1. Additional Mathematics Simultaneous Equations
Chapter 2 (page 30)

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
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 (If needed)
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
Subs 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. Simultaneous linear and quadratic equations
When one of the equations in a pair of simultaneous equations is
quadratic, we often end up with two pairs of solutions. For example,
y = x2 + 1 1
y = x + 3 2
Substituting equation into equation ,
x2 + 1 = x + 3
We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax2 + bx + c = 0.
Point out that we would get the same quadratic equation if we subtracted equations 1 and 2 to eliminate y.
If the quadratic equation had not factorized then we would have had to solve it by completing the square or by using the quadratic formula.
x2 – x – 2 = 0
factorize:
(x + 1)(x – 2) = 0
x = – or x = 2

13. Simultaneous linear and quadratic equations
We can substitute these values of x into one of the equations
y = x2 + 1 1
y = x + 3 2
to find the corresponding values of y.
It is easiest to substitute into equation 2 because it is linear.
When x = –1 we have,
When x = 2 we have,
You can check that these solutions also satisfy equation 1 by substituting them into the equation.
y = –1 + 3
y = 2 + 3
y = 2
y = 5
The solutions are x = –1, y = 2 and x = 2, y = 5.

14. Using graphs to solve equations
We can also show the solutions to
using a graph.
y = x2 + 1
y = x + 3 –1 –2 –3 –4 1 2 3 4 6 8 10
y = x2 + 1
y = x + 3
(2, 5)
The points where the two graphs intersect give the solution to the pair of simultaneous equations.
(–1,2)
Explain that graphs are a good way to demonstrate a solution but are not used to solve linear and quadratic simultaneous equations exactly. For this we should use an algebraic method.
It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs.

15. Simultaneous linear and quadratic equations
Look at this pair of simultaneous equations:
y = x + 1 1
x2 + y2 = 13 2
What shape is the graph given by x2 + y2 = 13?
The graph of x2 + y2 = 13 is a circular graph with its centre at the origin and a radius of √13.
We can solve this pair of simultaneous equations algebraically using substitution.
We can also plot the graphs of the equations and observe where they intersect.

16. Simultaneous linear and quadratic equations
y = x + 1 1
x2 + y2 = 13 2
Substituting equation 1 into equation 2 ,
x2 + (x + 1)2 = 13
x2 + x2 + 2x + 1 = 13
expand the bracket:
simplify:
2x2 + 2x + 1 = 13
2x2 + 2x – 12 = 0
subtract 13 from both sides:
Discuss how we can quickly expand (x + 1)2. Use A1.2 Multiplying out brackets slides 35 and 36 if needed.
x2 + x – 6 = 0
divide through by 2:
factorize:
(x + 3)(x – 2) = 0
x = – or x = 2

17. Simultaneous linear and quadratic equations
We can substitute these values of x into one of the equations
y = x + 1 1
x2 + y2 = 13 2
to find the corresponding values of y.
It is easiest to substitute into equation 1 because it is linear.
When x = –3 we have,
When x = 2 we have,
Verify verbally that these solutions also satisfy equation 2 by substituting them into the equation.
y = –3 + 1
y = 2 + 1
y = –2
y = 3
The solutions are x = –3, y = –2 and x = 2, y = 3.

18. Using graphs to solve equations