ACJC 2025 Equations & Inequalities Lecture Notes

1 3 EQUATIONS AND INEQUALITIES SYLLABUS ▪ Formulate an equation, a system of linear equations, or inequalities from a problem situation ▪ Solve an equation exactly or approximately using a graphing calculator or a graphing software ▪ Solve a system of linear equations using a graphing calculator or a graphing software ▪ Solve inequalities of the form f ( ) 0g( ) x x  where f( )x and g( )x are linear expressions or quadratic expressions ▪ Concept of x , and use of relations x a b a b x a b−   −   + and or x a b x a b x a b−    −  + ▪ Solve inequalities by graphical methods

ACJC 2025/26 H2 Mathematics (9758) 2 CONTENTS 1 Equations .......................................................................................... 3 1.1 Exact or approximate solutions of equations .......................... 3 1.2 Solutions of systems of linear equations using a graphing calculator ................................................................................. 6 1.3 Formulation of equations or a system of linear equations ...... 7 2 Inequalities ..................................................................................... 11 2.1 Some basic rules and results for manipulating inequalities .. 11 2.2 Polynomial inequalities ......................................................... 12 2.3 Inequalities involving rational functions ............................... 18 2.4 Inequalities involving the modulus function ......................... 23 Annex A: Practice Questions on Equations and Inequalities ................... 29

3 Equations and Inequalities 3 LECTURE 1 Lesson Outline • Solving an equation exactly or approximately using a graphing calculator • Solving a system of linear equations using a graphing calculator • Formulation and solution of equations or a system of linear equations from problem situations 1 EQUATIONS 1.1 Exact or Approximate Solutions of Equations Example 1 Solve the following equations exactly: (a) 322 9 10 3 0x x x− + − = (b) ( )cot 2 tan 1 0 2 π  = +   (c) 12e 3 exx +=− Solution (a) 322 9 10 3 0x x x− + − = ( ) 2( 1) 2 7 3 0 ( 1)(2 1)( 3) 0 11 or or 32 x x x x x x x  − − + =  − − − = = (b) cot 2 tan 1=+ 2 2 1 2 tan 1tan 1 2 tan tan 2 tan tan 1 0 (2 tan 1)(tan 1) 0     =+ =+ + − = − + = 11 1tan or tan 12 11tan , π tan ,22 3π 7πor 44   −− = =− =+ (c) 12e 3 exx +=− 12e e 3 e (2 e) 3 3e 2e 3ln 2e xx x x x ++= += = + =  + ■ The solutions should show the factorisation because exact solutions are required

ACJC 2025/26 H2 Mathematics (9758) 4 Example 2 Solve the following equations: (a) 3 200 4000 0xx+ − = (b) 22e x x= (c) ( ) 7ln 1 0xx− + = Solution (a) Here we shall use an APP to solve the polynomial equation 3 200 4000 0xx+ − = . 1) Press [APPS] and select [4:PlySmlt2]. Press [ENTER] to display the main menu. 2) Select [1: POLYNOMIAL ROOT FINDER]. 3) Select the order of the polynomial, in this example, the order is 3, as it is a cubic equation. 4) Leave the other options as default. Press [NEXT]. 5) Key in the coefficients of the equation accordingly, and press [SOLVE]. The only real root is 11.8 (to 3 s.f.).

3 Equations and Inequalities 5 (b) For the non-polynomial equation 22e x x= , we will use graph to solve it. 1) Press [Y=], type in the equation. Press [GRAPH] to display. Press [2ND][TRACE]. Select [2:zero]. 2) Move the cursor to define the “Left” and “Right” bounds. Do the same to find all x-intercepts. From the GC, 0.816x=− or 1.43x= or 8.61x= . (c) For ( ) 7ln 1 0xx− + = , we shall also use graph to solve this equation. 1) Press [Y=], type in the equation. Press [GRAPH] to display. Press [2ND][TRACE]. Select [2:zero]. 2) Move the cursor to define the “Left” and “Right” bounds. Do the same to find all x-intercepts. From the GC, 0.931 or 0 or 1.11 or 21.5x x x x=− = = = . ■ We are interested in the zeroes, or the x- intercepts, as they are the solutions of the equation.

