Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

CBSE Class 8 Maths Worksheet Chapter 2 Linear Equations in One Variable - PDF

ffImage
Last updated date: 28th May 2024
Total views: 187.5k
Views today: 3.87k

CBSE Maths Worksheet Chapter 2 Linear Equations in One Variable - Download Free PDF

Linear Equations Class 8 PDF is a crucial resource for CBSE students. It aids in giving them a good understanding of the chapter, mainly the questions asked in the exam and how to respond to them. Vedantu offers all CBSE Class 8 Maths Solutions divided into different exercises to assist students in test preparation and clear up any questions.

 

An equation containing variables is known as an algebraic equation. It expresses that the values of the expressions on either side of the equality sign are equal. You can comprehend the following after studying the chapter Linear Equations in One Variable, such as multiplication and division of algebraic expressions (Coefficient should be integers), Some common errors, identities, factorisation, the method of solving linear equations in one variable in contextual problems involving multiplication and division (word problems) (avoid complex coefficient in the equations), etc.


CBSE Class 8 Maths Worksheet Chapter 2 Linear Equations in One Variable - PDF will be uploaded soon

Popular Vedantu Learning Centres Near You
centre-image
Mithanpura, Muzaffarpur
location-imgVedantu Learning Centre, 2nd Floor, Ugra Tara Complex, Club Rd, opposite Grand Mall, Mahammadpur Kazi, Mithanpura, Muzaffarpur, Bihar 842002
Visit Centre
centre-image
Anna Nagar, Chennai
location-imgVedantu Learning Centre, Plot No. Y - 217, Plot No 4617, 2nd Ave, Y Block, Anna Nagar, Chennai, Tamil Nadu 600040
Visit Centre
centre-image
Velachery, Chennai
location-imgVedantu Learning Centre, 3rd Floor, ASV Crown Plaza, No.391, Velachery - Tambaram Main Rd, Velachery, Chennai, Tamil Nadu 600042
Visit Centre
centre-image
Tambaram, Chennai
location-imgShree Gugans School CBSE, 54/5, School road, Selaiyur, Tambaram, Chennai, Tamil Nadu 600073
Visit Centre
centre-image
Avadi, Chennai
location-imgVedantu Learning Centre, Ayyappa Enterprises - No: 308 / A CTH Road Avadi, Chennai - 600054
Visit Centre
centre-image
Deeksha Vidyanagar, Bangalore
location-imgSri Venkateshwara Pre-University College, NH 7, Vidyanagar, Bengaluru International Airport Road, Bengaluru, Karnataka 562157
Visit Centre
View More

Access Worksheet for Class 8 Maths Linear Equations in One Variable

1. What is the value of y, if y-2=7

  1. 8

  2. 9

  3. 10

  4. 7


2. What is the value of a, if 6a= 18

  1. 6

  2. 3

  3. 3

  4. 4


3. What is the value of X if $X+\dfrac{3}{7}=\dfrac{17}{7}$?

  1. 2

  2. 3

  3. 6

  4. 8


4. What is the value of $A$, If $7 A-16=9$

  1. $\dfrac{25}{7}$

  2. $\dfrac{17}{7}$

  3. $\dfrac{22}{7}$

  4. $\dfrac{1}{7}$


5. What is the value of $z$, If $\dfrac{z}{1.6}=1.5$

  1. $2.4$

  2. $1.6$

  3. $3.2$

  4. $4.2$


Fill in the blanks, For Question (6-10)

6. If $\dfrac{8 z-3}{3 z}=\mathbf{2}$, so $\mathbf{z}=\ldots \ldots$


7. If $\mathrm{y}=\dfrac{1}{2}$, then $\dfrac{5}{4}=\dfrac{y}{2}=\ldots \ldots$


8. $\dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{1}{2}\right)=\ldots \ldots$


9. If $5-\dfrac{3}{5}=\dfrac{2}{5} y-2$,then $y=\ldots \ldots$


10. If 6 is subtracted from the product of $x$ and 5, the result is 12. The value of $x$ is.....


11. Six times a number is 48. What is the number?


12. Sum of the two numbers is 96. If one exceeds the other by 16. Find the number.


13. Simplify $5(1+2 y)=(y-3) 3$


14. A positive number is 4  times another number. If 16  is added to both the numbers, then one of the new numbers becomes thrice  the other new number. What are the numbers?


15. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages


16. Two numbers are the ratio 10:6. If they differ by 36. What are the number?


17. Two equal sides of a triangle are each 4m less than three times the third side. Find the dimensions of the triangle, if its perimeter is 55m.


True / False( For question 18 and 19)

18. $\mathbf{y}=10$ for equation $\dfrac{y}{6}-\dfrac{y}{2}+\dfrac{y}{4}=\dfrac{3}{4}$


19. $\left(6-5 x^2\right)$ is a binomial?


Solve:

20. Solve the equation: $y+3=10$.


21. Solve the equation: $6=z+2$


22. Solve the equations: $\dfrac{3}{7}+x=\dfrac{17}{7}$


23. Solve the equation $6 x=12$


24. Solve the equation $\dfrac{t}{5}=10$.


25. Solve the equation $\dfrac{2 x}{3}=18$.


Answers to the Worksheet:

1. Correct option is (b)

If $y-2=7$

$\Rightarrow y=7+2$

$\Rightarrow y=9$


2. Correct option is (b)

If $6 a=18$

$\Rightarrow a=\dfrac{18}{6}$

$\Rightarrow a=3$


3. Correct option is (a)

If $X+\dfrac{3}{7}=\dfrac{17}{7}$

$X=\dfrac{17}{7}-\dfrac{3}{7}$

$\Rightarrow X=\dfrac{17-3}{7}$

$\Rightarrow X=\dfrac{14}{7}$

$\Rightarrow X=2$


4. Correct option is (a)

If $7 A-16=9$

$\Rightarrow 7 A=9+16$

$\Rightarrow 7 A=25$

$\Rightarrow A=\dfrac{25}{7}$


5. Correct option is (a)

If $\dfrac{z}{1.6}=1.5$

$\Rightarrow z=1.6 \times 1.5$

$\Rightarrow z=2.4$


6. If $\dfrac{8 z-3}{3 z}=2$

So, $8 z-3=2 \times 3 z$

$\Rightarrow 8 z-3=6 z$

$\Rightarrow 8 z-6 z=3$

$\Rightarrow 2 z=3$

$\Rightarrow \mathrm{z}=\dfrac{3}{2}$


7. If $y=\dfrac{1}{2}$

Put the value on then $\dfrac{5}{4}-\dfrac{y}{2}$

$= \dfrac{5}{4}-\dfrac{\dfrac{1}{2}}{2}$

$= \dfrac{5}{4}-\dfrac{1}{2 \times 2}$

$= \dfrac{5}{4}-\dfrac{1}{4}$

$= \dfrac{5-1}{4}$

$= \dfrac{4}{4}$

= 1


8.  $\dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{1}{2}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{1}{4}+\dfrac{2}{4}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{1+2}{4}\right)$

$= \dfrac{1}{4} \times\left(\dfrac{3}{4}\right)$

$= \dfrac{1 \times 3}{4 \times 4}$

$= \dfrac{3}{16}$


9. If $5-\dfrac{3}{5}=\dfrac{2}{5} y-2$

$\Rightarrow \dfrac{5 \times 5}{1 \times 5}-\dfrac{3}{5}=\dfrac{2}{5} y-\dfrac{2 \times 5}{1 \times 5}$

