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Solve the following linear equations

a. $x-11=7$

b. $z+8=9$

c. $11x=121$

a. $x-11=7$

$x = 7 + 11 =18$

b. $z+8=9$

$z = 9-8$

$z=1$

c. $11x =121$

$x= \frac {121}{11}$

$x=11$

$\frac {y}{11} = 11$

$\frac {y}{11} = 11$

$y = 11 \times 11$

y=121

$\frac {23x}{2}= 46$

$\frac {23x}{2}= 46$

$23x = 46 \times 2$

$23x = 92$

$x = \frac {92}{23}$

x=4

$1.2= \frac {z}{1.4}$

$1.2= \frac {z}{1.4}$

or

$ \frac {z}{1.4}=1.2$

$z = 1.4 \times 1.2$

z=1.68

$7x-12=16$

$7x-12=16$

$7x = 16 + 12$

$7x = 28$

$x = \frac {28}{7}$

$x= 4$

Solve:

$7z-5=16$

$7z-5=16$

$7z = 16 +5$

$7z = 21$

$z= \frac {21}{7}$

$z=3$

Solve:

$10+6p=22$

$10+6p=22$

$6p = 22 -10$

$6p = 12$

$p = \frac {12}{6}$

$p=2$

Solve

$11-5x+3x+4x=18$

$11-5x+3x+4x=18$

$11- 5x + 7x=18$

$11 +2x =18$

$2x=18-11$

$2x=7$

$ x = \frac {7}{2}$

Solve

$(x-2)+(x-3) +(x-9)= 1$

$(x-2)+(x-3) +(x-9)= 1$

$ x-2 + x-3 + x-9=1$

$3x -14 =1$

$3x=1+14$

$3x= 15$

$x= \frac {15}{3}$

x=5

Solve

$(2x-2)+(3x-3) +(9x-9)= 1$

$(2x-2)+(3x-3) +(9x-9)= 1$

$2x- 2 + 3x -3 + 9x -9 =1$

$14 x -14 =1$

$14x = 1 + 14$

$14x = 15$

$x = \frac {15}{14}$

This Linear equations in One variable class 8 important questions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. You can download this also using the below link

Download this assignment as pdf**Notes****Worksheets****Ncert Solutions**- class 8 maths chapter 2 exercise 2.1 solutions
- class 8 maths chapter 2 exercise 2.2 solutions
- NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
- NCERT Solutions for Linear equations Class 8 maths Chapter 2 Exercise 2.4
- class 8 maths chapter 2 exercise 2.5 solutions
- class 8 maths chapter 2 exercise 2.6 solutions

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