04[Anal Add Math CD]

1

Additional Mathematics SPM Chapter 4

© Penerbitan Pelangi Sdn. Bhd.

1. (a) y – x – 2 = 0 y = x + 2 .................... 1 x2 – 6y + 17 = 0 .......................... 2

Substitute 1 into 2, x2 – 6(x + 2) + 17 = 0 x2 – 6x – 12 + 17 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x – 1 = 0 or x – 5 = 0 x = 1 or x = 5

Substitute x = 1 into 1, y = 1 + 2 = 3

Substitute x = 5 into 1, y = 5 + 2 = 7 \ The solutions are x = 1, y = 3 and x = 5,

y = 7.

(b) 2y + x = 5 x = 5 – 2y ............................ 1 xy + y = 0 .................................... 2

Substitute 1 into 2, (5 – 2y)y + y = 0 5y – 2y2 + y = 0 2y2 – 6y = 0 2y(y – 3) = 0 2y = 0 or y – 3 = 0 y = 0 or y = 3

Substitute y = 0 into 1, x = 5 – 2(0) = 5

Substitute y = 3 into 1, x = 5 – 2(3) = –1

The solutions are x = 5, y = 0 and x = –1, y = 3.

(c) 3x = 1 – y y = 1 – 3x ........................... 1 5x + x2 = 48 – 3y ......................... 2

Substitute 1 into 2, 5x + x2 = 48 – 3(1 – 3x) = 48 – 3 + 9x x2 – 4x – 45 = 0 (x – 9)(x + 5) = 0 x – 9 = 0 or x + 5 = 0 x = 9 or x = –5

Substitute x = 9 into 1, y = 1 – 3(9) = –26

Substitute x = –5 into 1, y = 1 – 3(–5) = 16

The solutions are x = 9, y = –26 and x = –5, y = 16.

(d) y + 2x – 5 = 0 y = 5 – 2x ..................... 1 xy = 4 – x2 ...................... 2

Substitute 1 into 2, x(5 – 2x) = 4 – x2

5x – 2x2 = 4 – x2

x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0 x – 1 = 0 or x – 4 = 0 x = 1 or x = 4

Substitute x = 1 into 1, y = 5 – 2(1) = 3


The solutions are x = 1, y = 3 and x = 4, y = –3.

(e) y + 3x – 4 = 0 y = 4 – 3x ................ 1 y2 – 2xy + 4x2 = 3 ........................ 2

Substitute 1 into 2, (4 – 3x)2 – 2x(4 – 3x) + 4x2 = 3 16 – 24x + 9x2 – 8x + 6x2 + 4x2 – 3 = 0 19x2 – 32x + 13 = 0 (19x – 13)(x – 1) = 0

CHAPTER

4 Simultaneous Equations

2



19x – 13 = 0 or x – 1 = 0

x = 1319

or x = 1

Substitute x = 13–––19

into 1,

y = 4 – 31 1319 2

= 3719

Substitute x = 1 into 1, y = 4 – 3(1) = 1

The solutions are x = 1, y = 1 and x = 1319

,

y = 3719

.

(f) 3 – y = 2x y = 3 – 2x ........................... 1 x2 + y2 = 5 ................................... 2

Substitute 1 into 2, x2 + (3 – 2x)2 = 5 x2 + 9 – 12x + 4x2 – 5 = 0 5x2 – 12x + 4 = 0 (5x – 2)(x – 2) = 0 5x – 2 = 0 or x – 2 = 0 x =

2—5 or x = 2

Substitute x = 25 into 1,

y = 3 – 21 25 2

= 115


The solutions are x = 25 , y = 11

5 and x = 2,

y = –1.

(g) x—2 = 4 +

y—5

x—2 =

20 + y5

5x = 2(20 + y) 5x = 40 + 2y

y = 5x – 402 ...................... 1

xy + 32 = 4x ................................ 2

Substitute 1 into 2,

x1 5x – 402 2 + 32 = 4x

x(5x – 40) + 64 = 8x 5x2 – 40x + 64 = 8x 5x2 – 48x + 64 = 0

(5x – 8)(x – 8) = 0 5x – 8 = 0 or x – 8 = 0

x = 85 or x = 8


y = 51 8—

5 2 – 40–––––––––

2 = –16


y = 5(8) – 402

= 0

The solutions are x = 85 , y = –16 and x = 8,

y = 0.

