Page 103 - Maths Skills - 8

Linear Equations in One Variable 101

Example 4: Solve: 1 x 7 Example 5: Solve: 5x 20 3
4
3 3 3x 4

Solution: Given: 1 x 7 Solution: Given: 5x 20 3
4
3 3 3x 4

⇒ 1 x 7 4 [By transposition] ⇒ 5x + 20 = 3(3x + 4) [By cross
3 3 ⇒ 5x + 20 = 9x + 12 multiplication]

1 5 ⇒ 5x – 9x = 12 – 20
⇒ x
3 3 ⇒ – 4x = – 8

⇒ x 5 3 5 ⇒ 4x = 8
3 1 Hence, x = 2.

Hence, x = – 5

05. y 4 5
Example 6: Solve:
24. y 6 3

05. y 4 5
Solution: Given:
24. y 6 3
⇒ 3(0.5y – 4) = 5(2.4y + 6) [By cross multiplication]
⇒ 1.5y – 12 = 12y + 30

⇒ 1.5y – 12y = 30 + 12 [By transposition]
⇒ – 10.5y = 42

⇒ 10.5y = – 42 [Multiplying both sides by –1]

⇒ y 42 4
10 5.
Hence, y = – 4.

Exercise 7.1

1. Solve the following equations and verify the answers.
2
(i) 3x + 4 = 22 (ii) 3(x – 1) = 8 (iii) 3x + = 2x + 1
3

(iv) 5x 1 x 25 (v) 0.3(6 + t) = 0.4(8 – t) (vi) 2(3x – 1) = 16
4 4
2. Find a positive value of the variable used in the equation, for which the given equation is satisfied.
x 4 1 a 2 1
2
2
(i) (ii)
2
2
x 4 2 2 a 15 3

(iii) (12b ) ( b ) 2 (iv) (x 3 ) (x 3 ) 1

b
( ) 2 (12b (x 4 ) (x 4 ) 2
)