Concise Mathematics – Class 8 Selina Exercise 14(A) Solution by experts | 14(A) Selina Concise Solution Class 8 | Class 8 Selina Concise Ex 14(A) Page 165 – 166 Solution | RK Bansal Class 8 ICSE Book Solution Online.
Exercise 14(A)
Solve the following equations
(1) 20 = 6 + 2x
=> 6 + 2x = 20
=> 2x = 20 – 6
=> x = 14 / 2
Therefore x = 7
(2) 15 + x = 5x +3
=> x – 5x = 3 – 15
=> – 4x = -12
=> x = -12 / -4
=> x = 3
= -7x + 42 = 3x + 2
= – 7x – 3x = 2 – 42
= – 10x = – 40
= x = -40 / -10
= x = 4
(4) 3a – 4 = 2(4 – a)
=> 3a – 4 = 8 – 2a
=> 3a + 2a = 8 + 4
=> 5a = 12
=> a = 12/5
=> a = 2.4
(5) 3(b-4) = 2(4-b)
=> 3b – 12 = 8 – 2b
=> 3b + 2b = 8 + 12
=> 5b = 20
=> b = 20 / 5
=> b = 4
(8) 5 (8x + 3) = 9 (4x + 7)
=> 40x + 15 = 36x + 63
=> 40x – 36x = 63 – 15
=> 4x = 48
=> x = 48 / 4
=> x = 12
(9) 3 (x + 1) = 12 + 4(x – 1)
=> 3x + 3 = 12 + 4x – 4
=> 3x – 4x = 12 – 4 – 3
=> – 1x = 5
=> x = – 5
=> 15a – 1 = a + 27
=> 15a – a = 27 + 1
=> 14a = 28
=> a = 2
=> 52y + 104 = 455 + 25y
=> 52y – 25y = 455 – 104
=> 27y = 351
=> y = 351 / 27
=> y = 13
(16) 6 (6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
=> 36x – 30 – 35x + 40 = 48 – 12x + 1
=> 1x + 10 = 49 – 12x
=> 1x + 12x = 49 – 10
=> 13x = 39
=> x = 39 / 13 = 3
=> x = 3
(17) (x – 5) (x + 3) = (x – 7) (x + 4)
=>x2 – 5x + 3x – 15 = x2 – 7x + 4x – 28
=> x2 – 2x – 15 = x2 – 3x -28
=> x2 – 2x – x2 + 3x = -28 + 15
=> 1x = – 13
=> x = -13
(18) (x – 5)2 – (x+2)2 = -2
=> x2 – 10x + 25 – (x2 + 4x + 4) = -2
=> x2 – 10x + 25 – x2 – 4x – 4 = -2
=> – 14x = -2 -25 + 4
=> – 14x = – 23
=> x = = 1
(19) (x – 1) (x + 6) – (x – 2) (x – 3) = 3
=> x2 + 5x – 6 – (x2 – 5x + 6) = 3
=> x2 + 5x – 6 – x2 + 5x – 6 = 3
=> 10x – 12 = 3
=> 10x = 3 + 12
=> x = 15 / 10
=> x = 3/2
=> 3x2 – 9x + 6 = 3x2 – 13x +12
=> 3x2 – 9x – 3x2 + 14x = 12 – 6
=> 5x = 6
=> x = 6 / 5
=> x = 1.2
