1. System of Linear Equations Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T1 Solutions Systems of Linear Equations 1 Let y = ax3 + bx2 + cx + d, where a, b, c and d are real constants. When x = 0, y = 0,  d = 0 Given x = 2, y = 0, a(23) + b(22) + c(2) = 0 8a + 4b + 2c = 0 ------ (1) Given x = 2.55, y = – 0.0631, a(2.553) + b(2.552) + c(2.55) = – 0.0631 ------ (2) Differentiating y with respect to x, 2d 32d y ax bx cx    Given x = 0.785, d 0d y x  , 3(0.7852)a + 2(0.785)b + c = 0 ------ (3) Solving (1), (2) and (3) using GC, a = 0.0993, b = – 0.497, c = 0.596 (3 s.f.) Hence, y = 0.1x3 – 0.5x2 + 0.6x. 2 Let $x, $y and $z be the cost of a ticket for a senior citizen, adult and child respectively. 2 19 9 1982 10 3 908 7 4 778 x y z yz x y z        Using GC, 36 74 56 x y z    Thus, the cost of a ticket for a senior citizen is $36, for an adult is $74 and for a child is $56. 4(36) + 5(74) + 1(56) = 570 Therefore, the total cost for Group D = $570

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T2 3 420 6.4 17.5 25 100 102.4 420 3.8 18.5 19 120 92.6 420 8.6 17 23 90 121.2 a b c a b c abc                or 17.5 25 100 324 (1) 18.5 19 120 331.2 (2) 17 23 90 307.4 (3) a b c a b c abc             Using GC, 58 13 14,,5 5 25a b c   4 Let S, B and W be the number of strawberry, blueberry and walnut muffins purchased respectively. 30 1.6 1.75 2.2 53.40 2 2 0 S B W S B W S W S W           From GC, 12, 12, 6S B W   5 Let x, y, z be the price rate of electricity, gas and water respectively. 23 12 16 103 33 16 21 142 49 22 33 209 x y z xyz x y z        Solving 2.2x , 2.5y , 1.4z Her monthly utility bill for the month of July is        35 2.2 1.07 17 2.5 0.95 2 159.1656 1. 14 159. 7   6 Let the number of diagonals be 2d An Bn C   No of diagonals in a triangle = 0 No of diagonals in a quadrilateral = 2 Triangle (3 sides) Quadrilateral (4 sides) Pentagon (5 sides)

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T3 No of diagonals in a pentagon = 5 Therefore, 9 3 0 16 4 2 25 5 5 A B C A B C A B C          Solving using GC, 13, , 022A B C   Thus, 213 22d n n For a polygon of 200 sides, The number of diagonals = 213(200) (200) 1970022  7(i) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 800 7 21 39 20300 5 23 49 23900 m s t m s t m s t          Using GC, 200, 250, 350.m s t   Therefore the fisherman has 250 kg of salmon. (ii) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 600 7 21 39 20300 5 23 49 23900 m s t m s t m s t          Using GC, 460, 990, 70.m s t   Since the weight of all fishes must be non-negative, the fisherman’s claim is not possible. Or Since the weight of salmon and tuna is more than 600kg, the fisherman’s claim is not possible.

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T4 8 9 Let x, y, z be the number of apples, oranges and pineapples respectively. 50x y z   0.8 0.6 1.2 40x y z   1.6 0.3 1.2 53x y z   From GC, Adam bought 23 apples, 18 oranges and 9 pineapples. 10 Let the price of a X-box console be $x, a Kinect sensor be $y and a Game DVD be $z. 499 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30 x y z x y z x y z          Aug matrix = 1 1 1 499 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30     rref = 1 0 0 247 0 1 0 199 0 0 1 53    x = 247 , y = 199 , z = 53 Employees of Company P will pay $[0.9(247)+0.8(199)+0.85(53)] = $426.55 > $426.30. No, it will not be more attractive for employees to purchase all the 3 items from own company. 11 Let x, y and z be the digits on the hundreds, tens and ones position respectively.   8 (1) (100 10 ) 100 10 297 (2) x y z x y z z y x           

