ncert solutions for class 8 maths chapter 2 exercise 2.1
exercise 2.1 class 8 Linear Equations in One Variable involve complete answers for each question in the exercise 2.1. The solutions provide students a strategic methods to prepare for their exam. exercise 2.1 class 8 Linear Equations in One Variable questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. exercise 2.1 class 8 Linear Equations in One Variable prepared by our subject matter experts in very delicate, easy and creative way.
Solve: x-2 = 7.\(\)
Answer:\(\)
\(x − 2 = 7\)
\(x = 7 + 2 \)
\(x = 9\)
Question 2:
Solve: y + 3 = 10
Answer:
\(y + 3 = 10\)
\(y = 10 − 3\)
\(y= 7\)
Question 3:
Solve: 6 = z + 2
Answer:
\(6 = z + 2\)
\(6 − 2 = z\)
\(z = 4\)
Question 4:
Question 2:
Solve: y + 3 = 10
Answer:
\(y + 3 = 10\)
\(y = 10 − 3\)
\(y= 7\)
Question 3:
Solve: 6 = z + 2
Answer:
\(6 = z + 2\)
\(6 − 2 = z\)
\(z = 4\)
Question 4:
Solve: \(\displaystyle{3 \over 7} + x = {17\over 7}\)
Answer:
\(\displaystyle{3 \over 7} + x = \displaystyle{17\over 7}\)
\(x = \displaystyle{17\over 7} - {3 \over 7}\)
\(x=\displaystyle{14 \over 7}\)
\(x=2\)
Question 5:
Solve: 6x = 12
Answer:
\(6x = 12\)
\(x = \displaystyle{12\over 2}\)
\(x = 2\)
Question 6:
Solve: \(\displaystyle{t\over 5}=10\)
Answer:
\(\displaystyle{t\over 5}=10\)
\(\displaystyle{t}=10 × 5\)
\(t=50\)
Question 7:
Solve: \(\displaystyle{2x \over 3} = 18\)
Answer:
\(\displaystyle{2x \over 3} = 18\)
\(2x=18 × 3\)
\(x=\displaystyle{18 × 3 \over 2}\)
\(x = 27\)
Question 8:
Solve: \(1.6 = \displaystyle{y\over 1.5}\)
Answer:
\(1.6=\displaystyle{y\over 1.5}\)
\(y=2.4\)
Question 9:
Solve: \(7x − 9 = 16\)
Answer:
\(7x − 9 = 16\)
\(7x = 16 + 9\)
\(7x = 25\)
\(x = \displaystyle{25 \over 7}\)
Question 10:
Solve: \(14y − 8 = 13\)
Answer:
\(14y − 8 = 13\)
\(14y = 13 + 8\)
\(14y = 21\)
\(y= \displaystyle{21\over 14}\)
\( y=\displaystyle{3 \over 2}\)
Question 11:
Solve:\(17 + 6p = 9\)
Answer:
\(17 + 6p = 9\)
\(6p = 9 − 17\)
\(6p = −8\)
\(p = \displaystyle{-4\over 3}\)
Question 12:
Solve: \(\displaystyle{x\over 3} +1 =\displaystyle{7\over 15}\)
Answer:
\(\displaystyle{x\over 3} +1 = {7\over 15}\)
\(\displaystyle{x\over 3} = {7\over 15}-1\)
\(\displaystyle{x\over 3} = {7 -15 \over 15}\)
\(\displaystyle{x\over 3} = {-8 \over 15}\)
\(\displaystyle{x} = {-8×3 \over 15}\)
\(\displaystyle{x} = {-8 \over 5}\)
