18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Solving Sim. Equations Graphically
Solving Simple Sim. Equations by Substitution
Simultaneous Simultaneous EquationsEquations
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Solving Simple Sim. Equations by elimination
Solving harder type Sim. equations
S3 Credit
Short cut method using graphs
Choosing the Best Method
Using Sim. Equations to find formulae.
Using Sim. Equations to solve problems
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2. T
11. True or false a car reduced in value by 33 % in one year.
3 I f its initial price was £12 000. The new price is £3 000.
idy up the expression
- 6e + 5d - 7d +6e
3. Calculate a) 13
+ 4x
y b) 12x
- 5y
S3 Credit
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1. To solve simultaneous equations using graphical methods.
Simultaneous Equations
1.1. Interpret information from a Interpret information from a line graph.line graph.
2.2. Plot line equations on a Plot line equations on a graph.graph.
3.3. Find the coordinates were 2 Find the coordinates were 2 lines intersect ( meet)lines intersect ( meet)
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2 1 0 1 2 3 4
4
3
2
1
1
2
3
4
(1,3)
Q. Write down the coordinates where they meet.
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2 1.5 1 0.5 0 0.5 1 1.5 2
2
1.5
1
0.5
0.5
1
1.5
2
(-0.5,-0.5)
Q. Write down the coordinates where they meet.
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2 1.5 1 0.5 0 0.5 1 1.5 2
2
1.5
1
0.5
0.5
1
1.5
2
Q. Plot the lines.
y x
2 1y x Blue :
(1,1)
Q. Write down the coordinates where they meet.
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Credit Straight Lines
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Per day 0 1 2 3 4 5
Total Cost £ 0 60 120
Swinton Direct Car Hire
Per Day 0 1 2 3 4 5
Total Cost £ £100 120 140
Anrnold Palmer Car Hire
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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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths DepartmentDays
Tota
l C
ost
£
Arnold
Swinton
Summarise data !
Who should I hire the car from?
Up to 2 days Swinton
Over 2 days Arnold
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Key steps1. 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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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 2.1 & 2.2
Ch13 (page 253 )
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11. A house has increased in value by 12 % in one year.
2 I f its initial price was £96 000. How much is it now.
2. Find the length of the hypotenuse.
3. Write down the formula for the
area of the sector of a circle.
S3 Credit
6cm
8cm
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1. To use a quicker method (two points) for solving graphical methods.
Simultaneous Equations
1.1. Draw line graphs using two Draw line graphs using two points.points.
2.2. Find the coordinates where Find the coordinates where 2 lines intersect ( meet)2 lines intersect ( meet)
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww
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.com 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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Example : Solve graphically x - 2y = 4 and x + 2y = -2 First find x = 0 and y = 0 for line x – 2y = 4
x = 0 0 – 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 = 0 0 + 2y = -2 y = -1(0,-1)
y = 0 x + 2 x 0 = -2 x = -2 (-2,0)
0 1 2 3 4 5 6 7 8 9 10-1 x1
2
3
4
5
6
7
8
9
10
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-9 -8 -7 -6 -5 -4 -3 -2-10
( 0, -2)
(4, 0)
x + 2y = -2
x - 2y = 4 ( 0, -1)
(-2, 0)
Solution(1, -1.5)
Check !x – 2y1 – 2x(-1.5)= 1 + 3= 4
Check !x + 2y1 + 2x(-1.5)= 1 - 3 = -2
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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
2. Plot the two coordinates for each line and join them up. Extend each line if necessary so they cross over.
3. Read off solution where lines meet and check that it satisfies both equations.
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 3.1Ch13 (page 256 )
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2. Are the f ollowing equations the same . Why?
- 1 o1. True or false the angle for sin (0.8) = 53.1
(a) y + 2x = 8 and y = - 2x + 8
(b) 4x + 2y =10 and y = - 2x + 5
3. Calcu
- 9y
late a) - 5x b) - 19y
+ 5x
S3 Credit
18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww
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1. To solve pairs of equations by substitution.
Simultaneous Equations
1.1. Apply the process of Apply the process of substitution to solve simple substitution to solve simple simultaneous equations.simultaneous equations.
