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

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.To solve simultaneous equations using graphical methods. Simultaneous Equations 1.Interpret information from a line graph. 2.Plot line equations on a graph. 3.Find the coordinates were 2 lines intersect ( meet) Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department (1,3) Q. Write down the coordinates where they meet.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department (-0.5,-0.5) Q. Write down the coordinates where they meet.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Q. Plot the lines. (1,1) Q. Write down the coordinates where they meet.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Days Total Cost £ A r n o l d S w i n t o n Summarise data ! Who should I hire the car from? Up to 2 days Swinton Over 2 days Arnold

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 2.1 & 2.2 Ch13 (page 253 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit 6cm 8cm

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.To use a quicker method (two points) for solving graphical methods. Simultaneous Equations 1.Draw line graphs using two points. 2.Find the coordinates where 2 lines intersect ( meet) Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 = 0andy = 0

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example : Solve graphically x - 2y = 4andx + 2y = -2 First find x = 0 and y = 0 for line x – 2y = 4 x = 00 – 2y = 4y = -2(0,-2) y = 0x – 2 x 0 = 4x = 4(4,0) Next find x = 0 and y = 0 for line x + 2y = -2 x = y = -2y = -1(0,-1) y = 0x + 2 x 0 = -2x = -2(-2,0)

x ( 0, -2) (4, 0) x + 2y = -2x - 2y = 4 ( 0, -1) (-2, 0) Solution (1, -1.5) Check ! x – 2y 1 – 2 x (-1.5) = = 4 Check ! x + 2y x (-1.5) = = -2

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 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.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 3.1 Ch13 (page 256 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.To solve pairs of equations by substitution. Simultaneous Equations 1.Apply the process of substitution to solve simple simultaneous equations. Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations y = 2x y = x+1 by substitution

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 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 = = 2 Substitute y = 2x in equation 2 y = 2x y = x+1 The solution is x = 1 y = 2 or (1,2)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations y = x + 1 x + y = 4 by substitution (1.5, 2.5)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department At the point of intersection y coordinates are equal: x + 1 = -x + 4 Rearranging we get : 2x = x = 3 Finally : Sub into one of the equations to get y value y = x +1 = = 2.5 y = -x+4 = = 2.5 y = x +1 y =-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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 4.1 & 4.2 Ch13 (page 257 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.To solve simultaneous equations of 2 variables by elimination. Simultaneous Equations 1.Understand the term simultaneous equation. 2.Understand the process for solving simultaneous equation of two variables by elimination method. 3.Solve simple equations Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations x + 2y = 14 x + y = 9 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 The solution is x = 4 y = 5 Step 4:Check answers by substituting into both equations x = x = 4 x + 2y = 14 x + y = 9 ( = 14) ( = 9)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 2 Solve the equations 2x - y = 11 x - y = 4 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 The solution is x =7 y =3 Step 4:Check answers by substituting into both equations y = y = 3 2x - y = 11 x - y = 4 ( = 11) ( = 4)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 3 Solve the equations 2x - y = 6 x + y = 9 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 The solution is x = 5 y = 4 Step 4: Check answers by substituting into both equations y = y = 4 2x - y = 6 x + y = 9 ( = 6) ( = 9)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 5.1 & 5.2 Ch13 (page 260 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.To solve harder simultaneous equations of 2 variables. Simultaneous Equations 1.Apply the process for solving simultaneous equations to harder examples. Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations 2x + y = 9 x - 3y = 1 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 2x + y = 9 x -3y = 1 Step 1: Label the equations 2x + 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 The solution is x = 4 y = 1 Step 4: Check answers by substituting into both equations y = 1 2x + y = 9 x -3y = 1 ( = 9) ( = 1)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 2 Solve the equations 3x + 2y = 13 2x + y = 8 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 3x + 2y = 13 2x + y = 8 Step 1: Label the equations 3x + 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 The solution is x = 3 y = 2 Step 4: Check answers by substituting into both equations y = 2 3x + 2y = 13 2x + y = 8 ( = 13) ( = 8)

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 6.1 & 6.2 Ch13 (page 262 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit B A C xoxo yoyo

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.Investigate the best method of solving simultaneous equations for a given problem. Simultaneous Equations 1.Apply the most appropriate method for solving simultaneous equations for a given problem. Straight Lines S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Simultaneous Equations Straight Lines S3 Credit In this chapter you have solved Simultaneous Equations by 3 methods. Can you name them !!! Graphical Substitution Elimination Order of difficulty

13-Apr-15Created by Mr. Lafferty Maths Department Simultaneous Equations Straight Lines S3 Credit 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 4x + 3y + 8 = 0 and 3x - 5y = 23 y – 5x = 0 and x + y = -6 2y – 3x = 5 and 2x + 3y = 3

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department If we can arrange one of the equations into y =orx = SUBSTITUTION is easier ! Otherwise use ELIMINATION

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 7.1 Ch13 (page 264 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit B A C

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.Use simultaneous equations to find formulae. Simultaneous Equations 1.Apply the process for solving simultaneous equations to find formulae. Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department We can use simultaneous equations to find formulae of the form c = an + b

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Formula is :c = 8n + 6 Do a check ! Substituting :a = 8 b = 6 into equation (B) 4 x = 38

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 8.1 Ch13 (page 264 )

13-Apr-15Created by Mr. Lafferty Maths Department Starter Questions Starter Questions S3 Credit

13-Apr-15Created by Mr. Lafferty Maths Department Learning Intention Success Criteria 1.Use simultaneous equations to solve real life problems. Simultaneous Equations 1.Apply the process for solving simultaneous equations to solve real life problems. Straight Lines S3 Credit

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations 3x + 6y = x + 4y = 12 by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 6x + 12y = x + 4y = 12 Step 1: Label the equations 3x + 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

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 = x = The solution is x = 1.2 y = 1.2 Step 4: Check answers by substituting into both equations 3x = 3.6 3x + 6y = x + 4y = 12 ( = 10.8 ) ( = 12 ) x = 1.2

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 £ Calculate the cost of an adult price and a child price.

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Example 1 Solve the equations 4x + 6y = x + 4y = by elimination

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department 16x + 24y = x + 24y = Step 1: Label the equations 4x + 6y = 97.6(A) 7x + 4y = 126.6(B) Step 2: Decide what you want to eliminate Eliminate x by : -26x= x + 24y =390.4 (C) 42x + 24y =759.6 (D) x = (-369.2) ÷ (-26) = £14.20 Subtracting (A) x4 (B) x6

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created 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 = y = – The solution is x = adult price = £14.20 y = child price = £6.80 Step 4: Check answers by substituting into both equations 6y = x + 6y = x + 4y = ( = £97.60 ) ( = £ ) y = £6.80

Simultaneous Equations S3 Credit Straight Lines 13-Apr-15Created by Mr. Lafferty Maths Department Now try Ex 9.1 & 9.2 Ch13 (page 266 )