NCERT Solutions Class 8 Maths Chapter 2 - Linear Equations in One Variable

Linear Equations in One Variable is the second chapter of Class 8 for Maths. It covers various topics, including Introduction, Solving Equations having the Variable on both Sides, Reducing Equations to Simpler Form.

The chapter starts with an introduction to linear equations in one variable in section 2.1, and then covers key topics in sections 2.2 and 2.3.

1.0Download Class 8 Maths Chapter 2 Pdf with Answers

ALLEN's experts have crafted the Class 8 Linear Equations in One Variable solutions to enhance students' problem-solving skills. To get a clearer understanding of these NCERT solutions, students can download the below Class 8 maths chapter 2 pdf.

2.0NCERT Solutions Class 8 Maths Chapter 2: Important Topics

Introduction: Basics of linear equations in one variable.

Solving Equations with Variables on Both Sides: Techniques for balancing and solving equations.

Reducing Equations to Simpler Form: Simplifying complex equations for easier solutions.

3.0Class 8 Maths Chapter 2: Exercise Solutions

There are six exercises in chapter 2 (Linear Equations in One Variable) of Class 8 Maths. Students can find the split below:

Explore Linear Equations in One Variable and learn how to solve various problems only on NCERT Solutions For Class 8 Maths.

4.0NCERT Questions with Solutions Class 8 Maths Chapter 2 - Detailed Solutions

EXERCISE : 2.1

• Solve the following equations (1) $\mathrm{x}-2=7$ (2) $y+3=10$ (3) $6=\mathrm{z}+2$ (4) $\frac{3}{7}+x=\frac{17}{7}$ (5) $6x=12$ (6) $\frac{\mathrm{t}}{5}=10$ (7) $\frac{2\mathrm{x}}{3}=18$ (8) $1.6=\frac{\mathrm{y}}{1.5}$ (9) $7\mathrm{x}-9=16$ (10) $14y-8=13$ (11) $17+6p=9$ (12) $\frac{x}{3}+1=\frac{7}{15}$ Sol. (1) We have, $\mathrm{x}-2=7$ [Adding 2 on both sides] $\Rightarrow \mathrm{x}-2+2=7+2$ $\Rightarrow \mathrm{x}=9$ (2) $y+3=10$ [Subtracting 3 on both sides] $\Rightarrow \mathrm{y}+3-3=10-3$ $\Rightarrow \mathrm{y}=7$ (3) $6=\mathrm{z}+2$ [Subtracting 2 on both sides] $\Rightarrow 6-2=\mathrm{z}+2-2$ $\Rightarrow 4=\mathrm{z}$ or $\mathrm{z}=4$ (4) $\frac{3}{7}+x=\frac{17}{7}$ [Subtracting $\frac{3}{7}$ on both sides] $\Rightarrow x+\frac{3}{7}-\frac{3}{7}=\frac{17}{7}-\frac{3}{7}$ $\Rightarrow \mathrm{x}=\frac{14}{7}$ $\Rightarrow \mathrm{x}=2$ (5) $6x=12$ $\Rightarrow \frac{6\mathrm{x}}{6}=\frac{12}{6}$ [Dividing both sides by 6] $\Rightarrow \mathrm{x}=2$ (6) $\frac{\mathrm{t}}{5}=10$ [Multiplying both sides by 5] $\Rightarrow \frac{\mathrm{t}}{5}\times 5=10\times 5$ $\Rightarrow \mathrm{t}=50$ (7) $\frac{2\mathrm{x}}{3}=18$ [Multiplying both sides by 3/2] $\Rightarrow \frac{2}{3}\times \mathrm{x}\times \frac{3}{2}=18\times \frac{3}{2}$ $\Rightarrow \mathrm{x}=27$ (8) $1.6=\frac{y}{1.5}$ [Multiplying both sides by 1.5] $\Rightarrow 1.6\times 1.5=\frac{\mathrm{y}}{1.5}\times 1.5$ $\Rightarrow 2.4=\mathrm{y}$ or $\mathrm{y}=2.4$ (9) $7\mathrm{x}-9=16$ [Adding 9 on both sides] $\Rightarrow 7\mathrm{x}-9+9=16+9$ $\Rightarrow 7\mathrm{x}=25\Rightarrow \frac{7\mathrm{x}}{7}=\frac{25}{7}\Rightarrow \mathrm{x}=\frac{25}{7}$ (10) $14y-8=13$ [Adding 8 on both sides] $\Rightarrow 14y-8+8=13+8$ $\Rightarrow 14y=21$ [Dividing both sides by 14] $\Rightarrow \frac{14\mathrm{y}}{14}=\frac{21}{14}$ $\Rightarrow \mathrm{y}=\frac{3}{2}$ (11) $17+6\mathrm{p}=9$ [Subtracting 17 on both sides] $\Rightarrow 6\mathrm{p}+17-17=9-17$ $\Rightarrow 6\mathrm{p}=-8$ [Dividing both sides by 6] $\Rightarrow \frac{6\mathrm{p}}{6}=-\frac{8}{6}$ $\Rightarrow \mathrm{p}=-\frac{4}{3}$ (12) $\frac{\mathrm{x}}{3}+1=\frac{7}{15}$ [Subtracting 1 on both sides] $\Rightarrow \frac{\mathrm{x}}{3}+1-1=\frac{7}{15}-\frac{1}{1}$ $\Rightarrow \frac{\mathrm{x}}{3}=\frac{7-15}{15}\Rightarrow \frac{\mathrm{x}}{3}=-\frac{8}{15}$ [Multiplying both sides by 3] $\Rightarrow \frac{\mathrm{x}}{3}\times 3=-\frac{8}{15}\times 3$ $\Rightarrow \mathrm{x}=-\frac{8}{5}$

