## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - Free PDF

Suppose you are looking for the simplest and easiest way of understanding Class 8 Maths Chapter 2 concepts, then you must avail of Linear Equations in One Variable Class 8 CBSE Solutions created by the expert team of Vedantu. Keeping in mind the understanding level of class 8 students, the scholars have designed the NCERT Solutions for Class 8 Maths Chapter 2 after a lot of research. The solutions also adhere to the latest CBSE guidelines; hence students can expect to get good grades in maths after going through our Class 8 Maths NCERT Solution Chapter 2. You can also download NCERT Solutions for Class 8 Science on our website.

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## Download PDF of NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

A) $1$

B) $\dfrac{2}{3}$

C) $\dfrac{6}{5}$

D) $\dfrac{7}{3}$

A) $2$

B) $4$

C) $6$

D) $8$

$ x \times \dfrac{9}{{\sqrt 2 }} = 3\sqrt 2 $

I.Five years hence, the ratio of their ages would be \[9:5\]

II.Ten years back, the ratio of their ages was \[3:1\]

A.I alone sufficient while II alone not sufficient to answer

B.II alone sufficient while I alone not sufficient to answer

C.Either I or II alone sufficient to answer

D.Both I and II are not sufficient to answer

E.Both I and II are necessary to answer

A) x years

B) y years

C) z years

D) $\left( {x + y + z} \right)$ years

## Access NCERT solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

Exercise 2.1

1. Solve: $x-2=7$

Ans:

\[x-2=7\]

Transposing $2$ to R.H.S, we obtain

$x=7+2=9$

2. Solve:$y+3=10$

Ans:

$y+3=10$

Transposing $3$ to R.H.S, we obtain

$y=10-3=7$

3. Solve: $6=z+2$

Ans:

$6=z+2$

Transposing $2$ to L.H.S, we obtain

$6-2=z$

$z=4$

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

Ans:

$\frac{3}{7}+x=\frac{17}{7}$

Transposing $\frac{3}{7}$ to R.H.S, we obtain

$x=\frac{17}{7}-\frac{3}{7}=\frac{14}{7}=2$

5. Solve:$6x=12$

Dividing both sides by $6$ , we obtain

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

$x=2$

6. Solve: $\frac{t}{5}=10$

Ans:

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

Multiplying both sides by $5$, we obtain

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

$t=50$

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

Ans:

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

Multiplying both sides by$\frac{3}{2}$ , we obtain

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

$x=27$

8. Solve: $1.6=\frac{y}{1.5}$

Ans:

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

Multiplying both sides by$1.5$, we obtain

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

$2.4=y$

9. Solve:$7x-9=16$

Ans:

$7x-9=16$

Transposing$9$ to R.H.S, we obtain

$7x=16+9$

$7x=25$

Dividing both sides by$7$, we obtain

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

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

10. Solve: $14y-8=13$

Ans:

$14y-8=13$

Transposing $8$ to R.H.S, we obtain

$14y=13+8$

$14y=21$

Dividing both sides by$14$, we obtain

$\frac{14y}{14}=\frac{21}{14}$

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

11. Solve: $17+6p=9$

Ans:

$17+6p=9$

Transposing $17$ to R.H.S, we obtain

$6p=9-17$

$6p=-8$

Dividing both sides by$6$, we obtain

$\frac{6p}{6}=-\frac{8}{6}$

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

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

Ans:

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

Transposing $1$ to R.H.S, we obtain

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

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

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

Multiplying both sides by $3$ , we obtain

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

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

Exercise 2.2

1. If you subtract$\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$ , $\frac{1}{8}$you get . What is the number?

Ans:

Let the number be x. According to the question,

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}=\frac{1}{8}$

On multiplying both sides by$2$, we obtain

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}\times 2=\frac{1}{8}\times 2$

$x-\frac{1}{2}=\frac{1}{4}$

On transposing $\frac{1}{2}$ to R.H.S, we obtain

$x=\frac{1}{4}+\frac{1}{2}$

$\,\,\,\,=\frac{1+2}{4}=\frac{3}{4}$

Therefore, the number is$\frac{3}{4}$ .

2. The perimeter of a rectangular swimming pool is $154$ m. Its length is $2$ m more than twice its breadth. What are the length and the breadth of the pool?

Ans:

Let the breadth be x m. The length will be$\left( 2x+2 \right)$ m.

Perimeter of swimming pool $=2\left( l+b \right)=154$m

$2\left( 2x+2+x \right)=154$

$2\left( 3x+2 \right)=154$

Dividing both sides by$2$, we obtain

$\frac{2\left( 3x+2 \right)}{2}=\frac{154}{2}$

$3x+2=77$

On Transposing $2$ to R.H.S, we obtain

$3x=77-2$

$3x=75$

On dividing both sides by$3$, we obtain

$\frac{3x}{3}=\frac{75}{3}$

$x=25$

$2x+2=2\times 25+2=52$

Hence, the breadth and length of the pool are $25$ m and$52$ m respectively.

3. The base of an isosceles triangle is $\frac{4}{3}$ cm. The perimeter of the triangle is $4\frac{2}{15}$ cm. What is the length of either of the remaining equal sides?

Ans:

Let the length of equal sides be x cm.

Perimeter$=\text{x }cm+\,\text{x}\,cm+\,\text{Base}=4\frac{2}{15}\text{cm}$

$2x+\frac{4}{3}=\frac{62}{15}$

On Transposing $\frac{4}{3}$ to R.H.S, we obtain

$2x=\frac{62}{15}-\frac{4}{3}$

$2x=\frac{62-4\times 5}{15}=\frac{62-20}{15}$

$2x=\frac{42}{15}$

On dividing both sides by$2$, we obtain

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$

$x=\frac{7}{5}=1\frac{2}{5}$

Therefore, the length of equal sides is $1\frac{2}{5}$ cm.

4. Sum of two numbers is$95$. If one exceeds the other by $15$, find the numbers.

Ans:

Let one number be x. Therefore, the other number will be$\text{x}+15$. According to the question,

$x+x+15=95$

$2x+15=95$

On Transposing $15$ to R.H.S, we obtain

$2x=95-15$

$2x=80$

On dividing both sides by$2$, we obtain

$\frac{2x}{2}=\frac{80}{2}$

$x=40$

$x+15=40+15=55$

Hence, the numbers are$40\text{ and }55$.

