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Linear equations inequalities and simultaneous equations - Mathematics Year 9

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Introduction

What does it mean for one value to satisfy an equation, an interval to satisfy an inequality, or one point to satisfy two equations together? Keep these three solution types distinct.

Solve six equations, four inequalities and two simultaneous pairs. Every answer is checked in its original relationship; an original point diagram connects the first simultaneous solution with the intersection of two straight lines.

Solve six equations and check untouched originals

A solution makes the original equation true. Equivalent transformations preserve its solution set: adding or subtracting the same expression on both sides, and multiplying or dividing both sides by the same permitted nonzero number. Expanding brackets must distribute every term. An equation is not balanced by changing only the side that looks inconvenient.

Source equation. Start $3 (2 x - 1) = 4 x + 9$ . Expand to 6 x −3=4 x+9, subtract 4 x to get 2 x −3=9, add 3 to get 2 x=12, then divide by 2 to obtainx=6. The original left side is 3(12−1)=33 and right side 24+9=33. Checking only the last line 2 x=12 would not catch an earlier bracket error.

Six complete equation tasks.

Task	Original equation	Solution x	Original left at solution	Original right at solution
1	3(2 x −1)=4 x+9	6	33	33
2	5(x+2)=3 x −4	−7	−25	−25
3	(x −3)/4=2	11	2	2
4	7−2 x=3 x+12	−1	9	9
5	2(x+1)−3=x+8	9	17	17
6	x/3+2=x/2−1	18	8	8

For task 2, 5 x+10=3 x −4 leads 2 x=−14 andx=−7. The negative answer is valid: both original sides equal −25. In task 3, multiply both sides by 4 to obtainx −3=8, thenx=11. The fixed denominator 4 is nonzero; no variable-denominator case is being hidden.

For task 4, 7−2 x=3 x+12 becomes −5=5 x andx=−1. For task 5, expand 2 x+2−3=x+8, collect to 2 x −1=x+8, thenx=9. In each case the table records the original left and right values, not merely a repeated simplified equation.

Task 6 uses common denominator 6. Multiplying the complete equationx/3+2=x/2−1 by 6 gives 2 x+12=3 x −6. Subtract 2 x and add 6 to getx=18. Original 18/3+2=8 and 18/2−1=8. Multiplying only fraction terms while leaving constants unchanged would not be an equivalent transformation.

First-error analysis. The source bracket 3(2 x −1) expands to 6 x −3, not 6 x −1. The latter changes the constant before any equation solving occurs. Repair that first invalid line, then solve again and substitute in the untouched original equation. A plausible final number is not a substitute for preservation of equality.

Some equations have no solution or many solutions. For 2(x+1)=2 x+2, both sides are identical and every real x works. For 2(x+1)=2 x+5, the reduction 2=5 is false, so no real x works. Do not divide by a zero coefficient to invent a unique answer in either case. Those cases help explain why a solver must inspect the reduced relation rather than always output a single number.

Solve four inequality intervals with endpoint checks

A blank hundred-cell grid of equal squares. — This is a reference for equal spacing and counting,not a plotted number line. Draw an appropriately labelled number line for the inequality intervals rather than treating the blank cells as completed solutions.

An inequality solution is often an interval, not just its boundary. Adding the same amount preserves an inequality. Multiplying or dividing by a positive number preserves its direction; using a negative number reverses direction. For example −2<3 becomes 2>−3 after multiplying both sides by −1. This reversal follows from order on the number line, not a rule to flip a sign whenever a minus symbol appears anywhere.

Four interval tasks.

Task	Original inequality	Complete solution	Boundary	Boundary status
1	3 x+2<14	x<4	4	Excluded
2	−2 x+5≥11	x≤−3	−3	Included
3	4(x −1)≤2 x+6	x≤5	5	Included
4	7−x>2 x −5	x<4	4	Excluded

Task 1 gives 3 x<12 thenx<4. At boundary 4, the original sides are 14 and 14, and 14<14 is false. An open circle at 4 and shading left represents the interval. The single answer 4 is not a solution.

Task 2 gives −2 x≥6. Dividing by −2 reverses the relation tox≤−3. Boundary −3 gives 11≥11, true; sample −4 gives 13≥11, true; sample −2 gives 9≥11, false. A filled circle at −3 and shading left is appropriate. The erroneousx≥−3 would include values that fail the original inequality.

