Additional Mathematics Simultaneous Equations Chapter 2 (page 30)

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 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 (If needed) 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 Subs 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

Simultaneous linear and quadratic equations When one of the equations in a pair of simultaneous equations is quadratic, we often end up with two pairs of solutions. For example, y = x2 + 1 1 y = x + 3 2 Substituting equation 1 into equation 2 , x2 + 1 = x + 3 We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax2 + bx + c = 0. Point out that we would get the same quadratic equation if we subtracted equations 1 and 2 to eliminate y. If the quadratic equation had not factorized then we would have had to solve it by completing the square or by using the quadratic formula. x2 – x – 2 = 0 factorize: (x + 1)(x – 2) = 0 x = –1 or x = 2

Simultaneous linear and quadratic equations We can substitute these values of x into one of the equations y = x2 + 1 1 y = x + 3 2 to find the corresponding values of y. It is easiest to substitute into equation 2 because it is linear. When x = –1 we have, When x = 2 we have, You can check that these solutions also satisfy equation 1 by substituting them into the equation. y = –1 + 3 y = 2 + 3 y = 2 y = 5 The solutions are x = –1, y = 2 and x = 2, y = 5.

Using graphs to solve equations We can also show the solutions to using a graph. y = x2 + 1 y = x + 3 –1 –2 –3 –4 1 2 3 4 6 8 10 y = x2 + 1 y = x + 3 (2, 5) The points where the two graphs intersect give the solution to the pair of simultaneous equations. (–1,2) Explain that graphs are a good way to demonstrate a solution but are not used to solve linear and quadratic simultaneous equations exactly. For this we should use an algebraic method. It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs.

Simultaneous linear and quadratic equations Look at this pair of simultaneous equations: y = x + 1 1 x2 + y2 = 13 2 What shape is the graph given by x2 + y2 = 13? The graph of x2 + y2 = 13 is a circular graph with its centre at the origin and a radius of √13. We can solve this pair of simultaneous equations algebraically using substitution. We can also plot the graphs of the equations and observe where they intersect.

Simultaneous linear and quadratic equations y = x + 1 1 x2 + y2 = 13 2 Substituting equation 1 into equation 2 , x2 + (x + 1)2 = 13 x2 + x2 + 2x + 1 = 13 expand the bracket: simplify: 2x2 + 2x + 1 = 13 2x2 + 2x – 12 = 0 subtract 13 from both sides: Discuss how we can quickly expand (x + 1)2. Use A1.2 Multiplying out brackets slides 35 and 36 if needed. x2 + x – 6 = 0 divide through by 2: factorize: (x + 3)(x – 2) = 0 x = –3 or x = 2

Simultaneous linear and quadratic equations We can substitute these values of x into one of the equations y = x + 1 1 x2 + y2 = 13 2 to find the corresponding values of y. It is easiest to substitute into equation 1 because it is linear. When x = –3 we have, When x = 2 we have, Verify verbally that these solutions also satisfy equation 2 by substituting them into the equation. y = –3 + 1 y = 2 + 1 y = –2 y = 3 The solutions are x = –3, y = –2 and x = 2, y = 3.

Using graphs to solve equations