EXERCISE : 2.2

• If you subtract $1/2$ from a number and multiply the result by $1/2$, you get $1/8$. What is the number? Sol. Let the required number be x . Then, $$\begin{gathered} \left(x-\frac{1}{2}\right)\times \frac{1}{2}=\frac{1}{8} \\ \Rightarrow x-\frac{1}{2}=\frac{1}{8}÷\frac{1}{2}\Rightarrow x-\frac{1}{2}=\frac{1}{8}\times \frac{2}{1} \\ \Rightarrow x-\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \end{gathered}$$ Thus, the required number is $\frac{3}{4}$.
• 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? Sol. Let the breadth of rectangular swimming pool be x metres. Then, its length $=(2x+2)$ metres. Perimeter $=2$ (Length + Breadth) $=2(2x+2+x)=6x+4$ But it is given that the perimeter of the swiming pool is 154 m . $\therefore 6\mathrm{x}+4=154\Rightarrow 6\mathrm{x}=154-4$ $\Rightarrow 6\mathrm{x}=150\Rightarrow \mathrm{x}=\frac{150}{6}=25$ Thus, length $=(2\times 25+2)\mathrm{m}=52\mathrm{m}$ and breadth $=25\mathrm{m}$.
• The base of an isosceles triangle is $\frac{4}{3}\mathrm{cm}$. The perimeter of the triangle is $4\frac{2}{15}\mathrm{cm}$. What is the length of either of the remaining equal sides? Sol. Let each equal side of an isosceles triangle ABC be x cm . Then, its perimeter $=\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$ $=\left(x+\frac{4}{3}+\mathrm{x}\right)\mathrm{cm}=\left(2\mathrm{x}+\frac{4}{3}\right)\mathrm{cm}$