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Example 1
Solve the equations
y = 2x y = x+1
by substitution
3 2 1 0 1 2 3
3
2
1
1
2
3
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At the point of intersection y coordinates are equal:
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 = 1 + 1 = 2
Substitute y = 2x in equation
2
y = 2x y = x+1
The solution is x = 1 y = 2 or (1,2)
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Example 1
Solve the equations
y = x + 1 x + y = 4
by substitution
0 0.5 1 1.5 2 2.5 3 3.5 4
0.5
1
1.5
2
2.5
3
3.5
4
(1.5, 2.5)
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At the point of intersection y coordinates are equal:
x + 1 = -x + 4
Rearranging we get : 2x = 4 - 12x = 3
Finally : Sub into one of the equations to get y value
y = x +1 = 1.5 + 1 = 2.5
y = -x+4 = -1.5 + 4 = 2 .5
y = x +1y =-x+ 4
The solution is x = 1.5 y = 2.5 (1.5,2.5)
x = 3 ÷ 2 = 1.5
OR
Substitute y = x + 1 in equation 2
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 4.1 & 4.2
Ch13 (page 257 )
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2. Are the equations below the same. Why?
1. A house appreciates by 20% in one year.
I f its initial price was £60 000. How much is it now.
(a) y + x = 2 and y = - x - 2
(b)
+ 9w
x + y =5 and y = x - 5
3. Calculate a) 5y b) 12w
- 2y
S3 Credit
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1. To solve simultaneous equations of 2 variables by elimination.
Simultaneous Equations
1.1. Understand the term Understand the term simultaneous equation.simultaneous equation.
2.2. Understand the process Understand the process for solving simultaneous for solving simultaneous equation of two variables equation of two variables by elimination method.by elimination method.
3.3. Solve simple equationsSolve simple equations
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Example 1
Solve the equations
x + 2y = 14 x + y = 9
by elimination
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Step 1: Label the equationsx + 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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Step 3: Sub into one of the equations to get other variable
Substitute y = 5 in (B)
x + y = 9 (B)
x + 5 = 9
The solution is x = 4 y = 5
Step 4: Check answers by substituting into both equations
x = 9 - 5
x = 4
x + 2y = 14x + y = 9
( 4 + 10 = 14)( 4 + 5 = 9)
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Example 2
Solve the equations
2x - y = 11 x - y = 4
by elimination
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Step 1: Label the equations2x - 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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Step 3: Sub into one of the equations to get other variable
Substitute x = 7 in (B)
x - y = 4 (B)
7 - y = 4
The solution is x =7 y =3
Step 4: Check answers by substituting into both equations
y = 7 - 4
y = 3
2x - y = 11 x - y = 4
( 14 - 3 = 11)( 7 - 3 = 4)
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Example 3
Solve the equations
2x - y = 6 x + y = 9
by elimination
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Step 1: Label the equations2x - 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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Step 3: Sub into one of the equations to get other variable
Substitute x = 5 in (B)
x + y = 9 (B)
5 + y = 9
The solution is x = 5 y = 4Step 4:
Check answers by substituting into both equations
y = 9 - 5
y = 4
2x - y = 6 x + y = 9
( 10 - 4 = 6)( 5 + 4 = 9)
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 5.1 & 5.2
Ch13 (page 260 )
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2. Tidy up the expressions
1. A fridge is increased by 4%.
I f its old price was £240.
Explain why the new price is £249.60
(a) 2y - x - 3y +2x
(b) 4x - 5y + 6x +8y
3. Explain how to solve the equations : x + y = 6
x - y = 4
S3 Credit
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1. To solve harder simultaneous equations of 2 variables.
Simultaneous Equations
1.1. Apply the process for solving Apply the process for solving simultaneous equations to simultaneous equations to harder examples.harder examples.
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Example 1
Solve the equations
2x + y = 9 x - 3y = 1
by elimination
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2x + y = 9 x -3y = 1
Step 1: Label the equations2x + y = 9 (A)
x -3y = 1 (B)
Step 2: Decide what you want to eliminate
Eliminate y by :
7x =28
6x + 3y = 27 (C) x - 3y = 1 (D)
x = 28 ÷ 7 = 4
Adding
(A) x3(B) x1
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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
The solution is x = 4 y = 1
Step 4:
Check answers by substituting into both equations
y = 1
2x + y = 9 x -3y = 1
( 8 + 1 = 9)( 4 - 3 = 1)
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Example 2
Solve the equations
3x + 2y = 132x + y = 8
by elimination
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3x + 2y = 13 2x + y = 8
Step 1: Label the equations3x + 2y = 13 (A)
2x + y = 8 (B)
Step 2: Decide what you want to eliminate
Eliminate y by :
-x = -3
3x + 2y = 13 (C) 4x + 2y = 16 (D) x = 3
Subtract
(A) x1(B) x2
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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
The solution is x = 3 y = 2Step 4:
Check answers by substituting into both equations
y = 2
3x + 2y = 132x + y = 8
( 9 + 4 = 13)( 6 + 2 = 8)
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 6.1 & 6.2
Ch13 (page 262 )
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2
2 8 32 2
5 15 4
( - 2)( 3 4)2. Multiply out
o o
1. True or false
3. Find angle x and y for the
right- angled triangle below.
y y y
S3 Credit
B
A
C
xo
yo
108
6
18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww
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1. Investigate the best method of solving simultaneous equations for a given problem.