$\Rightarrow \dfrac{25}{5}-\dfrac{3}{5}=\dfrac{2}{5} y-\dfrac{10}{5}$

$\Rightarrow \dfrac{25-3}{5}=\dfrac{2 y-10}{5}$

$\Rightarrow 22=2 y-10$

$\Rightarrow 2 y=22+10$

$\Rightarrow 2 y=32$

$\Rightarrow y=\dfrac{32}{2}$

$\Rightarrow y=16$


10. If 6 is subtracted from the product of $x$ and 5, the result is 12.

$5x - 6 =12$

$5x = 12 - 6$

$5x = 6$

$x=\dfrac{6}{5}$


11. Let us know the number is $y$.

Then, $6 y=48$

$y=\dfrac{48}{6}$

$y=8$


12. According to the question,

Let the smaller number be $\mathrm{X}$.

The other number is $X+96$

According to the question,

$\Rightarrow X+X+96=16$

$\Rightarrow 2 X=96-16$

$\Rightarrow X=40$

The other number is $40+16=56$

So, the two numbers are $40$ and $56$.


13. Simplify $5(1+2 y)=(y-3) 3$

$\Rightarrow 5 \times 1+5 \times 2 y=y \times 3-3 \times 3$

$\Rightarrow 5+10 y=3 y-9$

$\Rightarrow 10 y-3 y=-9-5$

$\Rightarrow 7 y=-14$

$\Rightarrow y=\dfrac{-14}{7}$

$\Rightarrow y=-2$


14. Let the first number be $x$,then another number be $4 x$. 

After adding 16, the new numbers are $x+16$ and $4 x+16$.

According to the question,

$4 x+16=3(x+16)$

$4 x+16=3 x+48$

$4 x-3 x=48-16$

$x=32$

Other number $=4 x=4 \times 32=128$


15. Let Aman’s son’s age be x years.

Therefore, Aman’s age will be 3x years.

Ten years ago, their age was (x - 10) years and (3x - 10) years respectively.

According to the question,10 years ago,

Aman’s age = 5 × Aman’s son’s age 10 years ago

3x - 10 = 5(x - 10)

3x - 10 = 5x - 50

2x = 40

x = 20 and 3x = 60

Therefore, Aman and his son's age are 60 and 20 years respectively.


16 . Let the two numbers be $10 x$ and $6 x$.

According to the condition,

$10 x-6 x=36$

$4 x=36$

$x=9$

Two numbers are

$10 x=10 \times 9=90$

And $6 x=6 \times 9=54$

So, the two numbers are $90$ and $54$.


17. Given, two equal sides of a triangle are each 4 m less than three times the third side.

The perimeter of the triangle is 55 m.

We have to find the dimensions of the triangle.

Let the third side of the triangle be x m

Two equal sides = 3x - 4 m

We know, Perimeter = sum of all sides.

According to the question,

3x - 4 + 3x - 4 + x = 55

7x - 8 = 55

By transposing,

7x = 55 + 8

7x = 63

$x = \dfrac{63}{7}$

x = 9

Now, 3x - 4 = 3(9) - 4

= 27 - 4

= 23 m

Therefore, the dimensions of the triangle are 23 m, 23 m and 9 m.


18. False

$\Rightarrow \dfrac{y}{6}-\dfrac{y}{2}+\dfrac{y}{4}=\dfrac{3}{4}$

$\Rightarrow \dfrac{2 \times y}{2 \times 6}-\dfrac{6 \times y}{6 \times 2}+\dfrac{3 \times y}{3 \times 4}=\dfrac{3}{4}$

$\Rightarrow \dfrac{2 y}{12}-\dfrac{6 y}{12}+\dfrac{3 y}{12}=\dfrac{3}{4}$

$\Rightarrow \dfrac{2 y-6 y+3 y}{12}=\dfrac{3}{4}$

$\Rightarrow \dfrac{-y}{12}=\dfrac{3}{4}$

$\Rightarrow -y=\dfrac{12 \times 3}{4}$

$\Rightarrow -\mathrm{y}=3 \times 3$

$\Rightarrow -\mathrm{y}=9$

$\Rightarrow \mathrm{y}=-9$

 

19. True, A polynomial or algebraic expression with just two terms is called a binomial.


20. Given: $y+3=10$

$\Rightarrow y+3-3=10-3$ (subtracting 3 from each side)

$\Rightarrow y=7$ (Required solution)


21. We have $6=z+2$

$\Rightarrow 6-2=z+2-2$ (subtracting 2 from each side)

$\Rightarrow 4=z$

Thus, $z=4$ is the required solution.