5. Two numbers are in the ratio $5:3$. If they differ by$18$, what are the numbers?

Ans:

Let the common ratio between these numbers be x. Therefore, the numbers will be $5\text{x and }3\text{x}$respectively.

Difference between these numbers$=18$

$5\text{x}-3\text{x}=18$

$2x=18$

Dividing both sides by$2$,

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

$x=9$

First number $=5x=5\times 9=45$

Second number $=3x=3\times 9=27$

6. Three consecutive integers add up to$51$. What are these integers?

Ans:

Let three consecutive integers be$x,x+1\text{ and }x+2$.

Sum of these numbers$=x+x+1+x+2=51$

$3x+3=51$

On Transposing $3$ to R.H.S, we obtain

$3x=51-3$

$3x=48$

On dividing both sides by$3$, we obtain

$\frac{3x}{3}=\frac{48}{3}$

$x=16$

$x+1=17$

$x+2=18$

Hence, the consecutive integers are$16,17,\,\text{and }18$ .

7. The sum of three consecutive multiples of $8\,\,\,\text{is }888$. Find the multiples.

Ans:

Let the three consecutive multiples of $8\,\text{be }8x,\,8\left( x+1 \right),\,8\left( x+2 \right)$.

Sum of these numbers$=8x+8\left( x+1 \right)+8\left( x+2 \right)=888$

$8\left( x+x+1+x+2 \right)=888$

$8\left( 3x+3 \right)=888$

On dividing both sides by$8$, we obtain

$\frac{8\left( 3x+3 \right)}{8}=\frac{888}{8}$

$3x+3=111$

On Transposing $3$ to R.H.S, we obtain

$3x=111-3$

$3x=108$

On dividing both sides by $3$ , we obtain

$\frac{3x}{3}=\frac{108}{3}$

$x=36$

First multiple $=8x=8\times 36=288$

Second multiple $=8\left( x+1 \right)=8\times \left( 36+1 \right)=8\times 37=296$

Third multiple $=8\left( x+2 \right)=8\times \left( 36+2 \right)=8\times 38=304$

Hence, the required numbers are$288,296,\text{and }304$ .

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by $2,3\,\text{and }4$respectively, they add up to $74$. Find these numbers.

Ans:

Let three consecutive integers be $\text{x,}\,\text{x}+1\text{,}\,\text{x}+2$. According to the question,

$2x+3\left( x+1 \right)+4\left( x+2 \right)=74$

$2x+3x+3+4x+8=74$

$9x+11=74$

On Transposing $11$ to R.H.S, we obtain

$9x=74-11$

$9x=63$

On dividing both sides by$9$, we obtain

$\frac{9x}{9}=\frac{63}{9}$

$x=7$

$x+1=7+1=8$

$x+2=7+2=9$

Hence, the numbers are $7,8\,\text{and }9$.

9. The ages of Rahul and Haroon are in the ratio$5:7$. Four years later the sum of their ages will be $56$ years. What are their present ages?

Ans: Let common ratio between Rahul’s age and Haroon’s age be x. Therefore, the age of Rahul and Haroon will be $5x$ years and $7x$ years respectively. After $4$ years, the age of Rahul and Haroon will be $\left( 5x+4 \right)$ years and $\left( 7x+4 \right)$ years respectively.

According to the given question, after$4$ years, the sum of the ages of Rahul and Haroon is $56$ years.

$\therefore \left( 5x+4+7x+4 \right)=56$

$12x+8=56$

On Transposing $8$ to R.H.S, we obtain

$12x=56-8$

$12x=48$

On dividing both sides by $12$, we obtain

$\frac{12x}{12}=\frac{48}{12}$

$x=4$

Rahul’s age $=5x$ years $=\left( 5\times 4 \right)$ years $=20$ years

Haroon’s age $=7x$ years $=\left( 7\times 4 \right)$years $=28$ years

10. The number of boys and girls in a class are in the ratio$7:5$. The number of boys is $8$ more than the number of girls. What is the total class strength?

Ans: Let the common ratio between the number of boys and numbers of girls be x.

Number of boys $=7x$

Number of girls $=5x$

According to the given question,

Number of boys $=$ Number of girls $+8$

$\therefore 7x=5x+8$

On Transposing $5x$ to L.H.S, we obtain

$7x-5x=8$

$2x=8$

On dividing both sides by\[2\] , we obtain

\[\frac{2x}{2}=\frac{8}{2}\]

\[x=4\]

Number of boys \[=7x=7\times 4=28\]

Number of girls \[=5x=5\times 4=20\]

Hence, total class strength \[=28+20=48\] students

11. Baichung’s father is $26$ years younger than Baichung’s grandfather and $29$ years older than Baichung. The sum of the ages of all the three is $135$ years. What is the age of each one of them?

Ans:

Let Baichung’s father’s age be x years. Therefore, Baichung’s age and Baichung’s grandfather’s age will be \[\left( x-29 \right)\] years and \[\left( x+26 \right)\] years respectively. According to the given question, the sum of the ages of these 3 people is \[135\] years.

\[\therefore x+x-29+x+26=135\]

\[3x-3=135\]

On Transposing \[3\] to R.H.S, we obtain

\[3x=135+3\]

\[3x=138\]

On dividing both sides by \[3\], we obtain

\[\frac{3x}{3}=\frac{138}{3}\]

\[x=46\]

Baichung’s father’s age \[=x\] years \[=46\] years

Baichung’s age \[=\left( x-29 \right)\] years \[=\left( 46-29 \right)\] years = \[17\] years

Baichung’s grandfather’s age \[=\left( x+26 \right)\] years \[=\left( 46+26 \right)\] years \[=72\] years

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Ans:

Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age \[=4\times \] His present age

\[x+15=4x\]

On Transposing x to R.H.S, we obtain

\[15=4x-x\]

\[15=3x\]

On dividing both sides by\[3\] , we obtain

\[\frac{15}{3}=\frac{3x}{3}\]

\[5=x\]

Hence, Ravi’s present age \[=5\]years

13. A rational number is such that when you multiply it by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, you get$-\frac{7}{12}$ . What is the number?

Ans:

Let the number be x.