(h) 2y + 5 = –3x

y = –3x – 52 ............. 1

xy + 7 + 1—3

x = 0 ......................... 2


x1 –3x – 52 2 + 7 + 1—

3x = 0

–3x2 – 5x2

+ 7 + 1—3

x = 0

× 6, 3(–3x2 – 5x) + 42 + 2x = 0 –9x2 – 15x + 42 + 2x = 0 9x2 + 13x – 42 = 0 (9x – 14)(x + 3) = 0 9x – 14 = 0 or x + 3 = 0 x = 14

9 or x = –3

Substitute x = 149

into 1,

y = –31 14

9 2 – 5 –––––––––––

2

= – 14

3 – 5

2

= – 296

Substitute x = –3 into 1,

y = –3(–3) – 52

= 2

The solutions are x = 149

, y = – 296

and x = –3, y = 2.

3



(i) 3x – 5y = 4

x = 4 + 5y

3 ................................1

6yx + 3x

y = 11

× xy, xy1 6yx 2 + xy1 3x

y 2 = 11xy

6y2 + 3x2 = 11xy .............2


6y2 + 31 4 + 5y3 2

2 = 11y1 4 + 5y

3 2 6y2 + 33 (4 + 5y)2

9 4 = 44y + 55y2

3

6y2 + 16 + 40y + 25y2

3 = 44y + 55y2

3 × 3, 18y2 + 16 + 40y + 25y2 = 44y + 55y2

12y2 + 4y – 16 = 0 ÷ 4, 3y2 + y – 4 = 0 (3y + 4)(y – 1) = 0 3y + 4 = 0 or y – 1 = 0 y = – 4—

3 or y = 1

Substitute y = – 4—3

into 1,

x = 4 + 51– 4—

3 2––––––––––

3

= – 8—9

Substitute y = 1 into 1,

x = 4 + 5(1)–––––––3

= 3

The solutions are x = – 8—9

, y = – 4—3

and x = 3,

y = 1.

(j) 12x = 5 – 12y

x = 5 – 12y–––––––

12 ..................... 1

1—x

= 10 – 1—y

× xy, y = 10xy – x ..................... 2


y = 10y 1 5 – 12y–––––––

12 2 – 1 5 – 12y–––––––

12 2 × 12, 12y = 10y(5 – 12y) – (5 – 12y) 12y = 50y – 120y2 – 5 + 12y 120y2 – 50y + 5 = 0

÷ 5, 24y2 – 10y + 1 = 0 (4y – 1)(6y – 1) = 0 4y – 1 = 0 or 6y – 1 = 0

y = 1—4

or y = 1—6

Substitute y = 1—4

into 1,

x = 5 – 121 1—

4 2––––––––––

12

= 1—6


into 1,

x = 5 – 121 1—

6 2––––––––––

12

= 1—4

\ The solutions are x = 1—6

, y = 1—4

and x = 1—4

,

y = 1—6

.

2. (a) 1 + x – y = 0 y = 1 + x ......................... 1 2xy – x2 = 4 ............................... 2

Substitute 1 into 2, 2x(1 + x) – x2 = 4 2x + 2x2 – x2 = 4 x2 + 2x – 4 = 0

x = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

, where a = 1, b = 2, c = – 4

= –2 ± ABBBBBBBBB22 – 4(1)(– 4)––––––––––––––––––2(1)

= –2 ± ABB20–––––––––2

= 1.236 or –3.236

Substitute x = 1.236 into 1, y = 1 + 1.236 = 2.236

Substitute x = –3.236 into 1, y = 1 – 3.236 = –2.236

The solutions are x = 1.236, y = 2.236 and x = –3.236, y = –2.236.

(b) 2y = 3x – 4

x = 2y + 4––––––

3 .................... 1

y2 = 5xy + 1 .................... 2


y2 = 5y1 2y + 4––––––

3 2 + 1

3y2 = 5y(2y + 4) + 3 3y2 = 10y2 + 20y + 3 7y2 + 20y + 3 = 0

4



y = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

, where a = 7, b = 20, c = 3

= –20 ± ABBBBBBBBB202 – 4(7)(3)–––––––––––––––––––2(7)

= –20 ± ABBB316–––––––––––14

= –0.159 or –2.698

Substitute y = –0.159 into 1,

x = 2(– 0.159) + 4––––––––––––3

= 1.227

Substitute y = –2.698 into 1,

x = 2(–2.698) + 4––––––––––––3

= – 0.465

The solutions are x = 1.227, y = –0.159 and x = – 0.465, y = –2.698.

