Simultaneous Equations

Simultaneous equations are two (or more) equations that must be satisfied at the same time. The solution is the set of values that works in both equations. This lesson covers solving by elimination, substitution, and linear–quadratic simultaneous equations (Higher).

---

Key Vocabulary

Term	Meaning
Simultaneous equations	A set of equations with the same unknowns, solved together
Elimination	Adding or subtracting equations to remove one variable
Substitution	Replacing one variable with an expression from the other equation
Linear equation	An equation where the highest power of the variable is 1
Linear–quadratic pair	One linear and one quadratic equation solved simultaneously

---

Solving by Elimination

The idea: make the coefficients of one variable the same in both equations, then add or subtract to eliminate it.

Worked Example 1

Solve: 2x + 3y = 12 ... (1) 5x + 3y = 21 ... (2)

The y-coefficients are already the same (both 3y).

Subtract (1) from (2): (5x + 3y) − (2x + 3y) = 21 − 12

$3x = 9 \to x = 3$3x=9→x=3

Substitute x = 3 into (1): $6 + 3y = 12 \to 3y = 6 \to y = 2$6+3y=12→3y=6→y=2

Answer: x = 3, y = 2

Check in (2): 5(3) + 3(2) = 15 + 6 = 21 ✓

Worked Example 2

Solve: 3x + 2y = 11 ... (1) 2x − 5y = 1 ... (2)

Neither coefficient matches. Multiply (1) by 5 and (2) by 2 to make the y-coefficients match (both 10):

15x + 10y = 55 ... (1') 4x − 10y = 2 ... (2')

Add: $19x = 57 \to x = 3$19x=57→x=3

Substitute into (1): $9 + 2y = 11 \to 2y = 2 \to y = 1$9+2y=11→2y=2→y=1

Answer: x = 3, y = 1

Worked Example 3

Solve: 4x + 3y = 5 ... (1) 3x + 2y = 4 ... (2)

Multiply (1) by 2 and (2) by 3: 8x + 6y = 10 and 9x + 6y = 12.

Subtract: x = 2.

Substitute into (2): $6 + 2y = 4 \to 2y = -2 \to y = -1$6+2y=4→2y=−2→y=−1.

Answer: x = 2, y = −1

---

Solving by Substitution

Rearrange one equation to make one variable the subject, then substitute into the other equation.

Worked Example 4

Solve: y = 2x + 1 ... (1) 3x + 2y = 16 ... (2)

Substitute (1) into (2): 3x + 2(2x + 1) = 16

3x + 4x + 2 = 16

$7x = 14 \to x = 2$7x=14→x=2

y = 2(2) + 1 = 5

Answer: x = 2, y = 5

Worked Example 5

Solve: x + 3y = 10 ... (1) 2x − y = 6 ... (2)

From (1): x = 10 − 3y

Substitute into (2): 2(10 − 3y) − y = 6

20 − 6y − y = 6

20 − 7y = 6

$-7y = -14 \to y = 2$−7y=−14→y=2

x = 10 − 3(2) = 4

Answer: x = 4, y = 2

Choosing the Method: Substitution vs Elimination

flowchart TD
    PAIR[Pair of simultaneous equations] --> TYPE{Both equations linear?}
    TYPE -->|No: one is quadratic H| SUB1["Use SUBSTITUTION<br/>rearrange linear, sub into quadratic"]
    TYPE -->|Yes, both linear| FORM{"Is one already in<br/>y = ... or x = ... form?"}
    FORM -->|Yes| SUB2["Use SUBSTITUTION<br/>substitute that expression"]
    FORM -->|No| MATCH{"Coefficients of one<br/>variable match or easy to match?"}
    MATCH -->|Yes, same sign| SUBT["ELIMINATION:<br/>subtract equations"]
    MATCH -->|Yes, opposite signs| ADD["ELIMINATION:<br/>add equations"]
    MATCH -->|No: neither matches| MULT["Multiply equations<br/>to make coefficients match"]
    MULT --> SUBT
    MULT --> ADD
    SUB1 --> SOLVE["Solve resulting equation,<br/>then back-substitute for other variable"]
    SUB2 --> SOLVE
    SUBT --> SOLVE
    ADD --> SOLVE

---

Forming Simultaneous Equations from Context

Worked Example 6

At a café, 3 coffees and 2 teas cost £9.10. 5 coffees and 2 teas cost £13.10.