According to the given question,

\[\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}\]

On Transposing\[\frac{2}{3}\] to R.H.S, we obtain

\[\frac{5}{2}x=-\frac{7}{12}-\frac{2}{3}\]

\[\frac{5}{2}x=\frac{-7-\left( 2\times 4 \right)}{12}\]

\[\frac{5}{2}x=-\frac{15}{12}\]

On multiplying both sides by\[\frac{2}{5}\] , we obtain

\[x=-\frac{15}{12}\times \frac{2}{5}=-\frac{1}{2}\]

Hence, the rational number is \[-\frac{1}{2}\].

14. Lakshmi is a cashier in a bank. She has currency notes of denominations$\text{Rs }100,\,\text{Rs }50\,\,\,\text{and Rs}\,10$, respectively. The ratio of the number of these notes is$2:3:5$. The total cash with Lakshmi is$\text{Rs }4,00,000$. How many notes of each denomination does she have?

Ans:

Let the common ratio between the numbers of notes of different denominations be x. Therefore, numbers of \[\text{Rs }100\] notes, \[\text{Rs }50\] notes, and \[\text{Rs }10\]notes will be \[2x,3x,\,\text{and }5x\] respectively.

Amount of \[\text{Rs }100\]notes \[=\text{Rs}\left( 100\times 2\text{x} \right)=\text{Rs}\,200\text{x}\]

Amount of \[\text{Rs }50\]notes \[=\text{Rs}\left( 50\times 3\text{x} \right)=\text{Rs}\,150\text{x}\]

Amount of \[\text{Rs }10\]note\[=\text{Rs}\left( 10\times 5\text{x} \right)=\text{Rs}\,50\text{x}\]

It is given that the total amount is \[\text{Rs}\,400000\].

\[\therefore 200\text{x+}150\text{x+}50\text{x=400000}\]

\[\Rightarrow \text{400x=400000}\]

On dividing both sides by\[400\], we obtain

\[x=1000\]

Number of Rs 100 notes \[=2x=2\times 1000=2000\]

Number of Rs 50 notes \[=3x=3\times 1000=3000\]

Number of Rs 10 notes \[=5x=5\times 1000=5000\]

15. I have a total of $\text{Rs}\,300$in coins of denomination$\text{Re }1,\text{Re }2\,\text{and Re }5$. The number of $\text{Rs }2$coins is$3$ times the number of $\text{Rs }5$ coins. The total number of coins is$160$. How many coins of each denomination are with me?

Ans:

Let the number of \[\text{Rs }5\] coins be x.

Number of \[\text{Rs }2\] coins \[=3\times \]Number of \[\text{Rs }5\] coins\[=3\text{x}\]

Number of \[\text{Re }1\] coins \[=160-\] (Number of coins of \[\text{Rs }5\] and of\[\text{Rs 2}\])

Amount \[\text{Re }1\]of coins \[=\,\text{Rs }\left[ 1\times \left( 160-4x \right) \right]=\,\text{Rs}\,\left( 160-4x \right)\]

Amount of \[\text{Rs }2\] coins \[=\,\text{Rs}\,\,\left( 2\times 3x \right)=\,\text{Rs}\,\,\text{6x}\]

Amount of \[\text{Rs 5}\]coins \[=\,\text{Rs}\,\,\left( 5\times x \right)=\,\text{Rs}\,\,5\text{x}\]

It is given that the total amount is \[\text{Rs 300}\].

\[\therefore 160-4x+6x+5x=300\]

\[160+7x=300\]

On Transposing \[160\] to R.H.S, we obtain

\[7x=300-160\]

\[7x=140\]

On dividing both sides by \[7\], we obtain

\[\frac{7x}{7}=\frac{140}{7}\]

\[x=20\]

Number of \[\text{Re }1\] coins \[=160-4x=160-4\times 20=160-80=80\]

Number of \[\text{Rs }2\] coins\[=3x=3\times 20=60\]

Number of \[\text{Rs 5}\] coins \[=x=20\]

16. The organizers of an essay competition decide that a winner in the competition gets a prize of $\text{Rs 100}$ and a participant who does not win gets a prize of$\text{Rs 25}$. The total prize money distributed is $\text{Rs 3000}$. Find the number of winners, if the total number of participants is $63$ .

Ans:

Let the number of winners be x. Therefore, the number of participants who did not win will be \[63-x\].

Amount given to the winners \[=\text{Rs }\left( 100\times x \right)=~~\text{Rs 100}x\]

Amount given to the participants who did not win \[=\text{Rs }\left[ 25\left( 63-x \right) \right]\]

\[=\text{Rs }\left( 1575-25x \right)\]

According to the given question,

\[100x+1575-25x=3000\]

On Transposing \[1575\] to R.H.S, we obtain

\[75x=3000-1575\]

\[75x=1425\]

On dividing both sides by \[75\] , we obtain

\[x=19\]

Hence, number of winners \[=19\]

Exercise 2.3

1. Solve and check result: $3x=2x+18$

Ans:

\[3x=2x+18\]

On Transposing \[2x\] to L.H.S, we obtain

\[3x-2x=18\]

\[x=18\]

L.H.S \[=3x=3\times 18=54\]

R.H.S \[=2x+18=2\times 18+18=36+18=54\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

2. Solve and check result: $5t-3=3t-5$

Ans:

\[5t-3=3t-5\]

On Transposing \[3t\] to L.H.S and \[-3\] to R.H.S, we obtain

\[5t-3=-5-\left( -3 \right)\]

\[2t=-2\]

On dividing both sides by\[2\], we obtain

\[t=-1\]

L.H.S \[=5t-3=5\times \left( -1 \right)-3=-8\]

R.H.S \[=3t-5=3\times \left( -1 \right)-5=-3-5=-8\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

3. Solve and check result: $5x+9-5+3x$

Ans:

\[5x+9=5+3x\]

On Transposing \[3x\] to L.H.S and \[9\] to R.H.S, we obtain

\[5x-3x=5-9\]

\[2x=-4\]

On dividing both sides by\[2\], we obtain

\[x=-2\]

L.H.S \[=5x+9=5\times \left( -2 \right)+9=-10+9=-1\]

R.H.S \[=5+3x=5+3\times \left( -2 \right)=5-6=-1\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

4. Solve and check result: $4z+3=6+2z$

Ans:

\[4z+3=6+2z\]

