Solving Simultaneous When you solve simultaneous equations you

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Solving Simultaneous ►When you solve simultaneous equations you are finding where two lines intersect

Solving Simultaneous ►When you solve simultaneous equations you are finding where two lines intersect

Q. a) Find the equation of each line. b) Write down the coordinates of

Q. a) Find the equation of each line. b) Write down the coordinates of the point of intersection. 4 y=x+2 3 y = 2 x + 1 (1, 3) 2 1 0 1 2 3 4

Q. a) Find the equation of each line. b) Write down the coordinates where

Q. a) Find the equation of each line. b) Write down the coordinates where they meet. y=x 2 y=-x-1 1. 5 1 0. 5 2 1. 5 (-0. 5, -0. 5) 1 0. 5 0 0. 5 1 1. 5 2

Q. a) Plot the lines: y=x y = 2 x - 1 b) Write

Q. a) Plot the lines: y=x y = 2 x - 1 b) Write down the coordinates where they meet. 2 1. 5 1 (1, 1) 0. 5 2 1. 5 1 0. 5 0 0. 5 1 1. 5 2

Simultaneous Equations ► Elimination

Simultaneous Equations ► Elimination

Solving Simultaneous Equations: Elimination 3 x + 4 y = 26 (1) 7 x

Solving Simultaneous Equations: Elimination 3 x + 4 y = 26 (1) 7 x – y = 9 (2) “Solve simultaneously” Make either coefficient of x or y ‘same size’ Put x = 2 into equation (1) (3) = (2) x 4 28 x - 4 y = 36 (1) 3 x + 4 y = 26 (1)+(3) 31 x = 62 x= 2 (2, 5) 3 x + 4 y = 26 3 2 + 4 y = 26 6 + 4 y = 26 4 y = 20 y=5

Solving Simultaneous Equations: Elimination 5 x + 3 y = 11 (1) 2 x

Solving Simultaneous Equations: Elimination 5 x + 3 y = 11 (1) 2 x + y = 4 (2) Put x = 1 into equation (1) (3) = (2) x 3 6 x + 3 y = 12 (1) – (3) 5 x + 3 y = 11 x 5 x + 3 y = 11 5 1 + 3 y = 11 = 1 5 + 3 y = 11 x= 1 (1, 2) 3 y = 6 y=2

10 x + 3 y = 1 (1) 3 x + 2 y =

10 x + 3 y = 1 (1) 3 x + 2 y = – 3 (2) To eliminate a letter we must multiply both equations by a constant to get equal coefficients (3) = (1) × 2 20 x + 6 y = 2 Put x = 1 into equation (1) (4) = (2) × 3 9 x + 6 y = – 9 (3) – (4) 11 x = 11 Remember takexaway = 1 a negative gives a positive 10 x + 3 y = 1 10 1 + 3 y = 1 (1, – 3 ) 3 y = – 9 y=– 3

3 x + 2 y = 10 (1) 5 x – 3 y =

3 x + 2 y = 10 (1) 5 x – 3 y = 4 (2) To eliminate a letter we must multiply both equations by a constant to get equal coefficients (3) = (1) × 3 9 x + 6 y = 30 (4) = (2) × 2 10 x – 6 y = 8 (3) + (4) 19 x Put x = 2 into equation (1) 3 x + 2 y = 10 3 2 + 2 y = 10 = 38 x= 2 2 y = 4 y=2 (2, 2)

Elimination Set 1 1 3 5 7 9 11 3 x + 2 y

Elimination Set 1 1 3 5 7 9 11 3 x + 2 y = 10 5 x – 3 y = 4 7 x + 2 y = 11 2 x - 3 y = -4 3 x - 4 y = -6 2 x + 5 y = 19 8 x - 3 y = 2 5 x +2 y = 9 3 x - 7 y = 7 2 x + 3 y = -3 10 x - 3 y = -13 4 x +5 y = 1 2 4 6 8 10 12 2 x - 5 y = 7 3 x + 4 y = -1 6 x - 5 y = 12 4 x + 3 y = 8 9 x - 5 y = 14 2 x + 3 y = -1 4 x - 5 y = 18 5 x + 6 y = -2 5 x - 2 y = 11 4 x + 3 y = -5 9 x + 5 y = -1 4 x - 3 y = 10

Elimination (Set 2) 1 5 x + 3 y = 11 2 x +

Elimination (Set 2) 1 5 x + 3 y = 11 2 x + y = 4 3 8 x + 5 y = -2 3 x + 4 y = -5 5 9 x + 4 y = 1 3 x + 2 y = -1 7 10 x + 3 y = 1 3 x +2 y = -3 9 3 x + 7 y = -1 2 x + 3 y = 1 11 10 x + 7 y = 14 3 x +5 y = 10 2 7 x + 2 y = 17 3 x + y = 8 4 4 x + 5 y = 18 x+y=4 6 5 x + 6 y = 12 3 x + 5 y = 10 8 4 x + 7 y = 1 x + 3 y = -1 10 5 x + 2 y = 16 4 x + 5 y = 6 12 6 x + 5 y = 8 x + 3 y = -3

We can use straight line theory to work out real-life problems especially useful when

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 for two companies. Complete tables and plot values on the same graph. 160 180 200 180 240 300

Total Cost £ Summarise data ! Who should I hire the car from? d

Total Cost £ Summarise data ! Who should I hire the car from? d l o n r A S wi on t n Up to 2 days Swinton Over 2 days Arnold Days

5 pens and 3 rubbers cost £ 0· 99 while 1 pen and 2

5 pens and 3 rubbers cost £ 0· 99 while 1 pen and 2 rubbers cost £ 0· 31. Make two equations and solve them to find the cost of each. Let p = pen, r = rubber 5 p + 3 r = 99 p + 2 r = 31 p = 15, r = 8

3 plum trees and 2 cherry trees cost £ 161 while 2 plum trees

3 plum trees and 2 cherry trees cost £ 161 while 2 plum trees and 3 cherry trees cost £ 154. Make two equations and solve them to find the cost of each. Let p = plum tree, c = cherry tree 3 p + 2 c = 161 2 p + 3 c = 154 p = 35, c = 28

4 bottles of red wine and 5 bottles of white wine cost £ 35·

4 bottles of red wine and 5 bottles of white wine cost £ 35· 50 while one bottle of each cost £ 8. Find the cost of each type of wine. Let r = red wine, w = white wine 4 r + 5 w = 35∙ 5 r+w=8 r = £ 4. 50, w = £ 3. 50

Tickets on sale for a concert are £ 25 each for the front section

Tickets on sale for a concert are £ 25 each for the front section and £ 20 each for the back section of theatre. If 212 people attended the concert and the total receipts were £ 4980, how many of each price of ticket were sold ? Let f = front seat, b = back seat f + b = 212 25 f + 20 b = 4980 f = 148, b = 64

Meals in a restaurant are available at £ 28 person for the fish courses

Meals in a restaurant are available at £ 28 person for the fish courses menu and £ 30 person for the meat courses menu. There are 95 guests attending the function in the restaurant and the total bill came to £ 2 760. How many guests chose each menu ? Let f = fish course, m = meat course f + m = 95 28 f + 30 m = 2760 f = 45, m = 50

Calendars cost £ 9 and £ 5 each and are on sale in a

Calendars cost £ 9 and £ 5 each and are on sale in a card shop. If 760 calendars are sold and the total takings for them were £ 5 240, how many of each price of calendar were sold ? Let x = £ 9 calendars, y = £ 5 calendars x + y = 760 9 x + 5 y = 5240 x = 360, y = 400