01 · Two lines
3x + 2y = 12
x − y = −1
Hint
The second equation gives x = y − 1.
Worked solution
3(y − 1) + 2y = 12
5y = 15 ⇒ y = 3
x = 2
The solution is (2, 3).
Understand · explore · practise
Simultaneous equations
Solve pairs of linear and quadratic equations, keep coordinates paired, and connect algebraic answers with graph intersections.
Before you startRearranging, expanding and solving quadratics
01 / Eliminate
The same pair must satisfy both equations.
Solving simultaneously means finding values of x and y that make both equations true. A pair such as (2, 3) means x = 2 and y = 3.
For two linear equations, elimination removes one unknown. Multiply an entire equation so that one pair of coefficients becomes equal or opposite; then subtract or add the equations.
2x + 3y = 13
5x − 2y = 4
To eliminate y here, multiply the first equation by 2 and the second by 3. The y terms then cancel when the equations are added.
Can two lines fail to have one solution?
Yes. x + y = 3 and 2x + 2y = 8 are inconsistent parallel lines: subtraction gives 0 = 2. But x + y = 3 and 2x + 2y = 6 describe the same line, so infinitely many pairs work. Only distinct, non-parallel lines have exactly one intersection.
Make opposite coefficientsWorked example
4x + 6y = 26
15x − 6y = 12
Multiply every term, including the constant.
19x = 38 ⇒ x = 2
Add the equations to eliminate y.
2(2) + 3y = 13 ⇒ y = 3
Substitute into an original equation.
5(2) − 2(3) = 4 ✓
Check the other equation. The solution is (2, 3).
02 / Substitute
Replace an unknown with an equal expression.
Substitution is useful when one equation already isolates an unknown, or can do so without awkward fractions.
y = 3x − 4
2x + y = 11
Replace y in the second equation by the whole expression 3x − 4. Solve for x, then return to the first equation to find y.
Brackets preserve signs and powers. If y = x − 2, then y² = (x − 2)², not x² − 2².
Rearrange unfamiliar forms first
2(x + y) = x + 7
3y = x + 1
The first equation becomes x + 2y = 7. Substituting y = (x + 1)/3 gives:
3x + 2(x + 1) = 21
5x = 19
x = 19/5, y = 8/5
Fractional solutions are valid. Check them in the original equations.
One expression takes the place of yWorked example
2x + (3x − 4) = 11
Substitute into the other equation, not back into itself.
5x − 4 = 11 ⇒ x = 3
Collect terms and solve.
y = 3(3) − 4 = 5
Recover the paired y value.
2(3) + 5 = 11 ✓
Both equations hold at (3, 5).
03 / Linear & quadratic
A substitution can produce two pairs.
When one equation is linear and the other contains x², y² or xy, isolate a variable in the linear equation. Substitution usually leaves a quadratic in one unknown.
For x + y = 7 and xy = 10, write y = 7 − x. This gives x(7 − x) = 10 and hence x² − 7x + 10 = 0.
(x − 2)(x − 5) = 0
x = 2 ⇒ y = 5
x = 5 ⇒ y = 2
The answers are (2, 5) and (5, 2). The pair (2, 2) does not satisfy the original equations: you cannot mix an x from one solution with a y from the other.
The quadratic terms might cancel
For x + 2y = 5 and x² − 4y² = 15, factor the second left side as (x − 2y)(x + 2y). Since x + 2y = 5, it follows that x − 2y = 3. Adding the two linear equations gives x = 4, then y = 1/2.
Always simplify before deciding how many roots to expect. A substituted equation may be linear, an identity or a contradiction. For example, y = x and x² − y² = 0 share every point on that line.
Handle a mixed xy termWorked example
x − y = 2
x² + xy = 12
Use x = y + 2.
(y + 2)² + y(y + 2) = 12
Expand each bracket carefully.
2y² + 6y − 8 = 0
(y + 4)(y − 1) = 0
Divide by 2, then factorise.
y = −4 ⇒ x = −2
y = 1 ⇒ x = 3
Solutions: (−2, −4) and (3, 1).
04 / Exact answers
Keep each sign attached to its pair.
Some intersection coordinates are irrational. Retain exact surds until a decimal is requested, and substitute each root separately.
For x + y = 4 and xy = 1, y = 4 − x gives x² − 4x + 1 = 0, so x = 2 ± √3. Subtracting from 4 reverses the sign in y.
(x, y) = (2 + √3, 2 − √3)
or (2 − √3, 2 + √3)
In each pair the sum is 4 and the product is (2 + √3)(2 − √3) = 4 − 3 = 1. That checks both equations exactly.
Writing x = 2 ± √3 and y = 2 ± √3 without a pairing convention is ambiguous. Two explicit coordinate pairs are clearer.
A line through a circleWorked example
x² + y² = 25, y = 2x
Substitute y = 2x into the quadratic equation.
x² + (2x)² = 25
5x² = 25
The square applies to the 2 as well as x.
x = ±√5
Keep both real roots.