On Transposing \[2z\] to L.H.S and \[3\] to R.H.S, we obtain

\[4z-2z=6-3\]

\[2z=3\]

Dividing both sides by\[2\] , we obtain

L.H.S \[=4z+3=4\times \left( \frac{3}{2} \right)+3=6+3=9\]

R.H.S \[=6+2z=6+2\times \left( \frac{3}{2} \right)=6+3=9\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

5. Solve and check result: $2x-1=14-x$

Ans:

\[2x-1=14-x\]

Transposing x to L.H.S and $1$ to R.H.S, we obtain

\[2x+x=14+1\]

\[3x=15\]

Dividing both sides by \[3\], we obtain

\[x=5\]

L.H.S \[=2x-1=2\times \left( 5 \right)-1=10-1=9\]

R.H.S \[=14-x=14-5=9\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

6. Solve and check result: $8x+4=3\left( x-1 \right)+7$

Ans:

\[8x+4=3\left( x-1 \right)+7\]

\[8x+4=3x-3+7\]

Transposing \[3x\] to L.H.S and $4$ to R.H.S, we obtain

\[8x-3x=-3+7-4\]

\[5x=-7+7\]

\[x=0\]

L.H.S \[=8x+4=8\times \left( 0 \right)+4=4\]

R.H.S \[=3\left( x-1 \right)+7=3\left( 0-1 \right)+7=-3+7=4\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

7. Solve and check result: $x=\frac{4}{5}\left( x+10 \right)$

Ans:

\[x=\frac{4}{5}\left( x+10 \right)\]

Multiplying both sides by\[5\], we obtain

\[5x=4\left( x+10 \right)\]

\[5x=4x+40\]

Transposing \[4x\] to L.H.S, we obtain

\[5x=4x+40\]

\[x=40\]

L.H.S \[=x=40\]

R.H.S \[=\frac{4}{5}\left( x+10 \right)=\frac{4}{5}\left( 40+10 \right)=\frac{4}{5}\times 50=40\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

8. Solve and check result: $\frac{2x}{3}+1=\frac{7x}{15}+3$

Ans:

\[\frac{2x}{3}+1=\frac{7x}{15}+3\]

Transposing \[\frac{7x}{15}\] to L.H.S and $1$ to R.H.S, we obtain

\[\frac{2x}{3}-\frac{7x}{15}=3-1\]

\[\frac{5\times 2x-7x}{15}=2\]

\[\frac{3x}{15}=2\]

\[\frac{x}{5}=2\]

Multiplying both sides by\[5\] , we obtain

\[x=10\]

L.H.S \[=\frac{2x}{3}+1=\frac{2\times 10}{3}+1=\frac{2\times 10+1\times 3}{3}=\frac{23}{3}\]

R.H.S\[=\frac{7x}{15}+3=\frac{7\times 10}{15}+3=\frac{7\times 2}{3}+3=\frac{14}{3}+3=\frac{14+3\times 3}{3}=\frac{23}{3}\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

9. Solve and check result: $2y+\frac{5}{3}=\frac{26}{3}-y$

Ans:

\[2y+\frac{5}{3}=\frac{26}{3}-y\]

Transposing y to L.H.S and \[\frac{5}{3}\] to R.H.S, we obtain

\[2y+y=\frac{26}{3}-\frac{5}{3}\]

\[3y=\frac{21}{3}=7\]

Dividing both sides by$3$, we obtain

\[y=\frac{7}{3}\]

L.H.S \[=2y+\frac{5}{3}=2\times \frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}\]

R.H.S = \[\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}=\frac{19}{3}\]

L.H.S. = R.H.S. Hence, the result obtained above is correct.

10. Solve and check result: $3m=5m-\frac{8}{5}$

Ans:

\[3m=5m-\frac{8}{5}\]

Transposing \[5m\] to L.H.S, we obtain

\[3m-5m=-\frac{8}{5}\]

\[-2m=-\frac{8}{5}\]

Dividing both sides by\[-2\] , we obtain

\[m=\frac{4}{5}\]

L.H.S \[=3m=3\times \frac{4}{5}=\frac{12}{5}\]

R.H.S \[5m-\frac{8}{5}=5\times \frac{4}{5}-\frac{8}{5}=\frac{12}{5}\]

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

Exercise 2.4

1. Amina thinks of a number and subtracts $\frac{5}{2}$ from it. She multiplies the result by$8$. The result now obtained is $3$ times the same number she thought of. What is the number?

Ans:

Let the number be x.

According to the given question,

\[8\left( x-\frac{5}{2} \right)=3x\]

\[8x-20=3x\]

Transposing \[3x\]to L.H.S and \[-20\] to R.H.S, we obtain

\[8x-3x=20\]

\[5x=20\]

Dividing both sides by \[5\], we obtain

\[x=4\]

Hence, the number is \[4\].

2. A positive number is $5$ times another number. If $21$ is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Ans:

Let the numbers be \[x\] and\[5x\] . According to the question,

\[21+5x=2\left( x+21 \right)\]

\[21+5x=2x+42\]

Transposing \[2x\] to L.H.S and 21 to R.H.S, we obtain

\[5x-2x=42-21\]

\[3x=21\]

Dividing both sides by \[3\], we obtain

\[x=7\]

\[5x=5\times 7=35\]

Hence, the numbers are $7\,\text{and }35$ respectively.

3. Sum of the digits of a two digit number is$9$ . When we interchange the digits it is found that the resulting new number is greater than the original number by$27$. What is the two-digit number?

Ans:

Let the digits at tens place and ones place be \[x\,\,\text{and }9-x\] respectively.

Therefore, original number \[=10x+\left( 9-x \right)=9x+9\]

On interchanging the digits, the digits at ones place and tens place will be \[x\,\,\text{and }9-x\] respectively.

Therefore, new number after interchanging the digits \[=10\left( 9-x \right)+x\]

\[=90-10x+x\]

\[=90-9x\]

According to the given question,

New number = Original number\[+27\]

\[90-9x=9x+9+27\]

\[90-9x=9x+36\]

Transposing \[9x\]to R.H.S and \[36\] to L.H.S, we obtain

\[90-36=18x\]

\[54=18x\]

Dividing both sides by \[18\], we obtain

\[3=x\text{ and }9-x=6\]

Hence, the digits at tens place and ones place of the number are \[3\text{ and }6\] respectively. Therefore, the two-digit number is\[9x+9=9\times 3+9=36\]

4. One of the two digits of a two digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get$88$. What is the original number?