Simultaneous Equations
1.1. Apply the most appropriate Apply the most appropriate method for solving method for solving simultaneous equations for a simultaneous equations for a given problem.given problem.
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Simultaneous EquationsStraight Lines
S3 Credit
In this chapter you have solved Simultaneous Equationsby 3 methods.
Can you name them !!!Graphica
l
Substitution
Elimination
Order of difficulty
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Simultaneous EquationsStraight Lines
S3 Credit
We commonly use either substitution or elimination
Substitution Elimination
Try solving these simultaneous equations by both methodsand then decide which was easier.
y = x + 7 and
3x + 4y = 14
4x + 3y + 8 = 0 and
3x - 5y = 23y – 5x = 0
and x + y = -6
2y – 3x = 5and
2x + 3y = 3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww
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y = or x =
SUBSTITUTION is easier !
Otherwise
use ELIMINATION
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 7.1Ch13 (page 264 )
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2( - 2)( 2) 4
2 73 8
2. True or f alse
1. 1
3. Why is the triangle right angled at B.
x x x
S3 Credit
B
A
C
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1. Use simultaneous equations to find formulae.
Simultaneous Equations
1.1. Apply the process for solving Apply the process for solving simultaneous equations to find simultaneous equations to find formulae.formulae.
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We can use simultaneous equations to find formulae of the form
c = an + b
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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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Solve the equations
6a + b = 54 (A)4a + b = 38 (B)
by substitutingb = 38 - 4a into (A) we get
6a + 38 - 4a = 54
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6a + 38 - 4a = 54
2a = 16
a = 8
Substituting : a = 8 into equation (A) we get
6 x 8 + b = 54
b = 6
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Formula is : c = 8n + 6
Do a check !
Substituting : a = 8 b = 6 into equation (B)
4 x 8 + 6 = 38
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Now try Ex 8.1 Ch13 (page 264 )
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Starter QuestionsStarter Questionsw
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2 4 73 2 6
3 5 15
2. Tidy up the expressions
1. True or false
(a) 2y - x - 3y +2x
(b) 4x - 5y + 6x +8y
3. Explain how to solve the equations : x + y = 6
x - y = 4
S3 Credit
18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww
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1. Use simultaneous equations to solve real life problems.
Simultaneous Equations
1.1. Apply the process for solving Apply the process for solving simultaneous equations to simultaneous equations to solve real life problems.solve real life problems.
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated 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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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Example 1
Solve the equations
3x + 6y = 10.8 6x + 4y = 12
by elimination
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
6x + 12y = 21.6 6x + 4y = 12
Step 1: Label the equations3x + 6y = 10.8 (A)
6x + 4y = 12 (B)
Step 2: Decide what you want to eliminate
Eliminate x by :
8y = 9.6
6x + 12y =21.6 (C) 6x + 4y = 12 (D)
y = 9.6 ÷ 8 = 1.2
Subtracting
(A) x2(B) x1
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated 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 = 10.8 - 7.2
The solution is x = 1.2
y = 1.2
Step 4:
Check answers by substituting into both equations
3x = 3.6
3x + 6y = 10.8
6x + 4y = 12
( 3.6 + 7.2 = 10.8 )( 7.2 + 4.8 = 12 )
x = 1.2
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated 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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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Example 1
Solve the equations
4x + 6y = 97.60 7x + 4y = 126.60
by elimination
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
16x + 24y = 390.4 42x + 24y = 759.6
Step 1: Label the equations4x + 6y = 97.6 (A)
7x + 4y = 126.6 (B)
Step 2: Decide what you want to eliminate
Eliminate x by :
-26x =-369.2
16x + 24y =390.4 (C) 42x + 24y =759.6 (D)
x = (-369.2) ÷ (-26) = £14.20
Subtracting
(A) x4(B) x6
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Simultaneous EquationsS3
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18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Department
Step 3: Sub into one of the equations to get other variable
Substitute y = 14.20 in equation (A)
4 x 14.20 + 6y = 97.60
6y = 97.60 – 56.80
The solution is x = adult price = £14.20 y = child
price = £6.80
Step 4:
Check answers by substituting into both equations
6y = 40.80
4x + 6y = 97.607x + 4y = 126.60
( 56.80 + 40.80 = £97.60 )( 99.40 + 27.20 = £ 126.60 )
y = £6.80