1. Simultaneous Equations
S3 Credit
Solving Sim. Equations Graphically
Short cut method using graphs
Solving Simple Sim. Equations by Substitution
Solving Simple Sim. Equations by elimination
Solving harder type Sim. equations
Choosing the Best Method
Using Sim. Equations to find formulae.
Using Sim. Equations to solve problems
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2. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
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3. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
To solve simultaneous equations using graphical methods.
Interpret information from a line graph.
Plot line equations on a graph.
3. Find the coordinates were 2 lines intersect ( meet)
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4. Created by Mr. Lafferty Maths DepartmentQ.
Write down the
coordinates where
they meet.
(1,3)
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5. Created by Mr. Lafferty Maths Department
Q. Write down the
coordinates
where they meet.
(-0.5,-0.5)
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6. Created by Mr. Lafferty Maths Department
Q. Plot the lines.
(1,1)
Q. Write down the
coordinates
where they meet.
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7. Created by Mr. Lafferty Maths Department
We can use straight line theory to work out real-life problems
especially useful when trying to work out hire charges.
Q. I need to hire a car for a number of days.
Below are the hire charges charges for two companies.
Complete tables and plot values on the same graph.
160
180
200
180
240
300
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8. Who should I hire the car from?
Summarise data !
Who should I hire the car from?
Total Cost £
Arnold
Up to 2 days Swinton
Over 2 days Arnold
Swinton
Days
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9. Created by Mr. Lafferty Maths Department
Key steps
1. Make or Fill in x – y table
2. Plot points on the same graph ( pick scale carefully)
3. Identify intersection point ( where 2 lines meet)
4. Interpret graph information.
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10. Created by Mr. Lafferty Maths Department
Now try Ex 2.1 & 2.2
Ch13 (page 253 )
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11. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
8cm
6cm
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12. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
To use a quicker method (two points) for solving graphical methods.
Draw line graphs using two points.
Find the coordinates where
2 lines intersect ( meet)
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13. Created by Mr. Lafferty Maths Department
There is a quick way of sketching a straight line.
We need only find two points and then draw a line through them.
Normally the easier points to find are
x = 0 and y = 0
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14. Created by Mr. Lafferty Maths Department
Example : Solve graphically x - 2y = 4 and x + 2y = -2
First find x = 0 and y = 0 for line x – 2y = 4
x = – 2y = 4 y = -2 (0,-2)
y = 0 x – 2 x 0 = 4 x = 4 (4,0)
Next find x = 0 and y = 0 for line x + 2y = -2
x = y = -2 y = -1 (0,-1)
y = 0 x + 2 x 0 = -2 x = -2 (-2,0)
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15. x - 2y = 4 x + 2y = -2 Solution (1, -1.5) Check ! x – 2y 1 – 2x(-1.5)
( 0, -2)
(4, 0)
x + 2y = -2
( 0, -1)
(-2, 0) 1 2 3 4 5 6 7 8 9 10 -1 x -2 -3 -4 -5 -6 -7 -8 -9
-10
Solution
(1, -1.5)
Check !
x – 2y
1 – 2x(-1.5)
= 1 + 3
= 4
Check !
x + 2y
1 + 2x(-1.5)
= 1 - 3
= -2

16. Key points for quick method for graphical solution
1. Find two points that lie on each of the two lines.
Normally easy to find x = 0 and y =0 coordinates for both lines
Plot the two coordinates for each line and join them up.
Extend each line if necessary so they cross over.
Read off solution where lines meet and check that it satisfies both equations.
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17. Created by Mr. Lafferty Maths Department
Now try Ex 3.1
Ch13 (page 256 )
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18. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
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19. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
To solve pairs of equations by substitution.
1. Apply the process of substitution to solve simple simultaneous equations.
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20. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
y = 2x
y = x+1
by substitution
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21. Created by Mr. Lafferty Maths Department
At the point of intersection y coordinates are equal:
Substitute
y = 2x in equation 2
y = 2x
y = x+1
2x = x+1
Rearranging we get :
2x - x = 1
x = 1
Finally : Sub into one of the equations to get y value
y = 2x = 2 x 1 = 2 OR
y = x+1 = = 2
The solution is x = 1 y = 2 or (1,2)
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22. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
y = x + 1
x + y = 4
by substitution
(1.5, 2.5)
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23. Created by Mr. Lafferty Maths Department
The solution is x = 1.5 y = 2.5 (1.5,2.5)
At the point of intersection y coordinates are equal:
Substitute
y = x + 1 in equation 2
x + 1 = -x + 4
y = x +1
y =-x+ 4
2x = 4 - 1
Rearranging we get :
2x = 3
x = 3 ÷ 2 = 1.5
Finally : Sub into one of the equations to get y value
y = x +1 = = 2.5 OR
y = -x+4 = = 2 .5
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24. Created by Mr. Lafferty Maths Department
Now try Ex 4.1 & 4.2
Ch13 (page 257 )
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25. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
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26. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
To solve simultaneous equations of 2 variables by elimination.