(c) 2x – y = –3 y = 2x + 3 ............................ 1

1—x = x + y

× x, 1 = x2 + xy ........................... 2

Substitute 1 into 2, 1 = x2 + x(2x + 3) = x2 + 2x2 + 3x 3x2 + 3x – 1 = 0

\ a = 3, b = 3, c = –1

x = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

= –3 ± ABBBBBBBBB32 – 4(3)(–1)––––––––––––––––––

2(3)

= –3 ± ABB21–––––––––6

= 0.264 or –1.264

Substitute x = 0.264 into 1, y = 2(0.264) + 3 = 3.528

Substitute x = –1.264 into 1, y = 2(–1.264) + 3 = 0.472 The solutions are x = 0.264, y = 3.528 and

x = –1.264, y = 0.472.

3. y = x2 + 6 .............................. 1y – 4x – 2 = 0...................................... 2

Substitute 1 into 2,x2 + 6 – 4x – 2 = 0 x2 – 4x + 4 = 0 (x – 2)2 = 0 x = 2

Substitute x = 2 into 1,y = 22 + 6 = 10

\ D(2, 10)Height of gate = CD = 10 m

Width of gate = BC = 2 × 2 = 4 m

4. y = x2 – 6x + 11 ................... 1y + 2x – 7 = 0 ..................................... 2Substitute 1 into 2, x2 – 6x + 11 + 2x – 7 = 0 x2 – 4x + 4 = 0 (x – 2)2 = 0 x = 2The x-coordinate of point B is 2.

5. Area of rectangle = 420 (2x + 20)(y + 10) = 420 2xy + 20x + 20y + 200 = 420 xy + 10x + 10y = 110 ............ 1

Circumference of circle = 2π(28)

= 2 × 227

× 28

= 176 cm

Perimeter of rectangle = 176 2 × (2x + 20 + y + 10) = 176 2x + y + 30 = 88 y = 58 – 2x ...... 2Substitute 2 into 1, x(58 – 2x) + 10x + 10(58 – 2x) = 110 58x – 2x2 + 10x + 580 – 20x = 110 x2 – 24x – 235 = 0

\ a = 1, b = –24, c = –235

x = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

= –(–24) ± ABBBBBBBBBBBBBB(–24)2 – 4(1)(–235) ––– ––––––––––––––––––––––––2(1)

= 24 ± ABBBB1516–––––––––––2

= 12 ± ABBB379 = 12 + ABBB379 or 12 – ABBB379 = 31.47 or –7.47

Substitute x = 31.47 into 2,y = 58 – 2(31.47) = – 4.94

Substitute x = –7.47 into 2,y = 58 – 2(–7.47) = 72.94

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\ Possible length = 2x + 20 = 2(31.47) + 20 or 2(–7.47) + 20 = 82.94 cm or 5.06 cm

\ Possible width = y + 10 = – 4.94 + 10 or 72.94 + 10 = 5.06 cm or 82.94 cm

The possible lengths and widths are 82.94 cm and 5.06 cm or vice versa.

1. 2x + y = 6 y = 6 – 2x ................. 1 y2 – y + x = 13 ....................... 2Substitute 1 into 2, (6 – 2x)2 – (6 – 2x) + x = 1336 – 24x + 4x2 – 6 + 2x + x – 13 = 0 4x2 – 21x + 17 = 0 (4x – 17)(x – 1) = 0 4x – 17 = 0 or x – 1 = 0

x = 174 or x = 1


y = 6 – 21 174 2

= – 5—2Substitute x = 1 into 1,y = 6 – 2(1) = 4The solutions are x = 17

4 , y = – 5—2 and x = 1, y = 4.

2. x – 2y = 0 x = 2y ................1 y2 – xy + x + 3 = 0 ..................2

Substitute 1 into 2: y2 – (2y)(y) + 2y + 3 = 0 y2 – 2y2 + 2y + 3 = 0 –y2 + 2y + 3 = 0 y2 – 2y – 3 = 0 (y + 1)(y – 3) = 0 y + 1 = 0 or y – 3 = 0 y = –1 or y = 3

Substitute y = –1 into 1,x = 2(–1) = –2

Substitute y = 3 into 1,x = 2(3) = 6

The solutions are x = –2, y = –1 and x = 6, y = 3.

