Simultaneous Equations Elimination Method Substitution method Graphical Method Matrix Method

What are they? Simply 2 equations With 2 unknowns Usually x and y To SOLVE the equations means we find values of x and y that Satisfy BOTH equations [work in] At same time [simultaneously]

We have the same number of y’s in each Elimination Method We have the same number of y’s in each 2x – y = 1 A + If we ADD the equations, the y’s disappear B 3x + y = 9 5x = 10 Divide both sides by 5 x = 2 2 x 2 – y = 1 Substitute x = 2 in equation A 4 – y = 1 Answer x = 2, y = 3 y = 3

We have the same number of y’s in each Elimination Method We have the same number of y’s in each 5x + y = 17 A - B 3x + y = 11 If we SUBTRACT the equations, the y’s disappear 2x = 6 Divide both sides by 2 x = 3 5 x 3 + y = 17 Substitute x = 3 in equation A 15 + y = 17 Answer x = 3, y = 2 y = 2

We have the same number of x’s in each Elimination Method We have the same number of x’s in each 2x + 3y = 9 A - B 2x + y = 7 If we SUBTRACT the equations, the x’s disappear 2y = 2 Divide both sides by 2 y = 1 2x + 3 = 9 Substitute y = 1 in equation A 2x = 6 Answer x = 3, y = 1 x = 3

We have the same number of y’s in each Elimination Method We have the same number of y’s in each 4x - 3y = 14 A + B 2x + 3y = 16 If we ADD the equations, the y’s disappear 6x = 30 Divide both sides by 6 x = 5 20 – 3y = 14 Substitute x = 5 in equation A 3y = 6 Answer x = 5, y = 2 y = 2

Basic steps Look at equations Same number of x’s or y’s? If the sign is different, ADD the equations otherwise subtract tem Then have ONE equation Solve this Substitute answer to get the other CHECK by substitution of BOTH answers

What if NOT same number of x’s or y’s? 3x + y = 10 If we multiply A by 2 we get 2y in each B 5x + 2y = 17 A - 6x + 2y = 20 B 5x + 2y = 17 x = 3 In B 5 x 3 + 2y = 17 Answer x = 3, y = 1 15 + 2y = 17 y = 1

+ A 4x - 2y = 8 B 3x + 6y = 21 A 12x - 6y = 24 B 3x + 6y = 21 15x = 45 What if NOT same number of x’s or y’s? A 4x - 2y = 8 If we multiply A by 3 we get 6y in each B 3x + 6y = 21 A 12x - 6y = 24 + B 3x + 6y = 21 15x = 45 x = 3 In B 3 x 3 + 6y = 21 Answer x = 3, y = 2 6y = 12 y = 2

- A 3x + 7y = 26 B 5x + 2y = 24 A 15x + 35y = 130 B 15x + 6y = 72 29y …if multiplying 1 equation doesn’t help? A 3x + 7y = 26 B Multiply A by 5 & B by 3, we get 15x in each 5x + 2y = 24 A 15x + 35y = 130 - B 15x + 6y = 72 Could multiply A by 2 & B by 7 to get 14y in each 29y = 58 y = 2 In B 5x + 2 x 2 = 24 Answer x = 4, y = 2 5x = 20 x = 4

+ A 3x - 2y = 7 B 5x + 3y = 37 A 9x – 6y = 21 B 10x + 6y = 74 19x = 95 …if multiplying 1 equation doesn’t help? A 3x - 2y = 7 B Multiply A by 3 & B by 2, we get +6y & -6y 5x + 3y = 37 A 9x – 6y = 21 + B 10x + 6y = 74 Could multiply A by 5 & B by 3 to get 15x in each 19x = 95 x = 5 In B 5 x 5 + 3y = 37 Answer x = 5, y = 4 3y = 12 y = 4

Substitution Method Given the following equations : y = x + 3 (i) y = 2x (ii) Replace the y in equation (i) with 2x from equation (ii) 2x = x + 3 2x – x = 3 x = 3 Sub. x = 3 into either of the two original equations to find the value of y y = x + 3 (i) y = 3 + 3 y = 6 The answer is (3, 6)

Substitution Method Let : y = $ in hiring tool x = no. of days hiring tool y = 20x -----(1) y = 40 + 10x -----(2) Sub.(1) into (2) 20x = 40 + 10x 10x = 40 x = 4, y = 80 A tool hire firm offers two ways in which a tool may be hired: Plan A - $20 a day Plan B - A payment of $40 then $10 a day Find the number of days whereby there is no difference in the cost of hiring the tool from Plan A and Plan B.

Graphical Method 1. x + y = 6 2x + y = 8 x + y = 6 Let x = 0 y = 6 Coordinates (0, 6) Let y = 0 x = 6 Coordinates (6, 0) 2x + y = 8 Let x = 0 y = 8 Coordinates (0, 8) Let y = 0 2x = 8 x = 4 Coordinates (4, 0)

Graphical Method y = x + 3 y = 2x y = x + 3 Let x = 0 y = 3 Coordinates (0, 3) Let y = 0 x = -3 Coordinates (-3, 0) y = 2x Let x = 0 y = 0 Coordinates (0, 0) Let y = 4 2x = 4 x = 2 Coordinates (2, 0)