Ans:

Let the digits at tens place and ones place be \[x\text{ and }3x\] respectively.

Therefore, original number \[=10x+3x=13x\]

On interchanging the digits, the digits at ones place and tens place will be \[x\text{ and }3x\]respectively.

Number after interchanging \[=10x+3x=13x\]

According to the given question,

Original number $+$ New number\[=88\]

\[13x+31x=88\]

\[44x=88\]

Dividing both sides by \[44\], we obtain

\[x=2\]

Therefore, original number \[=13x=13\times 2=26\]

By considering the tens place and ones place as \[3x\text{ and }x\]respectively, the two-digit number obtained is \[62\].

Therefore, the two-digit number may be \[26\text{ or }62\].

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of this mother’s present age. What are their present ages?

Ans:

Let Shobo’s age be x years. Therefore, his mother’s age will be $6x$ years.

According to the given question,

After \[5\text{ years, Shobo }\!\!'\!\!\text{ s age}=\frac{\text{Shobo }\!\!'\!\!\text{ s mother }\!\!'\!\!\text{ s present age}}{3}\]

\[x+5=\frac{6x}{3}\]

\[x+5=2x\]

Transposing x to R.H.S, we obtain

\[5=2x-x\]

\[5=x\]

\[6x=6\times 5=30\]

Therefore, the present ages of Shobo and Shobo’s mother will be \[5\text{ years and }30\,\text{years}\] respectively.

6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio$11:4$. At the rate $\text{Rs }100$ per metre it will cost the village panchayat $\text{Rs }75,000$ to fence the plot. What are the dimensions of the plot?

Ans:

Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be $11x$ m and $4x$ m respectively.

Perimeter of the plot \[=2\] (Length + Breadth) \[=\left[ 2\left( 11x+4x \right) \right]\text{m}=30x\text{ m}\]

It is given that the cost of fencing the plot at the rate of \[\text{Rs }100\] per metre is\[\text{Rs 75,000}\].

\[\therefore 100\times \text{Perimeter}=75000\]

\[100\times 30x=75000\]

\[3000x=75000\]

Dividing both sides by \[3000\], we obtain

\[x=25\]

Length \[=11\times \text{m}=\left( 11\times 25 \right)\text{m}=275\,\text{m}\]

Breadth \[=4\times \text{m}=\left( 4\times 25 \right)\text{m}=100\,\text{m}\]

Hence, the dimensions of the plot are 2\[275\text{ m and }100\text{ m}\] respectively.

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him $\text{Rs 50}$ per metre and trouser material that costs him $\text{Rs 90}$ per metre. For every $2$ meters of the trouser material he buys $3$metres of the shirt material. He sells the materials at $12%\text{ and }10%$ profit respectively. His total sale is $\text{Rs 36660}$. How much trouser material did he buy?

Ans:

Let $2x$ m of trouser material and $3x$ m of shirt material be bought by him.

Per metre selling price of trouser material \[\text{=}~\text{Rs }\left( 90+\frac{90\times 12}{100} \right)\text{=Rs 100}\text{.80}\]

Per metre selling price of shirt material \[\text{=}~\text{Rs }\left( 50+\frac{50\times 12}{100} \right)\text{=Rs 55}\]

Given that, total amount of selling \[\text{=}~\text{Rs }36660\]

\[100.80\times \left( 2x \right)+55\times \left( 3x \right)=36660\]

\[201.60x+165x=36660\]

\[366.60x=36660\]

Dividing both sides by \[366.60\], we obtain

\[x=100\]

Trouser material \[=2\times m=\left( 2\times 100 \right)m=200m\]

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest $9$ are drinking water from the pond. Find the number of deer in the herd.

Ans:

Let the number of deer be x.

Number of deer grazing in the field \[=\frac{x}{2}\]

Number of deer playing nearby\[=\frac{3}{4}\times \text{Number of remaining deer}\]

\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{4}\times \left( x-\frac{x}{2} \right)=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}\]

Number of deer drinking water from the pond $=9$

\[x-\left( \frac{x}{2}+\frac{3x}{8} \right)=9\]

\[x-\left( \frac{4x+3x}{8} \right)=9\]

\[x-\frac{7x}{8}=9\]

\[\frac{x}{8}=9\]

Multiplying both sides by\[8\], we obtain

\[x=72\]

Hence, the total number of deer in the herd is \[72\].

9. A grandfather is ten times older than his granddaughter. He is also $54$ years older than her. Find their present ages

Ans:

Let the granddaughter’s age be x years. Therefore, grandfather’s age will be \[10x\] years.

According to the question,

Grandfather’s age$=$ Granddaughter’s age \[+\,54\] years

\[10x=x+54\]

Transposing x to L.H.S, we obtain

\[10x-x=54\]

\[9x=54\]

\[x=6\]

Granddaughter’s age \[=x\] years \[=6\] years

Grandfather’s age \[=10x\] years \[=\left( 10\times 6 \right)\] years \[=60\] years

10. 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.

Ans:

Let Aman’s son’s age be x years. Therefore, Aman’s age will be $3x$ years. Ten years ago, their age was \[\left( x-10 \right)\] years and \[\left( 3x-10 \right)\] years respectively.

According to the question,

\[10\]years ago, Aman’s age \[=5\times \]Aman’s son’s age \[10\] years ago

\[3x-10=5\left( x-10 \right)\]

\[3x-10=5x-50\]

Transposing \[3x\] to R.H.S and \[50\] to L.H.S, we obtain

\[50-10=5x-3x\]

\[40=2x\]

Dividing both sides by\[2\], we obtain

\[20=x\]

Aman’s son’s age \[=x\] years \[=20\] years

Aman’s age \[=3x\] years \[=\left( 3\times 20 \right)\] years \[=60\] years

Exercise 2.5

1. Solve the linear equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

Ans:

\[\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\]

L.C.M. of the denominators, \[2,3,4,\text{and 5,}\]is.

