model the ticket budget: State the whole-number domain and verify maximum 9.

Advanced

Transfer equation solving: Solve 5 times bracket 2 x+1 equals 3 x+26 and check.
Transfer inequality solving: Findx≤−3 from new data and test three positions.
Change a contextual boundary: Compare at most 35 with strictly less than 35.
Repair the first invalid step: Use a counterexample to diagnose missing reversal or distribution.

Learning Check

Eight equation outcomes: Retain full solution sets and original substitutions.
Six inequality sets: State inclusion and direction.
Source boundary checks: Test inside, boundary and outside values.
Context and wording: Use nonnegative whole ticket counts.
New-value transfer: Check different equations and inequalities independently.
Error and retained work: Correct the first invalid transformation.

Worked checks and assessment guidance

x 4, x 5, x 10, x 11, x 3, x 4, every real x, no real solution. The final two reduce to an identity and a contradiction, respectively. A vanished variable coefficient does not implyx 0.
x<−2; x≤5; x≤5; x<7; x≥−1; −1≤x<3. Negative division reverses direction. The final interval contains −1 but excludes 3; integer solutions are −1, 0, 1, 2 only when an integer domain is specified.
For −2 x+6>10, inputs −3, −2, 0 give 12, 10, 6; only the first satisfies strict greater-than 10. For −3 times bracketx −2≤9, inputs −1, 0, −2 give 9, 6, 12, confirmingx≥−1.
8+3 n≤35 givesn≤9 and maximum 9, with cost 35. Ten costs 38. Strictly less than 35 instead gives maximum 8. The real interval is narrowed by the explicit count domain, not by an unstated algebra convention.
5 times bracket 2 x+1 equals 3 x+26 givesx 3 and both sides 35.−4 x+3≥15 givesx≤−3; inputs −3, −4, −2 give 15, 19, 11, so the first two satisfy it and the last fails.
Dividing −2 x>4 by −2 must reverse the sign. Expanding 4 times bracket 2 x −1 gives 8 x −4. Keep eight equations, six inequalities, original checks, context and an actual new response; rereading answers is not proof of independent mastery.

Learning evidence

Retain eight complete equation outcomes, six inequality solution sets, original-statement and endpoint checks, a whole-number context, and unfamiliar transfer with first-error repair.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Algebra - Solving equations and inequalities items 17 and 22. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

ZR20 (Dienes-Material).jpg, Rhetos, CC0. Unchanged original; caption and learning activities are original.

2D Cartesian Coordinates.svg, H Padleckas, Public domain. Unchanged original; caption and learning activities are original.