3. 2x – y = 3 y = 2x – 3 ...........1 x2 – 2y2 – xy + 27 = 0 ...................2

Substitute 1 into 2, x2 – 2(2x – 3)2 – x(2x – 3) + 27 = 0x2 – 2(4x2 – 12x + 9) – 2x2 + 3x + 27 = 0 x2 – 8x2 + 24x – 18 – 2x2 + 3x + 27 = 0 x2 – 3x – 1 = 0

\ a = 1, b = –3, c = –1

x = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

= –(–3) ± ABBBBBBBBBBB(–3)2 – 4(1)(–1) ––––––––––––––––––––––

2(1)

= 3 ± ABB13––––––––2

= 3.303 or –0.303

Substitute x = 3.303 into 1,y = 2(3.303) – 3 = 3.606

Substitute x = –0.303 into 1,y = 2(–0.303) – 3 = –3.606

The solutions are x = 3.303, y = 3.606 and x = –0.303, y = –3.606.

4. r + s = 3 s = 3 – r ...............................1 r2 – 4s = 10 ...................................2

Substitute 1 into 2, r2 – 4(3 – r) = 10r2 – 12 + 4r – 10 = 0 r2 + 4r – 22 = 0

\ a = 1, b = 4, r = –22

r = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

= – 4 ± ABBBBBBBBBB42 – 4(1)(–22)––––––––––––––––––

2(1)

= – 4 ± ABBB104––––––––––2

= 3.099, –7.099

Substitute r = 3.099 into 1,s = 3 – 3.099 = – 0.099

Substitute r = –7.099 into 1,s = 3 – (–7.099) = 10.099

The solutions are s = –0.099, r = 3.099 and s = 10.099, r = –7.099.

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5. 2x – y = –5 .................1 2y – x = 3xy................2From 1, y = 2x + 5 ...........3

Substitute 3 into 2, 2(2x + 5) – x = 3x(2x + 5) 4x + 10 – x = 6x2 + 15x 6x2 + 12x – 10 = 0 3x2 + 6x – 5 = 0

Use formula x = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

, where

a = 3, b = 6, c = –5.

x = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

x = –6 ± ABBBBBBBBB62 – 4(3)(–5) –––––––––––––––––

2(3) = 0.63 or –2.63Substitute x = 0.63 into 3,y = 2(0.63) + 5 = 6.26Substitute x = –2.63 into 3,y = 2(–2.63) + 5 = –0.26The solutions are x = 0.63, y = 6.26 and x = –2.63, y = –0.26.

1. x = y................................1 xy – x = 0 ...............................2

Substitute 1 into 2, y2 – y = 0y(y – 1) = 0 y = 0 or y – 1 = 0 y = 1

From 1, y = 0, \ x = 0 y = 1, \ x = 1

The solutions are x = 0, y = 0 and x = 1, y = 1.

2. x + y = 3 y = 3 – x ......................1 2x2 + xy = 4 ............................2

Substitute 1 into 2, 2x2 + x(3 – x) = 4 2x2 + 3x – x2 – 4 = 0 x2 + 3x – 4 = 0 (x + 4)(x – 1) = 0 x + 4 = 0 or x – 1 = 0 x = –4 or x = 1

Substitute x = –4 into 1,y = 3 – (–4) = 7

Substitute x = 1 into 1,y = 3 – 1 = 2

The other solution is x = –4, y = 7.

3. x + y = 6 y = 6 – x ..........................1 x2 – y = –4 ..............................2

Substitute 1 into 2, x2 – (6 – x) = –4x2 – 6 + x + 4 = 0 x2 + x – 2 = 0 (x + 2)(x – 1) = 0 x + 2 = 0 or x – 1 = 0 x = –2 or x = 1

Substitute x = –2 into 1,y = 6 – (–2) = 8


The solutions are x = –2, y = 8 and x = 1, y = 5.

