Find the values: y = x + 3 x 0 - 2 y 0 - 3

We have, y = x + 3 When x = 0 , y = 0 + 3 ⇒ y = 3 When y = 0 , 0 = x + 3 ⇒ x = 0 - 3 = - 3 When x = - 2 , y = - 2 + 3 ⇒ y = 1 When y = - 3 , - 3 = x + 3 ⇒ x = - 3 - 3 = - 6 Therefore, the complete table is: x 0 - 3 - 2 - 6 y 3 0 1 - 3

Find the values: 2 x + y - 6 = 0 x 0 - 1 y 0 - 2

We have, 2 x + y - 6 = 0 When x = 0 , 2 0 + y - 6 = 0 ⇒ y - 6 = 0 ⇒ y = 6 When y = 0 , 2 x + 0 - 6 = 0 ⇒ 2 x - 6 = 0 ⇒ x = 6 2 = 3 When x = - 1 , 2 - 1 + y - 6 = 0 ⇒ - 2 + y - 6 = 0 ⇒ y - 8 = 0 ⇒ y = 8 When y = - 2 , 2 x - 2 - 6 = 0 ⇒ 2 x - 8 = 0 ⇒ x = 8 2 = 4 Therefore, the complete table is: x 0 3 - 1 4 y 6 0 8 - 2

Find the values: y = 3 x + 1 x - 1 0 1 2 y

We have, y = 3 x + 1 When x = - 1 , y = 3 - 1 + 1 ⇒ y = - 3 + 1 = - 2 When x = 0 , y = 3 0 + 1 ⇒ y = 1 When x = 1 , y = 3 1 + 1 ⇒ y = 3 + 1 ⇒ y = 4 When x = 2 , y = 3 2 + 1 ⇒ y = 6 + 1 ⇒ y = 7 Therefore, the complete table is: x - 1 0 1 2 y - 2 1 4 7

E M B I B E

Standard Eight Mathematics>Chapter 3 - Algebra>Exercise 3.9

Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 37: Exercise 3.9

Author:Tamil Nadu Board

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 37: Exercise 3.9

Attempt the free practice questions on Chapter 3: Algebra, Exercise 37: Exercise 3.9 with hints and solutions to strengthen your understanding. Standard Eight Mathematics solutions are prepared by Experienced Embibe Experts.