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Simultaneous equations and quadratic equations - Mathematics Year 10

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Introduction

One equation may accept many pairs; two equations require the same pair to satisfy both. A factorised quadratic asks which inputs make at least one factor zero, not whether every factor must vanish together.

Solve four simultaneous linear pairs and four factorisable quadratics. Compare elimination with substitution, check every original equation and connect quadratic roots to horizontal-axis intercepts.

Solve four simultaneous pairs by two methods

A simultaneous solution must make both equations true with one shared pair of values. For the source 2 x+y=11 and x−y=1, add the equations: the+y and−y cancel, leaving 3 x=12. Thusx=4. Substitution into x−y=1 gives 4−y=1, so y=3. Original checks are 8+3=11 and 4−3=1.

Substitution provides a second method. Fromx−y=1, writey=x −1. Replace the whole y term in the first equation: 2 x+(x −1)=11. This gives 3 x −1=11 and x=4, theny=3. Elimination is particularly efficient here because opposite y coefficients are already present; substitution makes the shared-value requirement explicit.

Four complete simultaneous pairs. Cover the answer column. First identify an efficient variable to eliminate or a simple subject to substitute; then check both original equations.

Task	First equation	Second equation	Shared solution
1	2 x+y=11	x−y=1	x=4, y=3
2	3 x+2 y=16	x+y=6	x=4, y=2
3	2 x −3 y=1	x+y=3	x=2, y=1
4	0.5 x+y=4	x−y=2	x=4, y=2

For row 2, double the second equation to 2 x+2 y=12. Subtract it from 3 x+2 y=16 to getx=4, theny=2. Check 12+4=16 and 4+2=6. Subtracting an entire equation requires subtracting both its variable terms and its constant; changing only one sign is not an equivalent elimination step.

For row 3, use y=3−x from the second equation. Substitute into 2 x −3 y=1 to obtain 2 x −3(3−x)=1. Expand to 5 x −9=1, so x=2 andy=1. The minus before the bracket reaches both terms; the original checks 4−3=1 and 2+1=3 agree.

For row 4, adding the originals cancels y and gives 1.5 x=6, hencex=4 andy=2. Alternatively multiply the first equation by 2 to remove its finite decimal, then use an integer-coefficient elimination. In the originals, .5×4+2=4 and 4−2=2. Keep exact decimal or fractional arithmetic rather than introducing rounding.

The actual original Corbettmaths elimination example uses variables p and q, with different numerical coefficients. Follow how a whole equation is multiplied or combined. Our x/y pairs are separate problems; the requirement to satisfy both equations remains the same regardless of the variable letters. Original: corbettmaths, Simultaneous Equations elimination - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Finding only one coordinate does not finish a pair. Likewise, a pair that satisfies only the first equation is not a solution of the system. Keep the two original substitution checks alongside the result rather than checking only the final eliminated equation.

Solve four quadratics without losing roots

For real numbers, a product is zero if at least one factor is zero. Thus $(x - 2) (x - 3) = 0 ⟹ x = 2 or x = 3.$ The word “or” is essential: there is no real x equal to both 2 and 3 at once. The zero-product rule does not say that a sum is zero only when each term is zero.

For the source x²−5 x+6=0, the signed integers −2 and −3 have product 6 and sum −5. Re-expansion confirmsx²−5 x+6=(x −2)(x −3). Set each factor to zero separately to obtain 2 or 3. Original checks 4−10+6=0 and 9−15+6=0 agree.

Four complete quadratic equations. Factorise first, then supply every distinct real root. A repeated factor gives a repeated root but only one distinct input.

Task	Original equation	Verified factorisation	Complete real solution set
1	x²−5 x+6=0	(x −2)(x −3)=0	x=2 orx=3
2	x²+x −12=0	(x+4)(x −3)=0	x=−4 orx=3
3	x²−8 x+16=0	(x −4)²=0	x=4, one distinct repeated root
4	x²−49=0	(x −7)(x+7)=0	x=−7 orx=7

For row 2, integers 4 and −3 give product −12 and sum 1. Roots −4 and 3 make the original 16−4−12 and 9+3−12 equal 0. In row 3, the factorx −4 is repeated. The equation has the single distinct root 4, with multiplicity 2; do not invent a negative root −4 simply because a square is present.