Task 3 expands 4 x −4≤2 x+6, then 2 x≤10, sox≤5. Boundary 5 is included. Task 4 gives 12>3 x, thereforex<4. It can be found by addingx and 5 to both sides without a negative division. Compare this route with rearranging to −3 x>−12 and dividing by −3; both lead to the same interval if the reversal is handled correctly.

For each solution, substitute the boundary and points on both sides. These sample checks help catch a wrong interval or endpoint, but the algebraic order-preserving/reversing argument establishes the whole real interval. When a context restricts a variable to integer counts, list only the admissible integers within the interval; do not invent that restriction for a problem stated over real numbers.

Distinguish the equation 3 x+2=14 from inequality 3 x+2<14. The former has one boundary value 4; the latter excludes that boundary and has infinitely many smaller values. A correct final statement needs both the value and relation symbol. Keep a first-error correction showing whether the problem was a missing reversal, wrong endpoint or lost original term.

Connect two simultaneous pairs with line intersections

Simultaneous equations share the same unknown values. A solution pair(x, y) must satisfy both original equations at once. If each equation represents a straight line, the pair is their common point. A point merely lying on one line is not enough.

Source pair. With $y = 2 x + 1, y = - x + 7$ , both expressions equal y, so set 2 x+1=−x+7. Then 3 x=6, x=2 andy=5. Independently check both originals: 2×2+1=5 and −2+7=5.

Original paired-line point diagram
A:y=2x+1; B:y=7−x; X:both
 y
 9 | .  .  .  .  A 
 8 | .  .  .  .  . 
 7 | B  .  .  A  . 
 6 | .  B  .  .  . 
 5 | .  .  X  .  . 
 4 | .  .  .  B  . 
 3 | .  A  .  .  B 
 2 | .  .  .  .  . 
 1 | A  .  .  .  . 
 0 | .  .  .  .  . 
   +---------------
     0  1  2  3  4  x
Each axis interval is one unit.

The diagram marks sample points rather than a finished continuous drawing. Plot A through(0, 1), (1, 3), (2, 5), (3, 7), (4, 9) and B through(0, 7), (1, 6), (2, 5), (3, 4), (4, 3), then draw the two straight lines through their own points. Their common X mark is(2, 5). The intersection agrees with the algebraic solution; pixel reading is not used as an exact numerical proof.

Two complete simultaneous pairs.

Pair	First original equation	Second original equation	x	y	Both-original check
1	y=2 x+1	y=−x+7	2	5	2×2+1=5 and −2+7=5
2	2 x+y=11	x−y=1	4	3	2×4+3=11 and 4−3=1

Compare Corbettmaths’ original elimination example using p and q with our separate x, y systems. Add or subtract entire equations, including their right sides. Check both original equations after finding a pair; the video numbers are not hidden givens for our two tasks. Original: corbettmaths, Simultaneous Equations elimination - Corbettmaths. The written explanations and native tasks work without sound or the external player.

For pair 2, add 2 x+y=11 andx−y=1 to eliminate y. This gives 3 x=12 andx=4. Substituting into x−y=1 givesy=3. A separate substitution method usesy=x −1, so 2 x+x −1=11 and again x=4. Both methods must check the same original pair after solving.

Choose substitution when a variable is already isolated. Elimination can be efficient when adding or subtracting complete equations removes a variable. An operation on an equation includes every term and its right side; deleting a variable from only one expression is not elimination.

Two distinct parallel lines with the same gradient and different intercepts have no common point. Identical equations describe the same line and have infinitely many common points. For exampley=2 x+1 versusy=2 x+3 has no solution, whiley=2 x+1 versus 2 y=4 x+2 describes the same line. These are different from the unique intersection in our source pair.

For an unfamiliar pairy=3 x −2 andy=x+6, equate 3 x −2=x+6 to obtainx 4, y 10; both originals agree. Retain six equation solutions, four full inequality intervals, two simultaneous pairs, the point diagram and at least one first-error correction. Numerical checks support, but do not replace, the general algebraic reasoning. Record actual later retrieval separately.

