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The linear equations in one variable are of the form $ax+b=0$ where $a \neq 0$ and $a,b$ are real numbers and $x$ is the variable. Also, these equations have degree 1 ie. the maximum power of $x$ is 1. These linear equations in one variable have applications in our day-to-day life, for example, if you want to purchase a pen and you have been told that the cost of 20 pens is 5 less than 125. In this case equation will be like $20x+5=125$ where $x$ is the cost of 1 pen. Linear equation in one variable in Class 8 mainly covers solving linear equations in one variable with variable on both sides and solving linear equations in the form of fractions. This chapter is an important part of the NCERT Syllabus for Class 8 Maths that lays a foundation for algebra. In this article, we provide comprehensive step-by-step solutions to all the exercise problems in the chapter.
These NCERT solutions are highly reliable as they are developed by the subject matter experts of Careers360. These solutions are of great help for the exam preparation as the students can practice the exercise problems and check the answers to evaluate their performance, and can focus on their weaker areas.
Linear Equations in One Variable:
The linear equation in one variable is an equation that involves only one variable and is in the form $ax+b=0$ where $a \neq 0 $, $x$ is a variable and $a,b$ are real numbers.
Linear equations with variables on both sides:
Transpose both the number and the variable to one side, maintaining the sign of the number.
Solve (add/subtract) the equation on both sides to simplify and find the variable's value.
Linear equations that are reducible to a simpler form:
If the equation is of the form $\frac{x+a}{c}=\frac{x+b}{d}$, cross-multiply the numerator and denominator to simplify it to a linear form, such as $(x + a) \times d = c \times (x + b)$
Solve this linear equation with variables on both sides to find the variable's value.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations with One Variable Exercise 2.1 Page No: 17 Number of Questions: 10 |
Q1 Solve the following equations and check your results 3x = 2x + 18
Answer: We have
3x = 2x + 18
Subtracting 2x from both sides
3x - 2x = 2x - 2x + 18
or x = 18
Check:- Put x = 18 in both LHS and the RHS.
LHS = 3x = 3(18) = 54
RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54
Thus, LHS = RHS
Q2 Solve the following equation and check your results 5t – 3 = 3t – 5
Answer: We have
5t - 3 = 3t - 5
Transposing 3t to the LHS and -3 to the RHS, we get:
5t - 3t = -5 + 3
or 2t = -2
Dividing both sides by 2 :
t = -1
Check:- Put t = -1 in the LHS we have,
5t - 3 = 5(-1) - 3 = -5 -3 = -8
Similarly, put t = -1 in the RHS:
3t - 5 = 3(-1) - 5 = -3 - 5 = -8
Hence, LHS = RHS
Q3 Solve the following equation and check your result 5x + 9 = 5 + 3x
Answer: We have
5x + 9 = 5 + 3x
Transposing 3x to the LHS and 9 to the RHS, we get:
5x - 3x = 5 - 9
or 2x = -4
Dividing both sides by 2 :
x = -2
Check:- Put x = - 2 in both LHS and RHS
LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1
RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1
Hence, LHS = RHS
Q4 Solve the following equation and check your result. 4z + 3 = 6 + 2z
Answer: We have
4z + 3 = 6 +2z
Transposing 2z to the LHS and 3 to the RHS, we get:
4z - 2z = 6 - 3
or 2z = 3
Dividing both sides by 2,
$z = \frac{3}{2}$
Check:- Put $z = \frac{3}{2}$ in LHS as well as RHS, we have
LHS :- 4z + 3
$= 4\times \frac{3}{2} + 3 = 9$
RHS :- 6 + 2z
$= 6 + 2\times \frac{3}{2} = 6 + 3 = 9$
Thus, LHS = RHS
Q5 Solve the following equation and check your result 2x – 1 = 14 – x
Answer: We have
2x - 1 = 14 - x
Transposing -x to the LHS and -1 to the RHS
2x + x = 14 + 1
or 3x = 15
Dividing both sides by 3, we get
x = 5
Check:- Put x = 5 in both the LHS and the RHS
LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9
RHS :- 14 - x = 14 - 5 = 9
Hence, LHS = RHS
Q6 Solve the following equation and check your result 8x + 4 = 3 (x – 1) + 7
Answer: We have
8x + 4 = 3 (x - 1) + 7
or 8x + 4 = 3x - 3 + 7 = 3x + 4
Transposing 3x to the LHS and 4 to the RHS, we get
8x - 3x = 4 - 4 = 0
5x = 0
Dividing both sides by 5:
x = 0
Check:- Putting x = 0 in both LHS and RHS:
LHS :- 8x + 4 = 8(0) + 4 = 4
RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4
Hence, LHS = RHS
Q7 Solve the following equations and check your results.
