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

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

(-0.5,-0.5) 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 1 0.5 0.5 1 1.5 2 (-0.5,-0.5) 0.5 1 1.5 2

(1,1) Q. a) Plot the lines: y = x y = 2x - 1 b) Write down the coordinates where they meet. 2 1.5 1 (1,1) 0.5 2 1.5 1 0.5 0.5 1 1.5 2 0.5 1 1.5 2

Simultaneous Equations Elimination

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

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

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

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

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

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

Q. I need to hire a car for a number of days. 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

Who should I hire the car from? Summarise data ! Who should I hire the car from? Arnold Total Cost £ Up to 2 days Swinton Over 2 days Arnold Swinton Days

5 pens and 3 rubbers cost £0·99 while 1 pen and 2 rubbers cost £0·31 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 5p + 3r = 99 p + 2r = 31 p = 15, r = 8

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 3p + 2c = 161 2p + 3c = 154 p = 35, c = 28

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 4r + 5w = 35∙5 r + w = 8 r = £4.50, w = £3.50

Tickets on sale for a concert are £25 each for the front section and £20 each for the back section of the 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 25f + 20b = 4980 f = 148, b = 64

Meals in a restaurant are available at £28 per person for the fish courses menu and £30 per 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 28f + 30m = 2760 f = 45, m = 50

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 9x + 5y = 5240 x = 360, y = 400