Multiplying both sides by \[60\], we obtain

\[60\left( \frac{x}{2}-\frac{1}{5} \right)=60\left( \frac{x}{3}+\frac{1}{4} \right)\]

\[\Rightarrow 30x-12=20x+15\] (Opening the brackets)

\[\Rightarrow 30x-20x=15+12\]

\[\Rightarrow 10x=27\]

\[\Rightarrow x=\frac{27}{10}\]

2. Solve the linear equation$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

Ans:

\[\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21\]

L.C.M. of the denominators, \[2,4,\text{ and }6\text{ is }12\].

Multiplying both sides by\[12\], we obtain

\[6n-9n+10n=252\]

\[\Rightarrow 7n=252\]

\[\Rightarrow n=\frac{252}{7}\]

\[\Rightarrow n=36\]

3. Solve the linear equation $x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

Ans:

\[x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}\]

L.C.M. of the denominators, \[2,3,\text{ and }6,\text{is }6\].

Multiplying both sides by \[6\], we obtain

\[6x+42-16x=17-15x\]

\[\Rightarrow 6x-16x+15x=17-42\]

\[\Rightarrow 5x=-25\]

\[\Rightarrow x=\frac{-25}{5}\]

\[\Rightarrow x=-5\]

4. Solve the linear equation $\frac{x-5}{3}=\frac{x-3}{5}$

Ans:

\[\frac{x-5}{3}=\frac{x-3}{5}\]

L.C.M. of the denominators, \[3\text{ and }5,\text{ is }15\].

Multiplying both sides by\[15\], we obtain

\[5\left( x-5 \right)=3\left( x-3 \right)\]

\[\Rightarrow 5x-25=3x-9\] (Opening the brackets)

\[\Rightarrow 5x-3x=25-9\]

\[\Rightarrow 2x=16\]

\[\Rightarrow x=\frac{16}{2}\]

\[\Rightarrow x=8\]

5. Solve the linear equation $\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

Ans:

\[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\]

L.C.M. of the denominators, \[3\text{ and }4,\text{is}\,12\].

Multiplying both sides by \[12\], we obtain

\[3\left( 3t-2 \right)-4\left( 2t+3 \right)=8-12t\]

\[\Rightarrow 9t-6-8t-12=8-12t\] (Opening the brackets)

\[\Rightarrow 9t-8t+12t=8+6+12\]

\[\Rightarrow 13t=26\]

\[\Rightarrow t=\frac{26}{13}\]

\[\Rightarrow t=2\]

6. Solve the linear equation$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

Ans:

\[m-\frac{m-1}{2}=1-\frac{m-2}{3}\]

L.C.M. of the denominators, \[2\text{ and }3,\text{ is}\,\text{ }6\].

Multiplying both sides by \[6\], we obtain

\[6m-3\left( m-1 \right)=6-2\left( m-2 \right)\]

\[\Rightarrow 6m-3m+3=6-2m+4\] (Opening the brackets)

\[\Rightarrow 6m-3m+2m=6+4-3\]

\[\Rightarrow 5m=7\]

\[\Rightarrow m=\frac{7}{5}\]

7. Simplify and solve the linear equation $3\left( t-3 \right)=5\left( 2t+1 \right)$

Ans:

\[3\left( t-3 \right)=5\left( 2t+1 \right)\]

\[\Rightarrow 3t-9=10t+5\] (Opening the brackets)

\[\Rightarrow -9-5=10t-3t\]

\[\Rightarrow -14=7t\]

\[\Rightarrow t=\frac{-14}{7}\]

\[\Rightarrow t=-2\]

8. Simplify and solve the linear equation$15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0$

Ans:

\[15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0\]

\[\Rightarrow 15y-60-2y+18+5y+30=0\] (Opening the brackets)

\[\Rightarrow 18y-12=0\]

\[\Rightarrow 18y=12\]

\[\Rightarrow y=\frac{12}{8}=\frac{2}{3}\]

9.Simplify and solve the linear equation $3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17$

Ans:

\[3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17\]

\[\Rightarrow 15z-21-18z+22=32z-52-17\] (Opening the brackets)

\[\Rightarrow -3z+1=32z-69\]

\[\Rightarrow -3z-32z=-69-1\]

\[\Rightarrow -35z=-70\]

\[\Rightarrow z=\frac{70}{~35}=2\]

10. Simplify and solve the linear equation $0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)$

Ans:

\[0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)\]

$\frac{1}{4}\left( 4f-3 \right)=\frac{1}{20}\left( 10f-9 \right)$

Multiplying both sides by\[20\], we obtain

\[5\left( 4f-3 \right)=10f-9\]

\[\Rightarrow 20f-15=10f-9\] (Opening the brackets)

\[\Rightarrow 20f-10f=-9+15\]

\[\Rightarrow 10f=6\]

\[\Rightarrow f=\frac{3}{5}=0.6\]

Exercise 2.6

1. Solve: $\frac{8x-3}{3x}=2$

Ans:

\[\frac{8x-3}{3x}=2\]

On multiplying both sides by\[3x\] , we obtain

\[8x-3=6x\]

\[\Rightarrow 8x-6x=3\]

\[\Rightarrow 2x=3\]

\[\Rightarrow x=\frac{3}{2}\]

2. Solve: $\frac{9x}{7-6x}=15$

Ans:

\[\frac{9x}{7-6x}=15\]

On multiplying both sides by\[7-6x\], we obtain

\[9x=15\left( 7-6x \right)\]

\[\Rightarrow 9x=105-90x\]

\[\Rightarrow 9x+90x=105\]

\[\Rightarrow 99x=105\]

\[\Rightarrow x=\frac{105}{99}=\frac{35}{33}\]

3. Solve:

$\frac{z}{z+15}=\frac{4}{9}$

Ans:

\[\frac{z}{z+15}=\frac{4}{9}\]

On multiplying both sides by \[9\left( z+15 \right)\], we obtain

\[9z=4\left( z+15 \right)\]

\[\Rightarrow 9z=4z+60\]

\[\Rightarrow 9z-4z=60\]

\[\Rightarrow 5z=60\]

\[\Rightarrow z=12\]

4. Solve: $\frac{3y+4}{2-6y}=\frac{-2}{5}$

Ans:

\[\frac{3y+4}{2-6y}=\frac{-2}{5}\]

On multiplying both sides by\[5\left( 2-6y \right)\], we obtain

\[5\left( 3y+4 \right)=-2\left( 2-6y \right)\]

\[\Rightarrow 15y+20=-4+12y\]

\[\Rightarrow 15y-12y=-4-20\]

\[\Rightarrow 3y=-24\]

\[\Rightarrow y=-8\]

5. Solve: $\frac{7y+4}{y+2}=-\frac{4}{3}$

Ans:

\[\frac{7y+4}{y+2}=-\frac{4}{3}\]

On multiplying both sides by\[3\left( y+2 \right)\], we obtain

\[3\left( 7y+4 \right)=-4\left( y+2 \right)\]

\[\Rightarrow 21y+12=-4y-8\]

\[\Rightarrow 21y+4y=-8-12\]

\[\Rightarrow 25y=-20\]

\[\Rightarrow y=-\frac{4}{5}\]

6. The ages of Hari and Harry are in the ratio $5:7$. Four years from now the ratio of their ages will be$3:4$. Find their present ages.