$x =\frac{4}{5}\times \left ( x+10 \right )$
Answer: We have
$x = \frac{4}{5}\left ( x + 10 \right )$
$= \frac{4}{5}x + 8$
Transposing $\frac{4}{5}x$ to the left-hand side:
$x - \frac{4}{5}x = 8$
or $\frac{5x-4x}{5} = 8$
or $\frac{x}{5} = 8$
Multiplying both sides by 5
x = 40
Check:- Put x = 40 in both the LHS and the RHS.
LHS :- x = 40
RHS :- $\frac{4}{5}\left ( x + 10 \right ) = \frac{4}{5}\left ( 40 + 10 \right )$
$= \frac{4}{5}\left ( 50 \right ) = 40$
Thus, LHS = RHS
Q8 Solve the following equation and check your result.
$\frac{2x}{3}+1 = \frac{7x}{15}+3$
Answer: We have
$\frac{2x}{3} + 1 = \frac{7x}{15} + 3$
Transposing $\frac{7x}{15}$ to the LHS and 1 to the RHS:
$\frac{2x}{3} - \frac{7x}{15} = 3 - 1 = 2$
or $\frac{10x - 7x}{15} = 2$
or $\frac{3x}{15} = 2$
or x = 10
Check:- Put x = 10 in both the LHS and the RHS
LHS :
$\frac{2x}{3} + 1 = \frac{2}{3}\times 10 + 1 = \frac{23}{3}$
RHS :
$\frac{7x}{15} + 3 = \frac{7}{15}\times 10 + 3 = \frac{14}{3} + 3 = \frac{23}{3}$
Thus, LHS = RHS
Q9 Solve the following equations and check your results.
$2y+\frac{5}{3} = \frac{26}{3} -y$
Answer: We have
$2y+ \frac{5}{3} = \frac{26}{3} - y$
Transposing -y to the LHS and $\frac{5}{3}$ to the RHS, we get
$2y + y = \frac{26}{3}- \frac{5}{3} = \frac{21}{3} = 7$
or 3y = 7
Dividing both sides by 3:
$y = \frac{7}{3}$ i
Check:- Put $y = \frac{7}{3}$ in both the LHS and the RHS:
LHS :
$2y + \frac{5}{3} = 2\times \frac{7}{3} + \frac{5}{3} = \frac{19}{3}$
RHS :
$\frac{26}{3} - \frac{7}{3} = \frac{19}{3}$
Hence, LHS = RHS
Q10 Solve the following equation and check your result.
$3m = 5m - \frac{8}{5}$
Answer: We have
$3m = 5m - \frac{8}{5}$
Transposing 5m to the LHS
$3m -5m = - \frac{8}{5}$
or $-2m = - \frac{8}{5}$
Dividing both sides by -2, we get
$m = \frac{4}{5}$
Check:- Put $m = \frac{4}{5}$ in the LHS and the RHS:
LHS:-
$3m = 3\times \left ( \frac{4}{5} \right ) = \frac{12}{5}$
RHS:-
$5m - \frac{8}{5} = 5 \times \frac{4}{5} - \frac{8}{5} = \frac{12}{5}$
Hence, LHS = RHS
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations with One Variable Exercise 2.2 Page No: 19 Number of Questions: 10 |
Q1 Solve the following linear equations.
$\frac{x}{2}-\frac{1}{5} = \frac{x}{3}+\frac{1}{4}$
Answer: The detailed solution to the above question is here.
We have
$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$
or $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$
or $\frac{3x - 2x}{6} = \frac{4 +5}{20}$
or $\frac{x}{6} = \frac{9}{20}$
Multiplying both sides by 6:
$x = \frac{9}{20}\times 6 = \frac{27}{10}$
Thus $x = \frac{27}{10}$
Q2 Solve the following linear equations.
$\frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21$
Answer: We have
$\frac{n}{2}- \frac{3n}{4}+\frac{5n}{6} = 21$
Multiplying both sides by 12.
(We multiplied both sides by 12 because it is the lowest common factor.)
6n - 9n + 10n = 252
or 7n = 252
or n = 36
Q3 Solve the following linear equations.
