Simultaneous Equations – Grade 9 Maths

Elimination Method

Add or subtract equations to cancel a variable

Substitution Method

Rearrange one equation, substitute into the other

Word Problems

Form and solve simultaneous equations from context

Linear & Quadratic

One linear and one quadratic equation together

Graphical Interpretation

Intersection of two lines gives the solution

1. Elimination Method

In the elimination method, you multiply one or both equations so that the coefficients of one variable are equal, then add or subtract the equations to eliminate that variable.

Solve: 2x + 3y = 12 and 4x − 3y = 6
Add the equations: 6x = 18 → x = 3
Substitute back: 2(3) + 3y = 12 → y = 2

Steps:
1. Make the coefficient of one variable the same in both equations (multiply if needed).
2. Add the equations if the signs are opposite; subtract if the signs are the same.
3. Solve the resulting single-variable equation.
4. Substitute back to find the other variable.
5. Check your answer in both original equations.

When you subtract equations, be careful to subtract every term including the right-hand side.

2. Substitution Method

In the substitution method, you rearrange one equation to make one variable the subject, then substitute that expression into the other equation.

Solve: y = 2x + 1 and 3x + y = 16
Substitute: 3x + (2x + 1) = 16 → 5x = 15 → x = 3
Then: y = 2(3) + 1 = 7

Steps:
1. Rearrange one equation to get x = … or y = …
2. Substitute into the other equation.
3. Solve the resulting equation.
4. Find the second variable by substituting back.
5. Verify in both original equations.

Substitution is particularly efficient when one equation already has a variable with coefficient 1 (e.g., y = 3x − 2).

3. Word Problems

Many real-world problems can be modelled using two simultaneous equations. The key is to define variables clearly and form equations from the given information.

Example: Two numbers add to 15 and their difference is 3. Find both numbers.
Let the numbers be x and y. Then: x + y = 15 and x − y = 3.
Adding: 2x = 18 → x = 9, so y = 6.

Always state what your variables represent. Check that both equations are satisfied by your solution.

4. One Linear and One Quadratic

When one equation is linear and one is quadratic, use substitution. Substitute the linear expression into the quadratic to get a quadratic equation, then solve by factorisation or the quadratic formula.

Solve: y = x + 1 and y = x² − 1
Substitute: x + 1 = x² − 1 → x² − x − 2 = 0
Factorise: (x − 2)(x + 1) = 0 → x = 2 or x = −1
Corresponding y values: y = 3 or y = 0

A linear and quadratic pair can have 0, 1, or 2 solutions depending on whether the line intersects, is tangent to, or misses the parabola.

Always give both solution pairs (x, y) when two solutions exist.

5. Graphical Interpretation

Each linear equation represents a straight line. The solution to a pair of simultaneous equations is the point of intersection of the two lines.

Unique solution: The lines intersect at one point — one pair (x, y) satisfies both.
No solution: The lines are parallel (same gradient, different intercept) — no intersection.
Infinitely many solutions: The lines are coincident (identical) — every point on the line is a solution.

You can read approximate solutions from a graph, but algebraic methods give exact values.

Example 1 — Elimination (equal coefficients)

Solve: 3x + 2y = 11 and 3x − y = 5

Subtract equation 2 from equation 1: (3x + 2y) − (3x − y) = 11 − 5 → 3y = 6 → y = 2

Substitute y = 2 into equation 2: 3x − 2 = 5 → 3x = 7 → x = 7/3

Check in equation 1: 3(7/3) + 2(2) = 7 + 4 = 11 ✓

Example 2 — Elimination (multiply first)

Solve: 2x + 3y = 13 and 5x − 2y = 1

Multiply eq1 by 2 and eq2 by 3: 4x + 6y = 26 and 15x − 6y = 3

Add: 19x = 29 → x = 29/19 ... (in simpler problems coefficients work out neatly)

For the worked example with integers: Solve 2x + y = 7 and x − y = 2. Add: 3x = 9 → x = 3. Then y = 1.

Example 3 — Substitution

Solve: x = 2y − 1 and 3x + y = 17

Substitute x = 2y − 1: 3(2y − 1) + y = 17 → 6y − 3 + y = 17 → 7y = 20 → y = 20/7

Simpler illustration: x − y = 1 and 2x + y = 8. From eq1: x = y + 1. Substitute: 2(y+1)+y=8 → 3y=6 → y=2, x=3.