Ans:

Let the common ratio between their ages be x. Therefore, Hari’s age and Harry’s age will be\[5x\]years and \[7x\] years respectively and four years later, their ages will be \[\left( 5x+4 \right)\] years \[\left( 7x+4 \right)\] years respectively.

According to the situation given in the question,

\[\frac{5x+4}{7x+4}=\frac{3}{4}\]

\[\Rightarrow 4\left( 5x+4 \right)=3\left( 7x+4 \right)\]

\[\Rightarrow 20x+16=21x+12\]

\[\Rightarrow 16-12=21x-20x\]

\[\Rightarrow 4=x\]

Hari’s age \[=5x\] years \[=\left( 5\times 4 \right)\] years \[=20\] years

Harry’s age \[=7x\] years \[=\left( 7\times 4 \right)\] years \[=28\] years

Therefore, Hari’s age and Harry’s age are \[20\] years and \[28\] years respectively.

7. The denominator of a rational number is greater than its numerator by $8$. If the numerator is increased by $17$ and the denominator is decreased by$1$, the number obtained is. Find the rational number.

Ans:

Let the numerator of the rational number be\[x\]. Therefore, its denominator will be\[x+8\]

The rational number will be\[\frac{x}{x+8}\]. According to the question,

\[\frac{x+17}{x+8-1}=\frac{3}{2}\]$$

\[\Rightarrow \frac{x+17}{x+7}=\frac{3}{2}\]

\[\Rightarrow 2\left( x+17 \right)=3\left( x+7 \right)\]

\[\Rightarrow 2x+34=3x+21\]

\[\Rightarrow 34-21=3x-2x\]

\[\Rightarrow 13=x\]

Numerator of the rational number \[=x=13\]

Denominator of the rational number \[=x+8=13+8=21\]

Rational number\[=\frac{13}{21}\]

### NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - PDF Download

Getting a PDF format for the answers to all the questions in Linear Equations in One Variable Class 8 is now simple. Just go to the official website of Vedantu and download the NCERT Solutions Class 8 Maths Chapter 2 PDF from there on your devices anywhere anytime. Once you have downloaded them, you do not have to worry about internet connection when you need to quickly revise some formulas or essential topics of Chapter 2 Class 8 Maths. The solutions are also printable so that a hard copy can come in handy while you are at the examination center.

### Chapter 2 - Linear Equations in One Variable

### 2.1 Introduction

In the introduction part of NCERT Maths Class 8 Chapter 2, you will be reminded of algebraic equations and expressions of earlier days. Those are of the format shown below:

Expresions - 5x + y, x + y, y + y2, etc.

Equations - 5x + y = 10, x + y = -2, y + y2= 9 etc.

From these examples, you can see that equations have an equality sign (=), which is not present in expressions. Algebraic expression involves variables, constants, and some mathematical operations like addition or multiplication. An equation is an expression that equates two expressions.

Students would learn the history of algebraic equations and their definitions. They would also get an idea of what is coming up in the chapter in this section.

### 2.2 Solving Equations With Linear Expression on One SIde and Numbers on the Other Side

This section of Maths NCERT Solutions Class 8 Chapter 2 starts by recalling the method for solving equations like the one shown below:

5x - 3 = 8

We solve this by adding 3 on both sides so we get

5x - 3 + 3 = 8 + 3

5x = 11, hence x = 2.2

The above example is a linear expression where the highest power of a variable is only 1. A linear equation that involves one variable can be marked on a straight line. A linear equation can be in one or more than one variable.

You are then presented with a few examples which are based on the method of solving linear equations as described above.

### 2.3 Some Applications

In this topic, students will go through some applications of linear equations which are in the form of puzzles. There are many wordy samples solved which involve some real-life situations like calculating age, counting money, etc.

### 2.4 Solving Equations that Have Variables on Both Sides

Till now, the chapter has dealt with equations where the value on the right-hand side of the equality sign has been only numbers. In this section, we will look into problems where there are variables on both sides of the equation. An example of such an equation is:

2x - 5 = x +3

Solution -> 2x = x + 3 + 5

Subtracting x from both sides we get 2x - x = x + 8 - x

So, x = 8

You are then given many solved examples of such equations and an exercise on it with 10 questions.

### 2.6 Reducing Equations to Simpler Forms

A complex linear equation that has fractions in it can be reduced into a simpler form by the process described below:

Take the LCM(Least common multiple) of the denominator.

Now multiply both RHS and LHS of the equation with this LCM.

By applying the above multiplication, the equation gets reduced to a form without the denominator in it.

Then you can apply the methods learned in the sections above to solve such problems.

We will explain this with an example:

x/2 - 1/5 = x/3 + 1/4 + 1

To reduce the above equation into a simpler form, let us take the LCM of the denominators, i.e. 2, 5, 3, and 4, which is 60. Now multiply each term on both sides of the equation with 60.

60* x/2 - 1/5 * 60 = x/3 * 60 + 1/4 * 60 + 1 * 60

30x - 12 = 20x + 15 + 60

30x - 12 = 20x + 75

Now we will move all the expressions with variables to the LHS and all the constant to the RHS, so we get:

30x - 20x = 75 + 12

10x = 87

X = 87/10 = 8.7

### NCERT Solutions Class 8 Maths Chapter 2 Exercises

Chapter 2 - Linear Equations in One Variable Exercises in PDF Format | |

12 Questions with Solutions | |

16 Questions with Solutions | |

10 Questions with Solutions | |

10 Questions with Solutions | |

10 Questions with Solutions | |

7 Questions with Solutions |

### Important Topics under NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

Chapter 2 of the class 8 maths syllabus includes Linear Equations in One Variable, which is a very important chapter in mathematics that is covered in class 8. The chapter on Linear Equations in One Variable is divided into 5 major sections. The following is a list of the important topics covered under Linear Equations in One Variable. We recommend that students go through every topic carefully to make sure they don’t miss out on learning the concept of Linear Equations in One Variable, properly.

