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
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 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 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
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 + 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 = – 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 = 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 = 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 + 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 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 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 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 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· 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 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 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 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
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