**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)**

=>x^{2} – 5x + 3x – 15 = x^{2} – 7x + 4x – 28

=> x^{2} – 2x – 15 = x^{2} – 3x -28

=> x^{2 }– 2x – x^{2} + 3x = -28 + 15

=> 1x = – 13

=> x = -13

**(18) (x – 5) ^{2} – (x+2)^{2} = -2**

=> x^{2} – 10x + 25 – (x^{2} + 4x + 4) = -2

=> x^{2} – 10x + 25 – x^{2} – 4x – 4 = -2

=> – 14x = -2 -25 + 4

=> – 14x = – 23

=> x = = 1

(19) (x – 1) (x + 6) – (x – 2) (x – 3) = 3

=> x^{2} + 5x – 6 – (x^{2} – 5x + 6) = 3

=> x^{2} + 5x – 6 – x^{2} + 5x – 6 = 3

=> 10x – 12 = 3

=> 10x = 3 + 12

=> x = 15 / 10

=> x = 3/2

=> 3x^{2} – 9x + 6 = 3x^{2} – 13x +12

=> 3x^{2} – 9x – 3x^{2} + 14x = 12 – 6

=> 5x = 6

=> x = 6 / 5

=> x = 1.2