Let c = cost of a coffee, t = cost of a tea.

3c + 2t = 9.10 ... (1) 5c + 2t = 13.10 ... (2)

Subtract (1) from (2): $2c = 4.00 \to c = 2.00$2c=4.00→c=2.00

Substitute into (1): $6.00 + 2t = 9.10 \to 2t = 3.10 \to t = 1.55$6.00+2t=9.10→2t=3.10→t=1.55

A coffee costs £2.00 and a tea costs £1.55.

---

Linear–Quadratic Simultaneous Equations [H]

When one equation is linear and the other is quadratic, use substitution.

Method

1. Rearrange the linear equation to make x or y the subject.
2. Substitute into the quadratic equation.
3. Solve the resulting quadratic.
4. Use the linear equation to find the corresponding values of the other variable.

Worked Example 7

Solve: y = x + 3 ... (1) $x^{2} + y^{2} = 29$x2+y2=29 ... (2)

Substitute (1) into (2): $x^{2} + (x + 3)^{2} = 29$x2+(x+3)2=29

$x^{2} + x^{2} + 6x + 9 = 29$x2+x2+6x+9=29

$2x^{2} + 6x - 20 = 0$2x2+6x−20=0

$x^{2} + 3x - 10 = 0$x2+3x−10=0

(x + 5)(x − 2) = 0

x = −5 or x = 2

When x = −5: y = −5 + 3 = −2

When x = 2: y = 2 + 3 = 5

Answer: x = −5, y = −2 or x = 2, y = 5

Worked Example 8

Solve: y = 2x − 1 ... (1) $y = x^{2} - 3x + 1$y=x2−3x+1 ... (2)

Set equal: $2x - 1 = x^{2} - 3x + 1$2x−1=x2−3x+1

$0 = x^{2} - 5x + 2$0=x2−5x+2

Using the quadratic formula: $x = (5 \pm \sqrt{25 - 8}) / 2 = (5 \pm \sqrt{17}) / 2$x=(5±25−8​)/2=(5±17​)/2

$x = (5 + \sqrt{17})/2 \approx 4.56$x=(5+17​)/2≈4.56 or $x = (5 - \sqrt{17})/2 \approx 0.44$x=(5−17​)/2≈0.44

Corresponding y-values: y = 2x − 1 for each x.

When $x \approx 4.56$x≈4.56: $y \approx 8.12$y≈8.12

When $x \approx 0.44$x≈0.44: $y \approx -0.12$y≈−0.12

Exact answers: $x = (5 \pm \sqrt{17})/2$x=(5±17​)/2, $y = 4 \pm \sqrt{17}$y=4±17​

---

Graphical Interpretation

• The solutions to simultaneous equations are the coordinates of the intersection points of the two graphs.
• Two linear equations $\to$→ the lines cross at one point (unless parallel or identical).
• A line and a circle/parabola $\to 0$→0, 1, or 2 intersection points.

Worked Example 9

The line y = 3x + k is tangent to the circle $x^{2} + y^{2} = 10$x2+y2=10. Find k. [H]

Substitute: $x^{2} + (3x + k)^{2} = 10$x2+(3x+k)2=10

$x^{2} + 9x^{2} + 6kx + k^{2} = 10$x2+9x2+6kx+k2=10

$10x^{2} + 6kx + (k^{2} - 10) = 0$10x2+6kx+(k2−10)=0

For a tangent, there is exactly one solution, so the discriminant = 0:

$(6k)^{2} - 4(10)(k^{2} - 10) = 0$(6k)2−4(10)(k2−10)=0

$36k^{2} - 40k^{2} + 400 = 0$36k2−40k2+400=0

$-4k^{2} + 400 = 0$−4k2+400=0

$k^{2} = 100$k2=100

k = 10 or k = −10

---

Common Mistakes and Misconceptions

Mistake	Correct approach
Adding when you should subtract (or vice versa)	Same signs $\to$→ subtract to eliminate; different signs $\to add$→add
Forgetting to find BOTH variables	After finding x, substitute back to find y
Only finding one pair of solutions for linear–quadratic	There are usually two pairs — find both
Arithmetic errors with negative numbers	Be especially careful when substituting negative values
Not checking both solutions in both original equations	Always verify your answers

---