5) 6) 7) 8) More help: Forming equations Example 8: Find three consecutive numbers so that their sum is 96. Solution: Let the first number be n, then the second is n + 1 and the third is n + 2. Therefore n + (n + 1) + (n + 2) = 96 3n + 3 = 96 3n = 93 n = 31 So the numbers are 31, 32 and 33. Exercise D: 1) Find 3 consecutive even numbers so that their sum is 108. 2) The perimeter of a rectangle is 79 cm. One side is three times the length of the other. Form an equation and hence find the length of each side. 3) Two girls have 72 photographs of celebrities between them. One gives 11 to the other and finds that she now has half the number her friend has. Form an equation, letting n be the number of photographs one girl had at the beginning. Hence find how many each has now. Chapter 3: SIMULTANEOUS EQUATIONS An example of a pair of simultaneous equations is 3x + 2y = 8 5x + y = 11 In these equations, x and y stand for two numbers. We can solve these equations in order to find the values of x and y by eliminating one of the letters from the equations.
In these equations it is simplest to eliminate y. We do this by making the coefficients of y the same in both equations. This can be achieved by multiplying equation by 2, so that both equations contain 2y: 3x + 2y = 8 10x + 2y = 22 2× = To eliminate the y terms, we subtract equation from equation . We get: 7x = 14 i.e. x = 2 To find y, we substitute x = 2 into one of the original equations. For example if we put it into : 10 + y = 11 y = 1 Therefore the solution is x = 2, y = 1. Remember: You can check your solutions by substituting both x and y into the original equations. Example: Solve 2x + 5y = 16 3x – 4y = 1 Solution: We begin by getting the same number of x or y appearing in both equation. We can get 20y in both equations if we multiply the top equation by 4 and the bottom equation by 5: 8x + 20y = 64 15x – 20y = 5 As the signs in front of 20y are different, we can eliminate the y terms from the equations by ADDING: 23x = 69 + i.e. x = 3 Substituting this into equation gives: 6 + 5y = 16 5y = 10 So… y = 2 The solution is x = 3, y = 2. If you need more help on solving simultaneous equations, you can download a booklet from the following website: Exercise: Solve the pairs of simultaneous equations in the following questions: 1) x + 2y = 7 2) x + 3y = 0 3x + 2y = 9 3x + 2y = -7
Example 2: Factorise 6x2 + x – 12. Solution: We need to find two numbers that multiply to make 6 × -12 = -72 and add to make 1. These two numbers are -8 and 9. Therefore, 6x2 + x – 12 = 6x2 - 8x + 9x – 12 = 2x(3x – 4) + 3(3x – 4) (the two brackets must be identical) = (3x – 4)(2x + 3) Difference of two squares: Factorising quadratics of the form Remember that = (x + a)(x – a). Therefore: Also notice that: and Factorising by pairing We can factorise expressions like using the method of factorising by pairing: = x(2x + y) – 1(2x + y) (factorise front and back pairs, ensuring both brackets are identical) = (2x + y)(x – 1) If you need more help with factorising, you can download a booklet from this website: Exercise B Factorise 1) 2) 3) 4) (factorise by taking out a common factor) 5) 6)
Make sure that the equation is rearranged so that the right hand side is 0. It usually makes it easier if the coefficient of x2 is positive. Example 1 : Solve x2 –3x + 2 = 0 Factorise (x –1)(x – 2) = 0 Either (x – 1) = 0 or (x – 2) = 0 So the solutions are x = 1 or x = 2 Note: The individual values x = 1 and x = 2 are called the roots of the equation. Example 2: Solve x2 – 2x = 0 Factorise: x(x – 2) = 0 Either x = 0 or (x – 2) = 0 So x = 0 or x = 2 Method 2: Using the formula Recall that the roots of the quadratic equation are given by the formula: Example 3: Solve the equation Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a = 2, b = 3 and c = -12. Substituting these into the quadratic formula gives: (this is the surd form for the solutions) If we have a calculator, we can evaluate these roots to get: x = 1.81 or x = -3.31 If you need more help with the work in this chapter, there is an information booklet downloadable from this web site: EXERCISE 1) Use factorisation to solve the following equations: a) x2 + 3x + 2 = 0 b) x2 – 3x – 4 = 0 c) x2 = 15 – 2x 2) Find the roots of the following equations: a) x2 + 3x = 0 b) x2 – 4x = 0