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T5  8 3 x y z xz     Using GC: 3 , 5 2 ,x z y z z z     Since x, y and z are non-negative integer values, 3 0 3 5 2 0 2.5 0 zz zz z          0 2.5z   When 0, 3, 5z x y   When 1, 4, 3z x y   When 2, 5, 1z x y   Alternative: using GC Hence, the possible original numbers are 350, 431 or 512. 12   223' 2  bxaxxf   32 2f x ax bx x c     Curve passes through (1,2) 0 (1)abc     Curve passes through (-1,3) 5 (2)abc     Curve passes through (2,2) 8 4 2 (3)a b c     5 31,26ab  , 8 3c   325 31 8 22 6 3f x x x x    

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T6 13 Let , and be the usual retail price of a packet of cashew nuts, macadamia nuts and almonds respectively. 4 6 7 57.05 4(0.7) 4 7 10 33.05 2.8 4 7 43.05 4(0.8) 3 7( 0.15) 33.05 5.45 3.2 3 x y z x y z x y z x y z x y z x y                   7 39.55 By G.C., 3.5, 4.9, 1.95 z xyz     Thus, the usual retail price of a packet of cashew nuts, macadamia nuts and almonds is $3.50, $4.90 and $1.95 respectively. 14 At the points of intersection, 2 ln( )a x bx cx When x =1, ln 0a b c   ------------ (1) When x =2, 2 4 ln ln 2a b c   ------------ (2) When x =3, 3 9 ln ln 3a b c   ------------ (3) Using GC, 1.124670a     2 1.124670 1.265a (to 3 d.p.) 0.144b (to 3 d.p.) ln 0.98083c  0.98083e 2.667c (to 3 d.p.) 15 21x y z   ------ (1) 30xz ------- (2) 0.05 0.8 3(0.9 )y z x 2.7 0.05 0.8 0x y z    ----- (3) Using GC, 2.1, 12.6 and 6.3x y z   16 Sub (1,1) and (2,2) into h( )yx . 1 ----- (1) 8 4 2 2 -----(2) a b c d a b c d         Since (2,2) is also the stationary point, h '(2) 0 . i.e. 12 4 0 ----- (3)a b c   Using the GC, 11 24 35 24 2 ad bd cd    

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T7 0 1 1 3 5 2 4 2 4 02 ab c dd d            6{ : 2 or 0} 5d d d     CJC Prelim 9758/2018/01/Q2 17. Marking Scheme:   32f lnx x a bx cx d      2f ' 3 ln 2x x a xb c   At 5 320, 3 27   , 32 5 5 5 320ln3 3 3 27 125ln 75 45 27 320 (1) a b c d a b c d                    2 553 ln 2 033 25ln 10 3 0 (2) a b c a b c                            32 Let g = f 1 1 ln 1 1x x x a b x c x d              2 g ' 3 1 ln 2 1x x a x b c     At  0, 12 , ln 12 (3)a b c d     3ln 2 0 (4)a b c    Using GC and solve, ln 1, 4, 5, 10a b c d    e, 4, 5, 10a b c d     18         42f f 2 0 16 4 2 0 4 2 16 ...(1) f 3 12 81 9 3 12 9 3 69 ...(2) f 4 26 256 16 4 26 16 4 230...(3) x x ax bx c a b c a b c a b c a b c a b c a b c                                   Solving (1), (2) and (3), 54, 217, 234a b c   2 6 5 0    

River Valley High School, Mathematics Department 2023 Systems of Linear Equations T8 NYJC Prelim 9758/2022/01/Q1 Q19 Suggested Answers (i) 32f ( )x ax bx cx d    At (1, 1), 1a b c d     -----------------------------(1) 2f '( ) 3 2x ax bx c   At minimum point (1, 1), 3 2 0a b c   --------------