4. 2y – x = 0 x = 2y ..............................1

1—x = x + y

× x, 1 = x(x + y) 1 = x2 + xy ....................2

Substitute 1 into 2, 1 = (2y)2 + (2y)(y) 1 = 4y2 + 2y2

6y2 = 1

y2 = 1—6

y = ±ABB1—6 = ±0.4082

Substitute y = 0.4082 into 1,x = 2(0.4082) = 0.8164

Substitute y = –0.4082 into 1,x = 2(–0.4082) = –0.8164

The solutions are x = 0.8164, y = 0.4082 and x = – 0.8164, y = – 0.4082.

7



5. y + x = 0 x = –y............................. 1 x2 + y2 = 8 .............................. 2

Substitute 1 into 2, (–y)2 + y2 = 8 2y2 = 8 y2 = 4 y = ±2

From 1,y = 2, \ x = –2y = –2, \ x = 2

Hence, the coordinates of A are (–2, 2) and the coordinates of B are (2, –2).

6. x + 2y = 4 x = 4 – 2y ............... 1 –x + y2 – 4 = 0 ....................... 2

Substitute 1 into 2, –(4 – 2y) + y2 – 4 = 0 – 4 + 2y + y2 – 4 = 0 y2 + 2y – 8 = 0 (y + 4)(y – 2) = 0 y + 4 = 0 or y – 2 = 0 y = –4 or y = 2

At B, substitute y = – 4 into 1,x = 4 – 2(– 4) = 12Hence, the coordinates of B are (12, –4).

7. 1 – y = 0 y = 1 ............. 1x2 + y2 – 2y + 3x – 3 = 0 ............. 2

Substitute 1 into 2,x2 + 1 – 2 + 3x – 3 = 0 x2 + 3x – 4 = 0 (x + 4)(x – 1) = 0 x + 4 = 0 or x – 1 = 0 x = – 4 or x = 1

The points of intersection are (– 4, 1) and (1, 1).

8. 3y + 2x = 7 y = 7 – 2x––––––

3 ........................ 1

x2 + xy – 6 = 0 ............................. 2


x2 + x1 7 – 2x3 2 – 6 = 0

× 3, 3x2 + x(7 – 2x) – 18 = 0 3x2 + 7x – 2x2 – 18 = 0 x2 + 7x – 18 = 0 (x + 9)(x – 2) = 0 x + 9 = 0 or x – 2 = 0 x = –9 or x = 2


y = 7 – 2(–9)––––––––3

= 253


y = 7 – 2(2)–––––––3

= 1

\ The solutions are x = 2, y = 1 and x = –9, y = 253 .

9. y – 3x + 5 = 0 y = 3x – 5 ................ 1 3x2 + 5x = 6 + y .................. 2

Substitute 1 into 2, 3x2 + 5x = 6 + 3x – 5 3x2 + 2x – 1 = 0 (3x – 1)(x + 1) = 0 3x – 1 = 0 or x + 1 = 0

x = 1—3 or x = –1

Substitute x = 1—3 into 1,

y = 31 1—3 2 – 5

= – 4

Substitute x = –1 into 1,y = 3(–1) – 5 = –8\ The solutions are x = 1—3 , y = – 4 and x = –1, y = –8.

10. 2y + x = 5 x = 5 – 2y ...................... 1 x2 + y2 = 10 ............................ 2

Substitute 1 into 2, (5 – 2y)2 + y2 = 1025 – 20y + 4y2 + y2 – 10 = 0 5y2 – 20y + 15 = 0÷ 5, y2 – 4y + 3 = 0 (y – 1)(y – 3) = 0 y – 1 = 0 or y – 3 = 0 y = 1 or y = 3

Substitute y = 1 into 1,x = 5 – 2(1) = 3

Substitute y = 3 into 1,x = 5 – 2(3) = –1

The solutions are x = 3, y = 1 and x = –1, y = 3.

8



11. 2y + 3x + 5 = 0

y = –1 3x + 52 2 ............1

xy + 1—3 x + 7 = 0............................2


x3– 1 3x + 52 24 + 1—3 x + 7 = 0

× 6, –3x(3x + 5) + 2x + 42 = 0 –9x2 – 15x + 2x + 42 = 0 9x2 + 13x – 42 = 0 (9x – 14)(x + 3) = 0 9x – 14 = 0 or x + 3 = 0 x = 14

9 or x = –3


y = –3 31 149 2 + 5

––––––––––2 4

= – 296


y = –3 3(–3) + 5 –––––––––2 4

= 2

The solutions are x = 149 , y = – 29

6 and x = –3, y = 2.