(√5, 2√5) and (−√5, −2√5)
The coordinates have the same sign because y = 2x.
05 / Intersections
Look for the same output at the same input.
Every point on a graph satisfies its equation. An intersection lies on both graphs, so its coordinates solve the simultaneous equations.
y = x² − 2, y = 2x + 1
x² − 2 = 2x + 1
(x − 3)(x + 1) = 0
The intersections are (−1, −1) and (3, 7). Move the input slider to see the vertical gap between the outputs shrink to zero at each crossing.
A sketch helps you understand the number and location of solutions. Use algebra for exact coordinates unless the question specifically asks for graphical estimates.
Same x. Same y?Intersections
y = x² − 2y = 2x + 1
At x = 1, the curve gives −1 and the line gives 3. They do not meet at this input. The intersections are (−1, −1) and (3, 7).
Watch the two graphs agree at an intersection
Pause, replay or seek freely. The notes explain the same idea and stay in view.
06 / Parameters
Substitute first, then count the roots.
For a line and a quadratic curve, count roots of the equation formed by substitution. If it is genuinely quadratic, use its discriminant.
y = x² − 2, y = 2x + c
x² − 2x − 2 − c = 0
D = 4(c + 3)
There are two intersections for c > −3, one tangency for c = −3, and none for c < −3. At tangency, x = 1 and y = −1.
Check an exceptional parameter
For kx² + y = 0 and y = x + 1, the intersection equation is kx² + x + 1 = 0. When k ≠ 0, D = 1 − 4k.
Two intersections: k < 1/4 with k ≠ 0. One repeated intersection: k = 1/4, at (−2, −1). None: k > 1/4. At k = 0 the first graph is the line y = 0, giving one intersection (−1, 0); the quadratic test does not apply.
Use a known solution to find constantsWorked example
y = px − 3, x² + y² = q
(2, 1) is a solution
Insert the given pair into both equations.
1 = 2p − 3 ⇒ p = 2
q = 2² + 1² = 5
Now the equations are fully specified.
x² + (2x − 3)² = 5
5x² − 12x + 4 = 0
Substitute again to find all intersections.
(5x − 2)(x − 2) = 0
(x, y) = (2, 1) or (2/5, −11/5)
Finding the constants does not finish a request for the second solution.
07 / Your turn
Check both equations for each pair.
Keep your working in two columns once the first variable has two possible values. This helps prevent mismatched coordinates.
02 · Rearrange first
2(x − y) = x + 1
x + 3y = 16
Hint
Expand the first equation to isolate x.
Worked solution
x − 2y = 1 ⇒ x = 1 + 2y
1 + 5y = 16 ⇒ y = 3
x = 7
The pair (7, 3) makes the first sides both equal 8 and the second sum equal 16.
03 · A mixed quadratic
x + y = 5
x² + xy + y² = 19
Hint
Substitute y = 5 − x and collect all three contributions.
Worked solution
x² + x(5 − x) + (5 − x)² = 19
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
Solutions: (2, 3) and (3, 2). Each gives 4 + 6 + 9 = 19 in some order.
04 · Pair the surds
x + y = 6
xy = 6
Hint
Use y = 6 − x and complete the square.
Worked solution
x² − 6x + 6 = 0
(x − 3)² = 3
The pairs are (3 + √3, 3 − √3) and (3 − √3, 3 + √3). Each product is 9 − 3 = 6.
05 · Both signs
x² + y² = 20
y = 2x
Hint
Substitution gives 5x² = 20.
Worked solution
x = ±2
x = 2 ⇒ y = 4
x = −2 ⇒ y = −4
Solutions: (2, 4), (−2, −4).
06 · A tangent line
Find c so y = −2x + c touches y = x² + 1 at exactly one point. Find that point.
Hint
Set the outputs equal and require a repeated root.
Worked solution
x² + 2x + 1 − c = 0
D = 4c = 0 ⇒ c = 0
(x + 1)² = 0 ⇒ x = −1
y = 2
The contact point is (−1, 2).
07 · Transform a power relation
2x = 8y−1
x² − 9y² = 18
Hint
Write 8 as 2³ to get a linear equation relating x and y.
Worked solution
x = 3y − 3
(3y − 3)² − 9y² = 18
−18y + 9 = 18
y = −1/2, x = −9/2
The squared terms cancel. The pair gives 81/4 − 9/4 = 18 and equal powers of 2.
08 / Recap
A solution is a pair, not two separate lists.
- Eliminate or substitute to reduce two unknowns to one.
- Substitute a whole expression inside brackets.
- Recover the matching second coordinate for each root.
- Check both original equations.
- Intersections explain the solutions geometrically.
- Check for cancellation or a zero leading coefficient before using the quadratic discriminant.
Section 1 of 8 · Eliminate