$x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}$
Answer: We have
$x+7 -\frac{8x}{3} = \frac{17}{6}- \frac{5x}{2}$
Multiplying both sides by 6, we get
6x + 42 - 16x = 17 - 15x
or -10x +42 = 17 - 15x
Transposing -15x to the LHS and 42 to the RHS :
-10x + 15x = 17 - 42
or 5x = -25
or x = -5
Q4 Solve the following linear equations.
$\frac{x-5}{3} = \frac{x-3}{5}$
Answer: We have
$\frac{x-5}{3} = \frac{x-3}{5}$
Multiply both sides by 15. (because the LCM of the denominator is 15 )
5x - 25 = 3x - 9
Transposing 3x to the LHS and -25 to the RHS.
5x - 3x = -9 + 25
or 2x = 16
x = 8
Q5 Solve the following linear equations.
$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$
Answer: We have
$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$
Multiplying both sides by 12, we get
9t - 6 - ( 8t + 12 ) = 8 - 12t
or 9t - 6 - 8t - 12 = 8 - 12t
or t - 18 = 8 - 12t
Transposing -18 to the RHS and -12t to the LHS:
12t + t = 8 + 18 = 26
or 13t = 26
Thus, t = 2
Q6 Solve the following linear equation.
$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$
Answer: We have
$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$
Multiplying both sides by 6 (as it is the LCM of the denominator)
6m - 3(m - 1) = 6 - 2(m - 2)
or 6m - 3m + 3 = 6 - 2m + 4
or 3m + 3 = 10 - 2m
Transposing -2m to the LHS and 3 to the RHS:
3m + 2m = 10 - 3
or 5m = 7
$m = \frac{7}{5}$
Q7 Simplify and solve the following linear equation. 3(t – 3) = 5(2t + 1)
Answer: Let us open the brackets :
LHS : 3(t - 3) = 3t - 9
RHS : 5(2t + 1) = 10t + 5
The equation is 3t - 9 = 10t + 5
or -9 = 10t - 3t + 5
or -9 - 5 = 7t
or -14 = 7t
Therefore t = -2
Q8 Simplify and solve the following linear equation. 15(y – 4) –2(y – 9) + 5(y + 6) = 0
Answer: Let us open the brackets
Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0
or 18y - 12 = 0
or 18y = 12
$y = \frac{12}{18} = \frac{2}{3}$
Q9 Simplify and solve the following linear equation. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Answer: Let us open the brackets.
LHS: 3(5z - 7) - 2(9z - 11)
= 15z - 21 - 18z + 22
= -3z + 1
RHS: 4(8z - 13) - 17
= 32z - 52 - 17
= 32z - 69
So the equation becomes : -3z + 1 = 32z - 69
or 69 + 1 = 32z + 3z
or 70 = 35z
z = 2
Q10 Simplify and solve the following linear equations. 0.25(4f – 3) = 0.05(10f – 9)
Answer: We have
0.25(4f – 3) = 0.05(10f – 9)
Let us open the brackets:
f - 0.75 = 0.5f - 0.45
or f - 0.5f = - 0.45 + 0.75
or 0.5f = 0.30
Dividing both sides by 0.5, we get
f = 0.6
Linear equations with variables on both sides:
Transpose both the number and the variable to one side, maintaining the sign of the number.
Solve (add/subtract) the equation on both sides to simplify and find the variable's value.
Linear equations that are reducible to a simpler form:
If the equation is of the form $\frac{x+a}{c}=\frac{x+b}{d}$, cross-multiply the numerator and denominator to simplify it to a linear form, such as $(x + a) \times d = c \times (x + b)$
Solve this linear equation with variables on both sides to find the variable's value.
The NCERT Subject-wise Solutions for Class 8 is an essential study resource that helps students get trustworthy solutions for each subject to make the exam preparation easier. Students can refer to those solutions using the links below.
Students can also refer to the NCERT Books and the NCERT Syllabus using the links below:
Multiplication and division of algebraic equations, identities, factorization, and the method of solving linear equations in one variable are the important topics of this chapter.
CBSE class 8 maths is base and foundation for the upcoming classes. Most of the topics are related to previous classes. It is basic maths and very easy.
No, CBSE doesn’t provide NCERT solutions for any class or subject.
Here you will get the detailed NCERT solutions for class 8 by clicking on the link.
Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.
NCERT textbook is the best book for CBSE class 8 maths. You don't need to buy any supplementary book. All you need to do is a rigorous practice of NCERT problems.
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