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Year 9 Maths

Simultaneous Equations

Solve pairs of linear equations using substitution and elimination to find the values of x and y that satisfy both equations at the same time.

What Are Simultaneous Equations?

Simultaneous equations are two equations with the same two unknowns. The solution is the pair of values that makes both equations true at the same time. Graphically, it is where the two lines intersect.

Example System

x + y = 10    (1)

x − y = 2     (2)

We need the x and y that satisfy both equations.

Method 1: Substitution

Rearrange one equation to express one variable in terms of the other, then substitute into the second equation. This method works well when a variable is already isolated.

Example: y = 2x + 1 and 3x + y = 16

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

Step 2: Simplify: 5x + 1 = 16 ⇒ 5x = 15 ⇒ x = 3

Step 3: Find y: y = 2(3) + 1 = 7

Solution: x = 3, y = 7

Method 2: Elimination

Add or subtract the equations to eliminate one variable. Multiply one or both equations first if the coefficients don't match.

Example: 2x + 3y = 13 and 2x + y = 7

Step 1: Subtract (2) from (1) to eliminate 2x:

(2x + 3y) − (2x + y) = 13 − 7 ⇒ 2y = 6 ⇒ y = 3

Step 2: Substitute y = 3 into (2): 2x + 3 = 7 ⇒ x = 2

Solution: x = 2, y = 3

Tip: When eliminating, the goal is to create equal (or opposite) coefficients for one variable. Multiply one equation by a constant if needed.

Key Vocabulary

Simultaneous equations

Two or more equations with shared unknowns solved together.

Substitution

Replacing one variable using an expression from another equation.

Elimination

Adding/subtracting equations to remove one variable.

Point of intersection

The coordinate where two lines cross; the graphical solution.

Worked Examples

1

Substitution: y = x + 3 and 2x + y = 12

Substitute: 2x + (x + 3) = 12 ⇒ 3x = 9 ⇒ x = 3, y = 6

Solution: x = 3, y = 6

2

Elimination: 3x + 2y = 16 and x + 2y = 10

Subtract (2) from (1): 2x = 6 ⇒ x = 3

Substitute: 3 + 2y = 10 ⇒ y = 3.5

Solution: x = 3, y = 3.5

3

Elimination with scaling: 3x + 2y = 11 and 5x − 4y = 1

Multiply (1) by 2: 6x + 4y = 22

Add to (2): 11x = 23 ⇒ x = 23/11 ≈ 2.09

Back-substitute to find y = (11 − 3 × 23/11) / 2 = (121/11 − 69/11)/2 = (52/11)/2 = 26/11 ≈ 2.36

Solution: x = 23/11, y = 26/11

Knowledge Check

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Key Concepts Summary

Year 9: Parabolas Year 9: Data Analysis