Example 4 — Word Problem

A cinema sells adult tickets for £8 and child tickets for £5. 20 tickets are sold for £130 in total. How many adult tickets were sold?

Define variables: Let a = adult tickets, c = child tickets.

Form equations: a + c = 20 and 8a + 5c = 130

From eq1: c = 20 − a. Substitute: 8a + 5(20 − a) = 130 → 3a + 100 = 130 → 3a = 30 → a = 10

Check: c = 10; 8(10) + 5(10) = 80 + 50 = 130 ✓

Example 5 — Linear and Quadratic

Solve simultaneously: y = x + 1 and y = x²

Substitute: x² = x + 1 → x² − x − 1 = 0 (use quadratic formula if needed)

For integer example: y = 2x and y = x² − 3. Substitute: x² − 3 = 2x → x² − 2x − 3 = 0 → (x−3)(x+1) = 0 → x = 3 or x = −1.

Solutions: (3, 6) and (−1, −2)

Simultaneous Equation Solver & Graph

Enter two equations in the form ax + by = c. The graph will plot both lines and mark their intersection.

Eq 1: a₁ b₁ c₁

Eq 2: a₂ b₂ c₂

Exercise 1 — Elimination Method (find x)

1. Solve by elimination. Find x: 2x + y = 5 and x + y = 3

2. Solve by elimination. Find x: 3x + 2y = 13 and x + 2y = 7

3. Solve by elimination. Find x: 4x + y = 9 and 2x + y = 7

4. Solve by elimination. Find x: 5x − y = 11 and x − y = -9

5. Solve by elimination. Find x: 3x + 4y = 23 and x + 4y = 13

6. Solve by elimination. Find x: 2x + 3y = 7 and 4x + 3y = 9

7. Solve by elimination. Find x: x + 5y = 12 and x + 3y = 8

8. Solve by elimination. Find x: 2x − y = 4 and x − y = 1

9. Solve by elimination. Find x: 3x + 2y = 10 and 3x − 2y = 10

10. Solve by elimination. Find x: 4x + y = 25 and 2x + y = 13

Exercise 2 — Elimination Method (find y)

1. Solve by elimination. Find y: x + y = 5 and x − y = 3

2. Solve by elimination. Find y: 2x + y = 8 and 2x − y = 4

3. Solve by elimination. Find y: 3x + 2y = 13 and 3x − y = 7

4. Solve by elimination. Find y: 4x + 3y = 9 and 4x + y = 5

5. Solve by elimination. Find y: x + 2y = 11 and x − 2y = 3

6. Solve by elimination. Find y: 5x + y = 17 and 5x − y = 13

7. Solve by elimination. Find y: 2x + 5y = 4 and 2x − 5y = 34

8. Solve by elimination. Find y: 3x + y = 10 and x + y = 4

9. Solve by elimination. Find y: 4x + 3y = 23 and 4x − y = 7

10. Solve by elimination. Find y: 2x + y = 9 and 2x − 2y = 3

Exercise 3 — Substitution Method (find x)

1. Solve by substitution. Find x: y = x − 1 and x + y = 5

2. Solve by substitution. Find x: y = 2x and x + y = 3

3. Solve by substitution. Find x: y = x + 2 and 2x + y = 10

4. Solve by substitution. Find x: y = 3x − 1 and x + y = 7

5. Solve by substitution. Find x: y = 4 − x and 3x − y = 8

6. Solve by substitution. Find x: y = 2x + 3 and 3x − y = 12

7. Solve by substitution. Find x: x = y + 2 and 2x + 3y = 6

8. Solve by substitution. Find x: y = x − 1 and 4x + y = 14

9. Solve by substitution. Find x: y = 5 + 2x and x + y = 4

10. Solve by substitution. Find x: y = x − 3 and 2x − y = 17

Exercise 4 — Word Problems

1. Two numbers add to 12 and their difference is 2. What is the larger number?

2. A pen costs £p and a ruler costs £r. Two pens and a ruler cost £18; a pen and two rulers cost £14. Find p (price of a pen in £).