22. We have $\dfrac{3}{7}+x=\dfrac{17}{7}$

$\Rightarrow \quad \dfrac{3}{7}-\dfrac{3}{7}+x=\dfrac{17}{7}-\dfrac{3}{7}$

(subtracting $\dfrac{3}{7}$ from each side)

$\Rightarrow x=\dfrac{17-3}{7}$

$\Rightarrow x=\dfrac{14}{7}$

$\Rightarrow x=2$

Thus, $x=2$ is the required solution.


23. We have $6 x=12$

$\Rightarrow 6 x \div 6=12 \div 6$ (dividing each side by 6 )

$\Rightarrow x=2$

Thus, $x=2$ is the required solution.


24. Given $\dfrac{t}{5}=10$

$\Rightarrow \dfrac{t}{5} \times 5=10 \times 5$ (multiplying both sides by 5 )

$\Rightarrow t=50$

Thus, $t=50$ is the required solution.


25. We have $\dfrac{2 x}{3}=18$

$\Rightarrow \dfrac{2 x}{3} \times 3=18 \times 3$ (multiplying both sides by 3 )

$\Rightarrow 2 x=54$

$\Rightarrow 2 x \div 2=54 \div 2$ (dividing both sides by 2 )

$\Rightarrow x=27$

Thus, $x=27$ is the required solution.


Benefits of Class 8 Maths Linear Equation in One Variable Worksheet

Linear Equations in One Variable Class 8 Worksheets will improve fundamental conceptual understanding. Our exercises contain complete solutions to all essential theorems, greatly simplifying the idea for learners.

 

The linear Equations in One Variable Class 8 Worksheets are a comprehensive resource for students in Class 8. Through the use of these Maths worksheets, educators can gain knowledge and understanding about the development of students as well as methods and techniques for teaching them effectively.

 

Examples of Class 8 Linear Equations in One Variable

Here are some common examples of Class 8 Linear Equations questions:

Q. A rational number has an eight times larger denominator than its numerator. The result is 3/2 if the numerator is raised by 17 and the denominator is lowered by 1. Find the rational number.

Solution:

The denominator will be (x + 8) if the numerator is x.

As given,

(x + 17)/(x + 8 - 1) = 3/2

⇒ (x + 17)/(x + 7) = 3/2

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 - 21 = 3x - 2x

⇒ 13 = x

The rational number is 13/21 (or x/(x + 8)).

 

Q.(3y + 4)/(2 – 6y) = -2/5

Solution:

(3y + 4)/(2 – 6y) = -2/5

⇒ 3y + 4 = -2/5 (2 – 6y)

⇒ 5(3y + 4) = -2(2 – 6y)

⇒ 15y + 20 = -4 + 12y

⇒ 15y – 12y = -4 – 20

⇒ 3y = -24

⇒ y = -8

 

Q. (7y + 4)/(y + 2) = -4/3

Solution:

(7y + 4)/(y + 2) = -4/3

⇒ 7y + 4 = -4/3 (y + 2)

⇒ 3(7y + 4) = -4(y + 2)

⇒ 21y + 12 = -4y – 8

⇒ 21y + 4y = -8 – 12

⇒ 25y = -20

⇒ y = -20/25 = -4/5

 

Q. The ratio between Hari and Harry's ages is 5:7. Their ages will be proportionately 3:4 in four years. What is their age right now?