Interactive Tasks

Quiz: Test Your Knowledge

Source 3(2 x −1)=4 x+9: solution? (6) (!3) (!−6) (!12)

5(x+2)=3 x −4: solution? (−7) (!7) (!−3) (!3)

x/3+2=x/2−1: solution? (18) (!6) (!−18) (!3)

3 x+2<14: solution? (x<4) (!x≤4) (!x>4) (!x=4)

−2 x+5≥11: solution? (x≤−3) (!x≥−3) (!x<3) (!x≥3)

What happens when dividing an inequality by a negative number? (Reverse its direction) (!Always replace it with equality) (!Leave direction unchanged) (!Discard the boundary)

Source simultaneous pair intersection? (2, 5) (!5, 2) (!2, 3) (!0, 1)

2 x+y=11 andx−y=1: solution? (4, 3) (!3, 4) (!2, 7) (!5, 1)

Distinct equal-gradient lines with different intercepts? (No common point) (!Exactly one common point) (!Every point shared) (!The origin always shared)

Identical straight-line equations have? (Infinitely many common points) (!No common points) (!Exactly one common point) (!Only integer points)

Memory Game

Equation solution	Value making the original equality true
Interval	Set of values between or beyond stated boundaries
Strict inequality	Relation excluding its equality boundary
Substitution	Replace a variable with an equivalent expression
Elimination	Combine whole equations to remove a variable
Intersection	Point belonging to both plotted lines

Drag and Drop

6	Source equation solution
−7	Second equation solution
x≤−3	Negative-division inequality interval
2, 5	Source pair common point
4, 3	Second simultaneous pair common point

Crossword Puzzle

EQUALITY	Relation preserved on both equation sides
INTERVAL	Whole range of inequality solutions
BOUNDARY	Endpoint tested for inclusion
SUBSTITUTION	Method using an already-isolated variable
ELIMINATION	Method cancelling a variable between equations
INTERSECTION	Point shared by two lines

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

Solve the source equation: Keep bracket expansion, balance and 33=33 original check.
Check all six proposed values: Substitute into both untouched equation sides.
Read strict and inclusive boundaries: Contrastx<4 withx≤5.
Check a point in both lines: Substitute 2, 5 into both source rules.

Standard

Solve six equations: Retain method and original left/right check.
Solve four inequalities: State full intervals, endpoint status and sample checks.
Solve two simultaneous pairs: Use substitution or elimination and check both originals.
Reconstruct the point diagram: Plot labelled A/B points and draw straight lines through their own points.

Advanced

Compare two solving routes: Use elimination and substitution for pair 2.
Analyse zero-or-many cases: Distinguish a false reduced equation, an identity, distinct parallel lines and coincident lines.
Transfer the new pair: Solve 3 x −2=x+6 and verifyx 4, y 10.
Retain actual reasoning: Explain one first error and record real later retrieval without fabricated graph accuracy.

Learning Check

Six equation solutions: Show balance and both original sides at each answer.
Four inequality intervals: Give boundary and whether equality belongs.
Two simultaneous pairs: Solve and verify both original equations.
Graphical connection: Use exact defined A/B points and common X.
First-error and exceptional cases: Correct a bracket or inequality error and classify solution types.
New transfer and portfolio: Solve the unfamiliar pair and retain all required tasks.

Worked checks and assessment guidance

x 6, −7, 11, −1, 9, 18. Original left/right values 33, −25, 2, 9, 17, 8 respectively. Task 6 multiplies the entire equation by 6, including both constants, then solves 2 x+12=3 x −6.
x<4 excludes 4; x≤−3 includes −3; x≤5 includes 5; x<4 excludes 4. Negative division in task 2 reverses≥to≤. Test boundary and points on each side; retain algebraic reasoning for the whole interval.
Sourcey 2 x+1 andy−x+7 gives 2, 5; both rules give 5.2 x+y 11 andx−y 1 gives 4, 3; checks 11 and 1. Elimination adds complete equations; substitutiony=x −1 independently agrees.
A points 0, 1; 1, 3; 2, 5; 3, 7; 4, 9 andB 0, 7; 1, 6; 2, 5; 3, 4; 4, 3 share 2, 5. Connect each own linear-rule set. The diagram is schematic; an exact algebraic check supports the intersection.
3(2 x −1)=6 x −3, not 6 x −1. Dividing −2 x≥6 by −2 givesx≤−3.2(x+1)=2 x+2 is identity; 2(x+1)=2 x+5 has none. Distinct parallel 2 x+1/2 x+3 have none; 2 y=4 x+2 is the same line as y=2 x+1.
y 3 x −2 andyx+6 give x 4, y 10, checked in both. Keep six equations, four inequality intervals, two originalpairs, diagram andfirst-error repair. No physical plotting accuracy or delayed mastery is invented.

Learning evidence

Retain six equations, four fully specified inequality intervals, two simultaneous pairs, a reconstructed point diagram, first-error correction and unfamiliar transfer.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

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

100 grid.PNG, The original uploader was Leightwing at English Wikiversity., Public domain. Unchanged original; caption and learning activities are original.