3. The sum of two numbers is 9 and their product is 18. What is the smaller number?

4. A cinema charges £a for adults and £c for children. 3 adults and 2 children pay £50. 1 adult and 4 children pay £32. Find a.

5. The perimeter of a rectangle is 24 cm. The length is 3 cm more than the width. What is the length (in cm)?

6. Two friends share £26. One has £18 more than the other. How much does the smaller share amount to?

7. A train travels from A to B in x hours at 60 km/h and returns at 40 km/h taking y hours. Total time is 5 hours. Find the distance AB (km). Hint: distance = 60x = 40y, x + y = 5.

8. Five apples and three bananas cost £3.30. Two apples and five bananas cost £2.70. Find the cost of an apple in pence.

9. The angles of a triangle are x, y, and 90°. Also x = y + 18. Find x (degrees).

10. x years ago a mother was 4 times her son's age. In 6 years she will be twice his age. The son is currently s years old. Find s. (Current ages: mother M, son s. M−x = 4(s−x) and M+6 = 2(s+6). If x=6: M = 4s−18 and M = 2s+6, so 4s−18=2s+6 → s=12, checking: s=12, M=30. Give the son's age in the simpler formulation: if mother is 3 times son now, and in 10 years she's twice as old — mother = 3s, 3s+10=2(s+10), s=10. Enter 11.)

Exercise 5 — One Linear, One Quadratic (find positive x)

1. Solve simultaneously. Find positive x: y = x + 0 and y = x² − 6

2. Solve simultaneously. Find positive x: y = x + 0 and y = x² − 2

3. Solve simultaneously. Find positive x: y = 2x + 0 and y = x² − 8

4. Solve simultaneously. Find positive x: y = x − 0 and y = x² − 0 (i.e. x = x², so x = 0 or x = 1)

5. Solve simultaneously. Find positive x: y = x and y = x² − 20

6. Solve simultaneously. Find positive x: y = 3 and y = x² − 6

7. Solve simultaneously. Find positive x: y = 2 and y = x² − 2

8. Solve simultaneously. Find positive x: y = 8 and y = x² − 8

9. Solve simultaneously. Find positive x: y = 0 and y = x² − 1

10. Solve simultaneously. Find positive x: y = x and y = x² − 30

Practice — 20 Questions

1. Eliminate to find x: 2x + y = 5, x + y = 3

2. Eliminate to find x: 3x + 2y = 13, x + 2y = 7

3. Eliminate to find x: 4x + y = 9, 2x + y = 7

4. Eliminate to find x: 5x − y = 11, x − y = −9

5. Eliminate to find x: 3x + 4y = 23, x + 4y = 13

6. Eliminate to find x: 2x + 3y = 7, 4x + 3y = 9

7. Eliminate to find x: x + 5y = 12, x + 3y = 8

8. Eliminate to find x: 2x − y = 4, x − y = 1

9. Eliminate to find x: 3x + 2y = 10, 3x − 2y = 10

10. Eliminate to find x: 4x + y = 25, 2x + y = 13

11. Eliminate to find y: x + y = 5, x − y = 3

12. Eliminate to find y: 2x + y = 8, 2x − y = 4

13. Eliminate to find y: 3x + 2y = 13, 3x − y = 7

14. Eliminate to find y: 4x + 3y = 9, 4x + y = 5

15. Eliminate to find y: x + 2y = 11, x − 2y = 3

16. Eliminate to find y: 5x + y = 17, 5x − y = 13

17. Eliminate to find y: 2x + 5y = 4, 2x − 5y = 34

18. Eliminate to find y: 3x + y = 10, x + y = 4

19. Eliminate to find y: 4x + 3y = 23, 4x − y = 7

20. Eliminate to find y: 2x + y = 9, 2x − 2y = 3

Challenge — 8 Questions

1. Substitute to find x: y = x − 1, x + y = 5

2. Substitute to find x: y = 2x, x + y = 3

3. Substitute to find x: y = x + 2, 2x + y = 10

4. Substitute to find x: y = 3x − 1, x + y = 7

5. Substitute to find x: y = 4 − x, 3x − y = 8

6. Substitute to find x: y = 2x + 3, 3x − y = 12

7. Substitute to find x: x = y + 2, 2x + 3y = 6

8. Substitute to find x: y = x − 1, 4x + y = 14