12. xy + 2x = –2........................... 1 2y + 2x = –2 y + x = –1 y = –1 – x .................... 2

Substitute 2 into 1, x(–1 – x) + 2x = –2 –x – x2 + 2x = –2 x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x – 2 = 0 or x + 1 = 0 x = 2 or x = –1

Substitute x = 2 into 2,y = –1 – 2 = –3

Substitute x = –1 into 2,y = –1 – (–1) = 0

The solutions are x = 2, y = –3 and x = –1, y = 0.

13. 6y + x = 15 x = 15 – 6y .................13y2 + x2 – 2x = 15 .........................2

Substitute 1 into 2, 3y2 + (15 – 6y)2 – 2(15 – 6y) = 153y2 + 225 – 180y + 36y2 – 30 + 12y – 15 = 0 39y2 – 168y + 180 = 0

÷ 3, 13y2 – 56y + 60 = 0 (13y – 30)(y – 2) = 0 13y – 30 = 0 or y – 2 = 0

y = 3013 or y = 2

Substitute y = 3013 into 1,

x = 15 – 61 3013 2

= 1513


\ The solutions are x = 1513 , y = 30

13 and x = 3,

y = 2.

14. A = 3B = C 1 + 4y = 3(y – 2x) = y + 1—x + 7

1 + 4y = 3(y – 2x) 1 + 4y = 3y – 6x y = –1 – 6x .................................. 1

1 + 4y = y + 1—x + 7

× x, x(1 + 4y) = x(y + 1—x + 7)

x + 4xy = xy + 1 + 7x 3xy – 6x – 1 = 0 .............................. 2

Substitute 1 into 2, 3x(–1 – 6x) – 6x – 1 = 0 –3x – 18x2 – 6x – 1 = 0 18x2 + 9x + 1 = 0 (6x + 1)(3x + 1) = 0 3x + 1 = 0 or 6x + 1 = 0 x = – 1—3 or x = – 1—6

Substitute x = – 1—6 into 1,

y = –1 – 61– 1—6 2 = 0

Substitute x = – 1—3 into 1,

y = –1 – 61– 1—3 2 = 1

The solutions are x = – 1—6 , y = 0 and x = – 1—3 , y = 1

9



15. Substitute x = p, y = q into 2y + x = 3y + 1 ........................... 1and xy + 2 = 2y + x ............................ 2

From 1, 2q + p = 3q + 1 p = q + 1 .................... 3From 2, pq + 2 = 2q + p .................. 4

Substitute 3 into 4, q(q + 1) + 2 = 2q + q + 1 q2 + q + 2 = 3q + 1 q2 – 2q + 1 = 0 (q – 1)2 = 0 q = 1

Substitute q = 1 into 3,p = 1 + 1 = 2

Hence, p = 2, q = 1.

16. Substitute x = 3p, y = 2q into

3y

–––2x + 4x–––y – 5 = 0 ........................1

and y—2 – 2—3 x – 2 = 0 ........................2

From 1, 3(2q)–––––2(3p)

+ 4(3p)–––––

2q – 5 = 0

q—p +

6p–––q – 5 = 0

× pq, q2 + 6p2 – 5pq = 0 ........................ 3

From 2, 2q–––2 – 2—3 (3p) – 2 = 0

q – 2p – 2 = 0 q = 2p + 2 ...... 4Substitute 4 into 3, (2p + 2)2 + 6p2 – 5p(2p + 2) = 0 4p2 + 8p + 4 + 6p2 – 10p2 – 10p = 0 –2p + 4 = 0 p = 2

Substitute p = 2 into 4,q = 2(2) + 2 = 6Hence, p = 2, q = 6.

