1. Simultaneous Equations
Elimination Method
Substitution method
Graphical Method
Matrix Method

2. What are they? Simply 2 equations
With 2 unknowns
Usually x and y
To SOLVE the equations means we find values of x and y that
Satisfy BOTH equations [work in]
At same time [simultaneously]

3. We have the same number of y’s in each
Elimination Method
We have the same number of y’s in each
2x – y = 1 A +
If we ADD the equations, the y’s disappear B
3x + y = 9 5x
= 10
Divide both sides by 5 x
= 2
2 x 2 – y = 1
Substitute x = 2 in equation A
4 – y = 1
Answer
x = 2, y = 3
y = 3

4. We have the same number of y’s in each
Elimination Method
We have the same number of y’s in each
5x + y = 17 A - B
3x + y = 11
If we SUBTRACT the equations,
the y’s disappear 2x
= 6
Divide both sides by 2 x
= 3
5 x 3 + y = 17
Substitute x = 3 in equation A
15 + y = 17
Answer
x = 3, y = 2
y = 2

5. We have the same number of x’s in each
Elimination Method
We have the same number of x’s in each
2x + 3y = 9 A - B
2x + y = 7
If we SUBTRACT the equations,
the x’s disappear 2y
= 2
Divide both sides by 2 y
= 1
2x + 3 = 9
Substitute y = 1 in equation A
2x = 6
Answer
x = 3, y = 1
x = 3

6. We have the same number of y’s in each
Elimination Method
We have the same number of y’s in each
4x - 3y = 14 A + B
2x + 3y = 16
If we ADD the equations,
the y’s disappear 6x
= 30
Divide both sides by 6 x
= 5
20 – 3y = 14
Substitute x = 5 in equation A
3y = 6
Answer
x = 5, y = 2
y = 2

7. Basic steps Look at equations Same number of x’s or y’s?
If the sign is different, ADD the equations otherwise subtract tem
Then have ONE equation
Solve this
Substitute answer to get the other
CHECK by substitution of BOTH answers

8. What if NOT same number of x’s or y’s?
3x + y = 10
If we multiply A by 2 we
get 2y in each B
5x + 2y = 17 A -
6x + 2y = 20 B
5x + 2y = 17 x
= 3
In B
5 x 3 + 2y = 17
Answer
x = 3, y = 1
15 + 2y = 17
y = 1

9. + A 4x - 2y = 8 B 3x + 6y = 21 A 12x - 6y = 24 B 3x + 6y = 21 15x = 45
What if NOT same number of x’s or y’s? A
4x - 2y = 8
If we multiply A by 3 we
get 6y in each B
3x + 6y = 21 A
12x - 6y = 24 + B
3x + 6y = 21
15x
= 45
x = 3
In B
3 x 3 + 6y = 21
Answer
x = 3, y = 2
6y = 12
y = 2

10. - A 3x + 7y = 26 B 5x + 2y = 24 A 15x + 35y = 130 B 15x + 6y = 72 29y
…if multiplying 1 equation doesn’t help? A
3x + 7y = 26 B
Multiply A by 5 & B by 3,
we get 15x in each
5x + 2y = 24 A
15x + 35y = 130 - B
15x + 6y = 72
Could multiply A by 2 & B
by 7 to get 14y in each
29y
= 58
y = 2
In B
5x + 2 x 2 = 24
Answer
x = 4, y = 2
5x = 20
x = 4

11. + A 3x - 2y = 7 B 5x + 3y = 37 A 9x – 6y = 21 B 10x + 6y = 74 19x = 95
…if multiplying 1 equation doesn’t help? A
3x - 2y = 7 B
Multiply A by 3 & B by 2,
we get +6y & -6y
5x + 3y = 37 A
9x – 6y = 21 + B
10x + 6y = 74
Could multiply A by 5 & B
by 3 to get 15x in each
19x
= 95
x = 5
In B
5 x 5 + 3y = 37
Answer
x = 5, y = 4
3y = 12
y = 4

12. Substitution Method Given the following equations :
y = x + 3 (i)
y = 2x (ii)
Replace the y in equation (i) with 2x from equation (ii)
2x = x + 3
2x – x = 3
x = 3
Sub. x = 3 into either of the two original equations to find the value of y
y = x + 3 (i)
y = 3 + 3
y = 6
The answer is
(3, 6)

13. Substitution Method Let :
y = $ in hiring tool
x = no. of days hiring tool
y = 20x -----(1)
y = x -----(2)
Sub.(1) into (2)
20x = x
10x = 40
x = 4, y = 80
A tool hire firm offers two ways in which a tool may be hired:
Plan A - $20 a day
Plan B - A payment of $40 then $10 a day
Find the number of days whereby there is no difference in the cost of hiring the tool from Plan A and Plan B.

14. Graphical Method 1. x + y = 6 2x + y = 8 x + y = 6 Let x = 0 y = 6
Coordinates (0, 6)
Let y = 0
x = 6
Coordinates (6, 0)
2x + y = 8
Let x = 0
y = 8
Coordinates (0, 8)
Let y = 0
2x = 8
x = 4
Coordinates (4, 0)

15. Graphical Method y = x + 3 y = 2x y = x + 3 Let x = 0 y = 3
Coordinates (0, 3)
Let y = 0
x = -3
Coordinates (-3, 0)
y = 2x
Let x = 0
y = 0
Coordinates (0, 0)
Let y = 4
2x = 4
x = 2
Coordinates (2, 0)