Row 4 is a difference of squares: x²−49=(x −7)(x+7). Both −7 and 7 square to 49, so both satisfy the original equation. The principal root expression√49 evaluates to 7; solvingx²=49 instead asks for all permitted values and includes both signs.

Do not divide an equation by a factor which could be zero without checking that lost branch. Dividing(x −2)(x −3)=0 by x −2 would assume x≠2 and erase a valid root. Factorisation plus the zero-product alternatives preserves both branches.

Factoring an expression is different from solving an equation. The identityx²−5 x+6=(x −2)(x −3) holds for every real x. Its roots arise only after the expression is set equal to zero. A product equal to 6 cannot be solved by setting both factors to 0; the special zero-product conclusion requires an actual zero on the other side.

Connect solutions to graphs and transfer methods

A root of a function is an input where its output is 0. Forf(x)=x²−5 x+6, the roots 2 and 3 correspond to graph points(2, 0) and(3, 0) on the horizontal axis. The following actual value table and schematic show selected points; the factorisation proves there are exactly these two real roots.

x	f(x)
0	6
1	2
2	0
3	0
4	2
5	6

y
6 ●                   ●
5 |
4 |
3 |
2 |   ●           ●
1 |
0 +-------●---●---------→ x
  0   1   2   3   4   5

Text alternative: sample outputs are 6, 2, 0, 0, 2, 6 at integer inputs 0 through 5. This is a sample-point schematic, not a straight-line interpolation or a claim that the parabola is flat between its roots. The continuous curve goes below the axis between 2 and 3; for examplef(2.5)=−.25.

For a linear system, the shared solution is an intersection of two straight lines. In the source, y=11−2 x andy=x −1 both give 3 at x=4. Distinct parallel lines can have no intersection: x+y=3 and 2 x+2 y=8 would require the same sum to be both 3 and 4. If the second equation were 2 x+2 y=6, it would represent the same line and allow infinitely many pairs satisfyingx+y=3.

For unfamiliar system transfer, x+y=9 and 2 x−y=6 add to 3 x=15, so x=5 andy=4. Both 5+4=9 and 10−4=6 check. For unfamiliar quadratic transfer, x²−7 x+10=0 factors as(x −5)(x −2)=0 and gives roots 5 or 2. Re-expansion and original substitution verify each root independently.

First-error repair. A learner acceptsx=4, y=7 for the first system becausex−y was not checked. The first equation already gives 8+7=15 rather than 11, and the second gives −3 rather than 1. A different learner setsx −2=0 and x −3=0 simultaneously and concludes no solution; replace the incorrect “and” with the zero-product alternative “or”.

Keep four complete linear pairs, four complete quadratic solution sets, all original-equation checks, a graph interpretation and new-value transfer. State whether the answer is a point, a finite root set, no solution or infinitely many solutions. Later retrieval is evidence only after a real new attempt, not because the solved table was viewed.

Interactive Tasks

Quiz: Test Your Knowledge

Source 2 x+y=11 and x−y=1. Shared solution? (x=4, y=3) (!x=3, y=4) (!x=4, y=7) (!x=5, y=1)

3 x+2 y=16 and x+y=6. Solution? (x=4, y=2) (!x=2, y=4) (!x=3, y=3) (!x=5, y=1)

2 x −3 y=1 and x+y=3. Solution? (x=2, y=1) (!x=1, y=2) (!x=3, y=0) (!x=0, y=3)

0.5 x+y=4 and x−y=2. Solution? (x=4, y=2) (!x=2, y=3) (!x=6, y=1) (!x=0, y=4)

Roots ofx²−5 x+6=0? (2 or 3) (!-2 or -3) (!2 and 3 simultaneously) (!5 or 6)

Roots ofx²+x −12=0? (-4 or 3) (!4 or -3) (!-3 or -4) (!3 or 4)

Distinct roots ofx²−8 x+16=0? (4 only) (!4 and -4) (!8 and 2) (!No real root)