Solution:

Let Harry's age be seven times that of Hari. After four years,

Age of Hari = 5 times + 4 Age of Harry =7 x + 4

As given,

(5x + 4)/(7x + 4) = ¾

⇒ 4(5x + 4) = 3(7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 21x - 20x = 16 - 12

⇒ x = 4

Hari age is 5x, or i.e., 5* 4 years = 20 years

Age of Harry = 7x = 7*4 = 28 years.

 

Q. 3(t – 3) = 5(2t + 1)

Solution:

3(t – 3) = 5(2t + 1)

⇒ 3t – 9 = 10t + 5

⇒ 3t – 10t = 5 + 9

⇒ -7t = 14

⇒ t = 14/-7

⇒ t = -2

 

Key Features of Linear Equations in One Variable Class 8 Worksheets

It is now easy to get the solutions to every question in Linear Equations in One Variable Class 8 in PDF format. Visit the Vedantu website and download the Class 8 Maths Chapter 2 CBSE Solutions PDF file on your devices from there, wherever you are.

 

Once downloaded, you can rapidly review some formulas or crucial concepts from Chapter 2 of Class 8 Math without worrying about an online connection. 

  • Linear Equation Class 8 worksheets have been well-formulated following the latest CBSE guidelines.

  • Class 8 Maths Chapter 2 Worksheet PDF is also printable, so having a paper copy on hand at the test centre would be helpful.

  • Solving many linear equations in one variable question and answers will help you grasp concepts quickly.

  • Class 8 Linear Equation questions will improve fundamental conceptual understanding. Our exercises contain complete solutions to all essential theorems, greatly simplifying the idea for learners.

You can download the free Class 8 Maths Chapter 2 Worksheet PDF of Linear Equations in One Variable. Math professionals have created linear equations in one variable CBSE Solutions at Vedantu. To assist students in completing each practice question in the book and preparing for the exam, all of the Linear Equations in One Variable's solved questions adhere to the most recent CBSE syllabus and norms.

 

The students can ace the course and the final exams by practising solutions from linear equations in one variable pdf. These answers were developed using the most recent CBSE curriculum.

FAQs on CBSE Class 8 Maths Worksheet Chapter 2 Linear Equations in One Variable - PDF

1. What primary subjects are addressed in Chapter 2 of the NCERT Solutions for Class 8 Math?

There are 6 problems in Chapter 2 of the NCERT Solutions for Class 8 Maths that deal with linear equations in one variable. The key ideas presented in this chapter are: 2.1 The Start 2.2 Solving Equations with Numbers and Linear Expressions on One Side 2.3 Some Applications Solving Equations with a Variable on Both Sides, Section 2.4 2.5 A Few Additional Uses 2.6 Simplification Equations Equations Reduced to Linear Form, Section 2.7.

2. What is the one-variable solution to a linear equation?

A linear equation with one variable has the usual form Ax + B = 0. Here, x is the variable, A is its coefficient, and B is its constant term.

3. How are linear equations with one variable solved?

The following are the steps to solving linear equations in one variable worksheet:

  • By adding and subtracting on both sides of the equation, keep the variable term on one side and the constants on the other.

  • Make the constant terms simpler.

  • By multiplying or dividing the same term into both sides of the equation.

  • Simplify, then type the response.

4. Can there be more than one variable in a linear equation?

Yes, there can be more than one variable in a linear equation. Such equations are referred to as linear equations in two or three variables. The two-variable linear equations have the form ax + by + c = 0, while the three-variable linear equation has the form ax + by + cz + d = 0. Here, the variables are x, y, and z; the coefficients are a, b, and c; and the constant is d. These linear equations are frequently utilized in linear programming to obtain the best solutions.

5. What is the power of a variable in a linear equation in one variable?

In a linear equation with just one variable, the variable's power is 1. For instance, the power of the variable "a" in the equation 3a + 4 = 11 is 1.