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. 

Question 1:

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:

Solve: ${\displaystyle \frac{3}{7}+x=\frac{17}{7}}$

Answer:
${\displaystyle \frac{3}{7}+x={\displaystyle \frac{17}{7}}}$

$x={\displaystyle \frac{17}{7}-\frac{3}{7}}$

$x={\displaystyle \frac{14}{7}}$

$x=2$

Question 5:

Solve: 6x = 12

Answer:

$6x=12$

$x={\displaystyle \frac{12}{2}}$

$x=2$

Question 6:

Solve: ${\displaystyle \frac{t}{5}=10}$

Answer:

${\displaystyle \frac{t}{5}=10}$

${\displaystyle t=10\times 5}$

$t=50$

Question 7:

Solve: ${\displaystyle \frac{2x}{3}=18}$

Answer:

${\displaystyle \frac{2x}{3}=18}$

$2x=18\times 3$

$x={\displaystyle \frac{18\times 3}{2}}$

$x=27$

Question 8:

Solve: $1.6={\displaystyle \frac{y}{1.5}}$

Answer:

$1.6={\displaystyle \frac{y}{1.5}}$

$y=2.4$

Question 9:

Solve: $7x-9=16$

Answer:

$7x-9=16$

$7x=16+9$

$7x=25$

$x={\displaystyle \frac{25}{7}}$

Question 10:

Solve: $14y-8=13$

Answer:

$14y-8=13$

$14y=13+8$

$14y=21$

 $y={\displaystyle \frac{21}{14}}$

$y={\displaystyle \frac{3}{2}}$

Question 11:

Solve:$17+6p=9$

Answer:

$17+6p=9$

$6p=9-17$

$6p=-8$

$p={\displaystyle \frac{-4}{3}}$

Question 12:

Solve: ${\displaystyle \frac{x}{3}+1={\displaystyle \frac{7}{15}}}$

Answer:

${\displaystyle \frac{x}{3}+1=\frac{7}{15}}$

${\displaystyle \frac{x}{3}=\frac{7}{15}-1}$

${\displaystyle \frac{x}{3}=\frac{7-15}{15}}$

${\displaystyle \frac{x}{3}=\frac{-8}{15}}$

${\displaystyle x=\frac{-8\times 3}{15}}$

${\displaystyle x=\frac{-8}{5}}$