Real roots ofx²−49=0? (-7 or 7) (!7 only) (!-7 only) (!49 or -49)

At a quadratic horizontal-axis intercept, the output is? (0) (!1) (!The root’s square) (!Always positive)

x+y=3 and 2 x+2 y=8 have? (No shared solution) (!One shared solution) (!Every pair) (!The whole linex+y=3)

Memory Game

Elimination	Combining equations to remove one variable
Substitution	Replacing a quantity by an equal expression
Simultaneous	Requiring the same values in every equation
Root	Input making a function output zero
Repeated root	A zero arising from a repeated factor
Intersection	Point belonging to both graphs

Drag and Drop

x=4, y=3	Shared point of the source linear pair
x=4, y=2	Shared point of the second linear pair
2 or 3	Roots of the first quadratic
4 only	Distinct root of the repeated-factor quadratic
-7 or 7	Roots of the difference-of-squares quadratic

Crossword Puzzle

ELIMINATION	Removing a variable by combining equations
SUBSTITUTION	Replacing by an equal expression
FACTOR	Member of a product
ROOT	Input giving zero output
INTERSECTION	Common graph point
QUADRATIC	Polynomial equation with highest power two

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 point: Substitute 4 and 3 into both originals.
Read zero-product alternatives: Explain why 2 or 3 does not mean both at once.
Read the point table: Identify both horizontal-axis intercepts.
Distinguish root tasks: Compare√49 with solvingx²=49.

Standard

Solve four simultaneous pairs: Choose elimination or substitution and check both originals.
Solve four quadratics: Factorise, re-expand and retain every root.
Explain repeated roots: Show why the third quadratic has one distinct solution.
Interpret solution type: Distinguish one point, no intersection and identical lines.

Advanced

Compare two source methods: Use both elimination and substitution.
Transfer system solving: Find 5 and 4 from new equations and verify.
Transfer quadratic solving: Factorx²−7 x+10 and check both roots.
Diagnose lost roots: Explain why cancellingx −2 loses a valid zero-product branch.

Learning Check

Four systems: Give both coordinates and both original checks.
Four quadratic sets: Factorise, check signs and state every distinct root.
Graph interpretation: Connect roots to actual outputs.
Special systems: State whether equations are compatible.
New transfer: Use different numerical data and check originals.
Error and evidence: Retain complete original checks and alternatives.

Worked checks and assessment guidance

Pairs x=4, y=3; x=4, y=2; x=2, y=1; x=4, y=2 satisfy each pair. Source adds to 3 x=12; second subtracts a doubled equation; third substitutes y=3−x; fourth can add directly to 1.5 x=6. One coordinate or one satisfied equation is insufficient.
Roots 2 or 3; −4 or 3; 4 only with multiplicity 2; −7 or 7. Verified products arex −2 with x −3, x+4 withx −3, repeatedx −4, and x −7 withx+7. Original substitution checks all roots.
f(x) values at 0 through 5 are 6, 2, 0, 0, 2, 6. Roots 2 and 3 give axis points (2, 0) and (3, 0). f(2.5)=−.25, so the sample schematic does not imply a flat segment. Source straight lines meet at (4, 3).
x+y 3 with 2 x+2 y 8 has no solution because the required sums 3 and 4 conflict. Replacing 8 by 6 gives the same line and infinitely many pairs. A system need not always have one intersection.
x+y 9 and 2 x−y 6 give 5/4, checked by 9 and 6. x²−7 x+10 factors asx −5 timesx −2 and gives 5 or 2. Re-expansion and substitution preserve both branches.
A zero product requires at least one zero factor, not every factor simultaneously. Cancelling a possibly zero factor can lose a root. Keep four pairs, four quadratic sets, graph interpretation, first-error repair and actual new retrieval work.

Learning evidence

Keep four complete simultaneous pairs, four quadratic root sets, original substitution checks, graph interpretation 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 18 to 21. 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.

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

Illustration for the area of a rectangle.svg, Stephan Kulla, CC0. Unchanged original; caption and learning activities are original.