17. Perimeter = 3615 + (y – 3) + (x + 9) = 36 x + y + 21 = 36 y = 15 – x ........... 1

Use Pythagoras’s theorem,PR2 = PQ2 + QR2

152 = (y – 3)2 + (x + 9)2 ..................... 2

Substitute 1 into 2, 225 = (15 – x – 3)2 + x2 + 18x + 81 0 = (12 – x)2 + x2 + 18x + 81 – 225 0 = 144 – 24x + x2 + x2 + 18x + 81 – 225

2x2 – 6x = 0 2x(x – 3) = 0 2x = 0 or x – 3 = 0 x = 0 or x = 3




18. x – 3y = 1 x = 1 + 3y ............. 1xy + 2y2 – 3 = 0 ..................... 2

Substitute 1 into 2,y(1 + 3y) + 2y2 – 3 = 0 y + 3y2 + 2y2 – 3 = 0 5y2 + y – 3 = 0

\ a = 5, b = 1, c = –3

y = –b ± ABBBBBBb2 – 4ac––––––––––––––2a

= –1 ± ABBBBBBBBB12 – 4(5)(–3)

––––––––––––––––––2(5)

= –1 ± ABB61––––––––

10 = 0.681, –0.881

Substitute y = 0.681 into 1,x = 1 + 3(0.681) = 3.043

Substitute y = –0.881 into 1,x = 1 + 3(– 0.881) = –1.643

The solutions are x = 3.043, y = 0.681 and x = –1.643, y = – 0.881.

19. Volume = 6 2xy = 6 xy = 3.......................................... 1

Total length of edges = 24 4x + 4y + 4(2) = 24 4x + 4y = 16 x + y = 4 y = 4 – x .............. 2

Substitute 2 into 1, x(4 – x) = 3 4x – x2 = 3 x2 – 4x + 3 = 0(x – 1)(x – 3) = 0 x – 1 = 0 or x – 3 = 0 x = 1 or x = 3

10






20. Substitute y = 0 into y = –x2 + 4x, –x2 + 4x = 0 x(–x + 4) = 0 x = 0 or –x + 4 = 0 x = 4

Hence, C(4, 0).

Equation of straight line is

y = 1 4 – 0 –––––0 – 4 2x + 4

y = –x + 4 ........................................... 1y = –x2 + 4x ........................................ 2

1 = 2, –x + 4 = –x2 + 4x x2 – 5x + 4 = 0 (x – 1)(x – 4) = 0 x – 1 = 0 or x – 4 = 0 x = 1 or x = 4At B, substitute x = 1 into 1,y = –1 + 4 = 3

The coordinates of B are (1, 3).

1. x2 + y2 = 5 .......................................... 1 x + 2y = 3 x = 3 – 2y .................................. 2

Substitute 2 into 1, (3 – 2y)2 + y2 = 5 9 – 12y + 4y2 + y2 – 5 = 0 5y2 – 12y + 4 = 0 (5y – 2)(y – 2) = 0

y = 2—5

or y = 2


into 2,

x = 3 – 21 2—5 2

= 3 – 4—5

= 11–––5

Substitute y = 2 into 2,x = 3 – 2(2) = –1For x = 11–––

5, y = 2—

5,

–2p = 11–––5

and 3k = 2—5

p = – 11–––10

k = 2–––15

For x = –1, y = 2, –2p = –1 and 3k = 2

p = 1—2

k = 2—3

The solutions are p = 1—2

, k = 2—3

and p = – 11–––10

,

k = 2–––15

.

2. Let the length, breadth and height of the cuboid are x cm, x cm and y cm respectively.

8x + 4y = 104 2x + y = 26 y = 26 – 2x ............................... 1

4xy + 2x2 = 440 ................................... 2

Substitute 1 into 2, 4x(26 – 2x) + 2x2 = 440÷ 2, 2x(26 – 2x) + x2 = 220 52x – 4x2 + x2 – 220 = 0 –3x2 + 52x – 220 = 0 3x2 – 52x + 220 = 0 (3x – 22)(x – 10) = 0

x = 22–––3

or x = 10

Substitute x = 22–––3

into 1,

y = 26 – 21 22–––3 2

= 26 – 44–––3

= 34–––3

Substitute x = 10 into 1,y = 26 – 2(10) = 6

Volume = x2y or Volume = x2y

= 1 22–––3 2

2

1 34–––3 2

= 609 13–––27

cm3

= 102 × 6 = 600 cm3

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3. x + y = 18 y = 18 – x .................................... 1

x2 + y2 = 194 ....................................... 2

Substitute 1 into 2, x2 + (18 – x)2 = 194 x2 + 324 – 36x + x2 – 194 = 0 2x2 – 36x + 130 = 0 x2 – 18x + 65 = 0 (x – 5)(x – 13) = 0 x = 5 or 13



Therefore, the two numbers are 5 and 13.