Understand the term simultaneous equation.
Understand the process for solving simultaneous equation of two variables by elimination method.
3. Solve simple equations
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27. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
x + 2y = 14
x + y = 9
by elimination
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28. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
x + 2y = 14 (A)
x + y = 9 (B)
Step 2: Decide what you want to eliminate
Eliminate x by subtracting (B) from (A)
x + 2y = 14 (A)
x + y = 9 (B)
y = 5
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29. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute y = 5 in (B)
x + y = 9 (B)
x + 5 = 9
x = 9 - 5
The solution is x = 4
y = 5
x = 4
Step 4: Check answers by substituting into both equations
x + 2y = 14
x + y = 9
( = 14)
( = 9)
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30. Created by Mr. Lafferty Maths Department
Example 2
Solve the equations
2x - y = 11
x - y = 4
by elimination
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31. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
2x - y = 11 (A)
x - y = 4 (B)
Step 2: Decide what you want to eliminate
Eliminate y by subtracting (B) from (A)
2x - y = 11 (A)
x - y = 4 (B)
x = 7
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32. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute x = 7 in (B)
x - y = 4 (B)
7 - y = 4
y = 7 - 4
The solution is x =7
y =3
y = 3
Step 4: Check answers by substituting into both equations
2x - y = 11
x - y = 4
( = 11)
( = 4)
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33. Created by Mr. Lafferty Maths Department
Example 3
Solve the equations
2x - y = 6
x + y = 9
by elimination
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34. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
2x - y = 6 (A)
x + y = 9 (B)
Step 2: Decide what you want to eliminate
Eliminate y by adding (A) from (B)
2x - y = 6 (A)
x + y = 9 (B)
3x = 15
x = 15 ÷ 3 = 5
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35. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute x = 5 in (B)
x + y = 9 (B)
5 + y = 9
y = 9 - 5
The solution is x = 5
y = 4
y = 4
Step 4:
Check answers by substituting into both equations
2x - y = 6
x + y = 9
( = 6)
( = 9)
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36. Created by Mr. Lafferty Maths Department
Now try Ex 5.1 & 5.2
Ch13 (page 260 )
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37. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
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38. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
To solve harder simultaneous equations of 2 variables.
1. Apply the process for solving simultaneous equations to harder examples.
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39. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
2x + y = 9
x - 3y = 1
by elimination
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40. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
2x + y = 9 (A)
x -3y = 1 (B)
Step 2: Decide what you want to eliminate
Adding
Eliminate y by :
2x + y = 9
x -3y = 1
(A) x3
6x + 3y = 27 (C)
x - 3y = 1 (D)
(B) x1
7x =28
x = 28 ÷ 7 = 4
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41. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute x = 4 in equation (A)
2 x 4 + y = 9
y = 9 – 8
y = 1
The solution is x = 4
y = 1
Step 4:
Check answers by substituting into both equations
2x + y = 9
x -3y = 1
( = 9)
( = 1)
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42. Created by Mr. Lafferty Maths Department
Example 2
Solve the equations
3x + 2y = 13
2x + y = 8
by elimination
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43. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
3x + 2y = 13 (A)
2x + y = 8 (B)
Step 2: Decide what you want to eliminate
Subtract
Eliminate y by :
3x + 2y = 13
2x + y = 8
(A) x1
3x + 2y = 13 (C)
4x + 2y = 16 (D)
(B) x2
-x = -3
x = 3
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44. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute x = 3 in equation (B)
2 x 3 + y = 8
y = 8 – 6
y = 2
The solution is x = 3
y = 2
Step 4:
Check answers by substituting into both equations
3x + 2y = 13
2x + y = 8
( = 13)
( = 8)
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45. Created by Mr. Lafferty Maths Department
Now try Ex 6.1 & 6.2
Ch13 (page 262 )
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46. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit A xo 10 8 yo B C 6
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47. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
Investigate the best method of solving simultaneous equations for a given problem.
1. Apply the most appropriate method for solving simultaneous equations for a given problem.
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48. Simultaneous Equations
S3 Credit
Straight Lines
In this chapter you have solved Simultaneous Equations
by 3 methods.
Can you name them !!!