Introduction

Solving equations where linear expressions are on one side and numbers are on the other side of the equation

Applications of linear equations in one variable

Solving equations with variables on both the sides

Reducing the given equations to a simpler form

## Benefits of NCERT Solutions for Class 8 Maths Chapter 2 - Linear Equations in One Variable

The NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable by Vedantu will be of great help for students to score above 90 in their Class 8 Maths exams. The solutions we provide are best-in-class and can help students understand the answers to the questions asked in this chapter of the NCERT textbook.

### Quick Revision

The following is a list of key points that have been covered in the solutions and are required to be remembered while studying the chapter on Linear Equations in One Variable.

The multiplication and division of algebraic expressions

Identities

Factorisation

Some Common Errors

Methods to solve linear equations in one variable in contextual problems that involve multiplication and division operations

### Important Formula Related to Linear Equations in One Variable

The standard form of a linear equation in one variable is ax + b = 0, where, a ≠ 0 and x is the variable.

### Key Learnings from NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

Division and multiplication of algebraic expressions with integers as coefficients

Common errors faced by students

Factorisation

Introduction to identities

Process of solving linear equations in one variable

### Importance of Class 8 Maths Chapter 2 Linear Equations in One Variable Linear Equations

When there are two or more integers and one of them is unknown, then linear equations in one variable can be used for calculating the value of this integer. It is possible to easily find the value of an unknown integer using the expression in the equation form. We encourage students to learn as much as they can from this chapter to be able to solve problems linear equations easily in exams.

### Key Features of NCERT Solutions for Class 8 Maths Chapter 2

The Class 8 Maths Chapter 2 Solution provided by Vedantu is the most reliable online resource for the revision and preparation of CBSE exams. The key benefits of these solutions are:

The solutions are simple to understand and broken into steps so that students can grasp the concept perfectly.

You can also download the solutions in PDF format or print them for a group study, making exam time revisions quick and convenient.

The solutions are based entirely on the CBSE curriculum; hence you are well prepared for your exams once you go through our solutions.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

**1. Give an example of an equation that is not linear but can be reduced to a linear form.**

Sometimes we come across equations that are not linear as per the definition of linear equations but can be reduced to a linear form and then the method of solving linear equations can be applied to them to solve them. The example below illustrates one such equation and how to solve it:

(X + 1)/(2x + 6) = 3/8

To reduce this nonlinear equation into a linear equation we multiply both sides by the denominator of LHS which is 2x + 6.

(X + 1)/(2x + 6) * (2x + 6) = 3/8 * (2x + 6)

X + 1 = 6x + 18/8 - This is a linear equation now. We can solve it by multiplying both sides with LCM of denominators which is 8

8 * (x + 1) = 8 * (6x + 18/8)

8x + 8 = 6x + 18

Now moving all variable to LHS and all constant to RHS we get:

8x - 6x = 18 - 8

2x = 10

X = 5

**2. Mention some of the important features of a linear equation.**

A linear equation is characterized by the following key properties:

The highest power of the variable involved in a linear equation is 1.

The linear equation can have one or two variables in it.

A linear equation has an equality sign. The expression on the left side of the equality is called LHS (left-hand side), and the expression on the right side of the equality sign is termed as RHS (right-hand side).

The two expressions, LHS and RHS, are equal for only certain values of the variables. These values of the variables are the solutions of the linear equation.

The points of a linear equation with just one variable can be marked on the number line.

**3. What are the sub-topics covered in Chapter 2 of Class 8 Maths?**

The concepts covered in Chapter 2 of Class 8 Maths are :

2.1 Introduction

2.2 Solving equations which have a Linear expression on one side and numbers on the other side

2.3 Some applications

2.4 Solving equations having the variables on both sides

2.5 Some more applications

2.6 Reducing equations to a simpler form

2.7 Equations reducible to linear form.

**4. How many exercises are there in Chapter 2 of Class 8 Maths?**

Chapter 2, “Linear Equations In One Variable”, consists of 6 exercises -

Exercise 2.1 contains 12 questions.

Exercise 2.2 contains 16 questions.

Exercise 2.3 contains 10 questions.

Exercise 2.4 contains 10 questions.

Exercise 2.5 contains 10 questions.

Exercise 2.6 contains 10 questions.

Exercise 2.7 contains 7 questions.

Students can practise all the questions in the NCERT solutions designed by the experts at Vedantu to get well versed in this chapter. Download these solutions on the Vedantu website or the app for free of cost to prepare for the exams. These questions will help the students to secure a perfect score in the exams.

**5. Do I need to practice all the questions given in NCERT Solutions?**

Yes, you need to practice all the questions given in the NCERT Solutions. NCERT Solutions prepared by the experts at Vedantu will help the students clear all their doubts in the chapter and practice more to get a perfect score in the exam. You can download the solutions of this chapter by clicking on NCERT Solutions for Class 8 Maths Chapter 2 to ace your exams. These solutions are available on Vedantu (vedantu.com) free of cost. You can download it through the Vedantu app as well.

**6. Are NCERT Solutions Chapter 2 Class 8 Maths important for your exams?**

Yes, NCERT Solutions Chapter 2 Class 8 Maths are important for your exams since most of the questions are asked from this chapter. Vedantu’s NCERT Solutions Chapter 2 Class 8 Maths help students get in-depth knowledge on all the concepts. These solutions help students in revisions and also help them practice all the problems solved by the experts before the exams.

**7. Where to download NCERT Solutions for Chapter 2 Class 8 Maths?**

Students of Class 8 can download the NCERT Solutions for Chapter 2 using the link NCERT Solutions for Class 8 Maths Chapter 2 from Vedantu’s app or website (vedantu.com). Since the solutions are based on the CBSE syllabus, students will be able to practice all the problems. The experts at Vedantu prepare these solutions, keeping in mind the students so that they can excel in their exams.