13-Apr-15Created by Mr. Lafferty Maths Department Solving Sim. Equations Graphically Solving Simple Sim. Equations by Substitution Simultaneous Equations

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


Starter QuestionsStarter Questionsw

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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

18 Apr 202318 Apr 2023 Created by Mr. Lafferty Maths DepartmentCreated by Mr. Lafferty Maths Departmentww

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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)

Straight LinesS3

Credit

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Simultaneous EquationsS3

Credit Straight Lines


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)


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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)


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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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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.


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)

Straight LinesS3

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w.m

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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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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


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1. To solve pairs of equations by substitution.


1.1. Apply the process of Apply the process of substitution to solve simple substitution to solve simple simultaneous equations.simultaneous equations.

Straight LinesS3

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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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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.


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

Straight LinesS3

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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)


Eliminate y by subtracting (B) from (A)2x - y = 11 (A) x - y = 4 (B)

x = 7

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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)


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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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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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.


1.1. Apply the process for solving Apply the process for solving simultaneous equations to simultaneous equations to harder examples.harder examples.

Straight LinesS3

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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)


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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Substitute x = 4 in equation (A)

2 x 4 + y = 9

y = 9 – 8

The solution is x = 4 y = 1

Step 4:


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)


Eliminate y by :

-x = -3

3x + 2y = 13 (C) 4x + 2y = 16 (D) x = 3

Subtract

(A) x1(B) x2

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Substitute x = 3 in equation (B)

2 x 3 + y = 8

y = 8 – 6

The solution is x = 3 y = 2Step 4:


y = 2

3x + 2y = 132x + y = 8

( 9 + 4 = 13)( 6 + 2 = 8)

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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


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1. Investigate the best method of solving simultaneous equations for a given problem.


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.

Straight LinesS3

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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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w.m

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y = or x =

SUBSTITUTION is easier !

Otherwise

use ELIMINATION

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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.


1.1. Apply the process for solving Apply the process for solving simultaneous equations to find simultaneous equations to find formulae.formulae.

Straight LinesS3

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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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Now try Ex 8.1 Ch13 (page 264 )

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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


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1. Use simultaneous equations to solve real life problems.


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.

Straight LinesS3

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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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Example 1

Solve the equations

3x + 6y = 10.8 6x + 4y = 12

by elimination

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6x + 12y = 21.6 6x + 4y = 12

Step 1: Label the equations3x + 6y = 10.8 (A)

6x + 4y = 12 (B)


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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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:


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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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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Example 1

Solve the equations

4x + 6y = 97.60 7x + 4y = 126.60

by elimination

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16x + 24y = 390.4 42x + 24y = 759.6

Step 1: Label the equations4x + 6y = 97.6 (A)

7x + 4y = 126.6 (B)


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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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:


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

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Ch13 (page 266 )

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Documents

13-Apr-15Created by Mr. Lafferty Maths Department Solving Sim. Equations Graphically Solving Simple Sim. Equations by Substitution Simultaneous Equations