Graphical
Substitution
Order of difficulty
Elimination
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49. Simultaneous Equations
S3 Credit
Straight Lines
We commonly use either substitution or elimination
Substitution
Elimination
Try solving these simultaneous equations by both methods
and then decide which was easier.
y = x + 7
and
3x + 4y = 14
2y – 3x = 5
and
2x + 3y = 3
4x + 3y + 8 = 0
and
3x - 5y = 23
y – 5x = 0
and
x + y = -6
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50. y = or x = www.mathsrevision.com
If we can arrange one of the equations into
y = or x =
SUBSTITUTION is easier !
Otherwise
use ELIMINATION
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51. Created by Mr. Lafferty Maths Department
Now try Ex 7.1
Ch13 (page 264 )
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52. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit A B C
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53. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
Use simultaneous equations to find formulae.
1. Apply the process for solving simultaneous equations to find formulae.
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54. c = an + b We can use simultaneous equations
to find formulae of the form
c = an + b
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55. Created by Mr. Lafferty Maths Department
Example : The cost of hiring a bike is related to
the number of days hire ( n days ) by the formula
c = an + b
Stuart hires a bike for 6 days cost is £54.
John paid £38 for 4 days hire.
Find the values for a and b.
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56. Created by Mr. Lafferty Maths Department
Solve the equations
6a + b = 54 (A)
4a + b = 38 (B)
by substituting
b = a into (A) we get
6a a = 54
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57. Created by Mr. Lafferty Maths Department
6a a = 54
2a = 16
a = 8
Substituting : a = 8 into equation (A) we get
6 x 8 + b = 54
b = 6
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58. Created by Mr. Lafferty Maths Department
Do a check !
Substituting : a = 8 b = 6 into equation (B)
4 x = 38
Formula is : c = 8n + 6
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59. Created by Mr. Lafferty Maths Department
Now try Ex 8.1
Ch13 (page 264 )
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60. Created by Mr. Lafferty Maths Department
Starter Questions
S3 Credit
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61. Created by Mr. Lafferty Maths Department
Simultaneous Equations
S3 Credit
Straight Lines
Learning Intention
Success Criteria
Use simultaneous equations to solve real life problems.
1. Apply the process for solving simultaneous equations to solve real life problems.
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62. Created by Mr. Lafferty Maths Department
A jeweller uses two different arrangements of beads and pearls
The first arrangement consists of 3 beads and 6 pearls.
It has overall length of 10.8 cm.
The second arrangement consists of 6 beads and 4 pearls.
It has overall length of 12 cm.
Find the length of one bead and the length of one pearl.
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63. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
3x + 6y = 10.8
6x + 4y = 12
by elimination
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64. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
3x + 6y = 10.8 (A)
6x + 4y = 12 (B)
Subtracting
Step 2: Decide what you want to eliminate
Eliminate x by :
6x + 12y = 21.6
6x + 4y = 12
(A) x2
6x + 12y = 21.6 (C)
6x + 4y = 12 (D)
(B) x1
8y = 9.6
y = 9.6 ÷ 8 = 1.2
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65. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute y = 1.2 in equation (A)
3x + 6 x 1.2 = 10.8
3x =
3x = 3.6
The solution is
x = 1.2
y = 1.2
x = 1.2
Step 4:
Check answers by substituting into both equations
3x + 6y = 10.8
6x + 4y = 12
( = 10.8 )
( = 12 )
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66. Created by Mr. Lafferty Maths Department
One evening 4 adults and 6 children visited the sports centre.
The total collected in entrance fees was £97.60
The next evening 7 adults and 4 children visited the sports centre.
The total collected in entrance fees was £126.60
Calculate the cost of an adult price and a child price.
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67. Created by Mr. Lafferty Maths Department
Example 1
Solve the equations
4x + 6y =
7x + 4y =
by elimination
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68. Created by Mr. Lafferty Maths Department
Step 1: Label the equations
4x + 6y = (A)
7x + 4y = (B)
Subtracting
Step 2: Decide what you want to eliminate
Eliminate x by :
16x + 24y = 390.4
42x + 24y = 759.6
(A) x4
16x + 24y = (C)
42x + 24y = (D)
(B) x6
-26x =-369.2
x = (-369.2) ÷ (-26) = £14.20
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69. Created by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute y = in equation (A)
4 x y = 97.60
6y = – 56.80
6y = 40.80
The solution is x = adult price = £14.20
y = child price = £6.80
y = £6.80
Step 4:
Check answers by substituting into both equations
4x + 6y = 97.60
7x + 4y =
( = £97.60 )
( = £ )
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70. Created by Mr. Lafferty Maths Department
Now try Ex 9.1 & 9.2
Ch13 (page 266 )
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