4. y = 1—2

x + 5 ................................ 1

x2 + y2 = 100 ....................................... 2


x2 + 1 1—2

x + 522 = 100

x2 + 1—4

x2 + 5x + 25 – 100 = 0

5—4

x2 + 5x – 75 = 0

× 4, 5x2 + 20x – 300 = 0÷ 5, x2 + 4x – 60 = 0 (x + 10)(x – 6) = 0 x = –10 or 6


y = 1—2

(–10) + 5

= –5 + 5 = 0

Substitute x = 6 into 1,y = 1—

2(6) + 5

= 8Therefore, A(–10, 0) and B(6, 8)

Midpoint of AB = 1 –10 + 6–––––––2

, 0 + 8–––––2 2

= (–2, 4)

5. AC = 9 cm (x – 1) + (y + 2) = 9 x + y + 1 = 9 y = 8 – x ..................... 1

(x – 1)(y + 2) = 20 .............................. 2

Substitute 1 into 2, (x – 1)(8 – x + 2) = 20 (x – 1)(10 – x) = 20 10x – x2 – 10 + x = 20 11x – x2 = 30 x2 – 11x + 30 = 0 (x – 6)(x – 5) = 0 x = 5 or x = 6


Substitute x = 6 into 1, y = 8 – 6 = 2


6. Area of rectangle ABCD = 48 (x + 2)(y + 3) = 48 xy + 3x + 2y + 6 = 48 xy + 3x + 2y = 42 ........................ 1

Perimeter of ∆ABD = 24 y + 3 + x + 2 + 10 = 24 y + x = 9 y = 9 – x .............................2

Substitute 2 into 1, x(9 – x) + 3x + 2(9 – x) = 42 9x – x2 + 3x + 18 – 2x – 42 = 0 x2 – 10x + 24 = 0 (x – 4)(x – 6) = 0 x = 4 or x = 6



Therefore, x = 4, y = 5 or x = 6, y = 3.

7. x + y = 20 y = 20 – x ...................................... 1

x2 + y2 = 208 ....................................... 2

Substitute 1 into 2, x2 + (20 – x)2 = 208 x2 + 400 – 40x + x2 – 208 = 0 2x2 – 40x + 192 = 0÷ 2, x2 – 20x + 96 = 0 (x – 8)(x – 12) = 0 x = 8 or x = 12

12





Since y . x, therefore the solution is x = 8, y = 12.

8. Area of rectangle = 420 (x + 10)(2y + 20) = 420÷ 2, (x + 10)(y + 10) = 210 xy + 10x + 10y + 100 – 210 = 0 xy + 10x + 10y – 110 = 0 .....................1

Perimeter of the circle = Perimeter of the rectangle 2pr = 2(x + 10) + 2(2y + 20)

21 22–––7 2(14) = 2(x + 10) + 2(2y + 20)

÷ 2, 44 = x + 10 + 2y + 20 44 = x + 2y + 30 x = 14 – 2y ....................2

Substitute 2 into 1, (14 – 2y)y + 10(14 – 2y) + 10y – 110 = 0 14y – 2y2 + 140 – 20y + 10y – 110 = 0 –2y2 + 4y + 30 = 0÷ (–2), y2 – 2y – 15 = 0 (y – 5)(y + 3) = 0 y = 5 or y = –3


Substitute y = –3 into 2,x = 14 – 2(–3) = 20

The solutions are x = 4, y = 5 and x = 20, y = –3.

9. (a) When x = 0, y = 02 – 6(0) + 8 y = 8

The coordinates of A are (0, 8).

Area of ∆OAC = 32

1—2

(OA)(OC) = 32

1—2

(8)(OC) = 32

OC = 8

The coordinates of C are (8, 0).

Equation of the line AC is

x—8

+ y—8

= 1

x + y = 8

(b) x + y = 8 y = 8 – x ................................ 1

y = x2 – 6x + 8 ............................. 2

Substitute 1 into 2, 8 – x = x2 – 6x + 8 x2 – 5x = 0 x(x – 5) = 0 x = 0 (rejected) or x = 5

Substitute x = 5 into 1, y = 8 – 5 = 3

Therefore, the coordinates of B are (5, 3).

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