ACJC 2025/26 H2 Mathematics (9758) 6 1.2 Solutions of S ystems of Linear Equations Using G raphing Calculator Consider the following: Case 1 Case 2 Case 3 28 2 3 4 xy xy += −= 28 4 2 16 xy xy += += 28 27 xy xy += += Solving, 74 4 1, 2y y x=  = = A unique solution exists. Simplifying, 4 2 16 28 xy xy +=  + = Since the 2 lines coincide, infinitely many solutions exist. Since the 2 lines are parallel, no solutions exist. Note There are 3 possibilities for the solution of the above system of linear equations. • There is only one unique solution (when the two lines intersect at one point). • There are infinitely many solutions (when the two lines coincide). • There are no solutions (when the two lines are parallel). Graphically, the solution may be obtained by drawing the graphs of the equations and finding the point(s) of intersection. In general, any system of linear equations will have one of the above three types of solutions. y x y x y x

3 Equations and Inequalities 7 Example 3 Solve the system of linear equations 29 2 4 3 1 3 6 5 0. x y z x y z x y z + + = + − = + − = Solution 1) Press [APPS] and select [4:PlySmlt2]. Press [ENTER] to display the main menu. 2) Select [2: SIMULTANEOUS EQN SOLVER]. 3) Adjust the settings accordingly as depicted on the screen. Press [NEXT]. 4) Enter the values of the coefficients accordingly. Press [SOLVE]. The solution is 1, 2, 3x y z= = = . ■ 1.3 Formulation of equations or a system of linear equations Example 4 The diagram shows the graph of a polynomial of degree n. Explain why n cannot be an even number. Given that 3n= , find the equation of the graph if it passes through the points ( )1, 3−− and ( )3,13 , and has a minimum point at ( )1,1 . Solution From the given diagram, it is observed that as ,xy→ → , and as ,xy→− →− . Hence n cannot be an even number. y x (–1, –3) (1, 1) (3, 13)

ACJC 2025/26 H2 Mathematics (9758) 8 Let 32y ax bx cx d= + + + . At ( )1, 3−− : 3 (1)a b c d− + − + =− −−− At ( )1,1 : 1 (2)a b c d+ + + = −−− At ( )3,13 : 27 9 3 13 (3)a b c d+ + + = −−− 2d 32d y ax bx cx = + + . At ( )1,1 : 3 2 0 (4)a b c+ + = −−− By GC, 1, 2, 1, 1a b c d= =− = = .  Equation of the graph is 32 21y x x x= − + + . ■ Example 5 Mr. Spongebob went to the supermarket on 3 separate occasions to buy crabs, lobsters and bamboo clams. He observed that while the price of crabs and bamboo clams remained constant, the price of lobsters consistently increased by 10% compared to the immediate previous v isit. The amount of crabs, lobsters and bamboo clams that he bought by weight for each visit as well as the total amount spent is shown in the table below. 1st visit 2nd visit 3rd visit Crab (kg) 3.20 5.60 4.50 Lobster (kg) 1.50 1.20 2 Bamboo Clam (kg) 7 6.50 6.50 Total amount paid in $ 277.50 347.00 395.18 What is the price per kilogram for crab, lobster and bamboo clam during Mr. Spongebob’s third visit to the supermarket? [VJC 2009 Prelim] Solution Let the prices per kg for crab, lobster and bamboo clam be c, l and b respectively on the first visit. Therefore 2 3.2 1.5 7 277.5 (1) 5.6 1.2(1.1 ) 6.5 347 (2) 4.5 2(1.1 ) 6.5 395.18 (3) c l b c l b c l b + + = −−− + + = −−− + + = −−− By GC, 236.198, 79.998 ( 1.1 96.798), 5.953c l l b= =  = = .  The prices per kg for crab, lobster and bamboo clam during the 3rd visit are $36.20, $96.80 and $5.95 respectively. ■

3 Equations and Inequalities 9 Example 6 The diagram below shows part of an arch bridge. The arch takes the shape of a circular arc. When the water level is at AB, the portion of the arch that is above the water level has a span of 30 m and a rise of 10 m. By taking coordinate axes Ox and Oy as shown, and the highest point of the arch to be directly ab