• But it is given that the perimeter of the triangle is $4\frac{2}{15}\mathrm{cm}=\frac{62}{15}\mathrm{cm}$ $\therefore 2x+\frac{4}{3}=\frac{62}{15}\Rightarrow 30x+20=62$ [On multiplying both sides by 15] $\Rightarrow 30\mathrm{x}=62-20\Rightarrow 30\mathrm{x}=42$ $\Rightarrow \mathrm{x}=\frac{42}{30}=\frac{7}{5}=1\frac{2}{5}$ Thus, each equal side is of $1\frac{2}{5}\mathrm{cm}$.
• Sum of two numbers is 95 . If one exceeds the other by 15 , find the numbers. Sol. Let the two numbers be x and $(\mathrm{x}+15)$. Then $x+(x+15)=95$ $\Rightarrow 2\mathrm{x}+15-15=95-15$ $\Rightarrow 2\mathrm{x}=80\Rightarrow \frac{2\mathrm{x}}{2}=\frac{80}{2}$ $\Rightarrow \mathrm{x}=40$ $\Rightarrow \mathrm{x}+15=40+15=55$ Hence, the numbers are 40 and 55 .
• Two numbers are in the ratio $5:3$. If they differ by 18 , what are the numbers? Sol. Let the numbers be $5x$ and $3x$. Then, $5\mathrm{x}-3\mathrm{x}=18$ $\Rightarrow 2\mathrm{x}=18\Rightarrow \mathrm{x}=\frac{18}{2}=9$ $\Rightarrow 5\mathrm{x}=9\times 5=45$ and $3\mathrm{x}=3\times 9=27$ Hence, the numbers are 45 and 27.
• Three consecutive integers add up to 51. What are these integers? Sol. Let the three consecutive numbers be $x$, $(x+1)$ and $(x+2)$. Then, $x+(x+1)+(x+2)=51$ $\Rightarrow 3\mathrm{x}+3=51$ $\Rightarrow 3\mathrm{x}=51-3=48$ $\Rightarrow \mathrm{x}=\frac{48}{3}=16$ Hence, the three consecutive numbers are 16,17 and 18.
• The sum of three consecutive multiples of 8 is 888 . Find the multiples. Sol. Let the three consecutive multiple of 8 be $8\mathrm{x},8(\mathrm{x}+1)$ and $8(\mathrm{x}+2)$. It is given that the sum of three consecutive multiples of 8 is 888 . $\therefore 8\mathrm{x}+8(\mathrm{x}+1)+8(\mathrm{x}+2)=888$ $\Rightarrow 8\mathrm{x}+8\mathrm{x}+8+8\mathrm{x}+16=888$ $\Rightarrow 24\mathrm{x}+24=888$ $\Rightarrow 24\mathrm{x}=888-24$ $\Rightarrow 24\mathrm{x}=864$ $\Rightarrow \mathrm{x}=\frac{864}{24}=36$ Hence, the three consecutive multiples of 8 are $8\times 36,8\times (36+1)$ and $8\times (36+2)$ i.e., 288,296 and 304 .
• Three consecutive integers are such that when they are taken in increasing order and multiplied by 2,3 and 4 respectively, they add up to 74. Find these numbers. Sol. Let the three consecutive integers be x , $(x+1)$ and $(x+2)$. It is given that $2\times x+3\times (x+1)+4\times (x+2)=74$ $\Rightarrow 2\mathrm{x}+3\mathrm{x}+3+4\mathrm{x}+8=74$ $\Rightarrow 9x+11=74$ $\Rightarrow 9x=74-11$ $\Rightarrow 9\mathrm{x}=63$ $\Rightarrow \mathrm{x}=\frac{63}{9}=7$ Hence, the numbers are 7, 8 and 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 ? Sol. Let ages of Rahul and Haroon be 5 x and 7 x years. Four years later : Rahul's age $=5\mathrm{x}+4$ and Haroon's age $=7\mathrm{x}+4$ According to the given condition, we have $(5x+4)+(7x+4)=56$ $\Rightarrow 12\mathrm{x}+8=56$ $\Rightarrow 12\mathrm{x}=56-8$ $\Rightarrow 12x=48$ $\Rightarrow \mathrm{x}=\frac{48}{12}=4$ $\therefore$ Rahul's age $=5\times 4=20$ years And Haroon's age $=7\times 4=28$ years
• 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? Sol. Let the number of girls be $x$. Number of boys will be $=x+8$ Ratio between boys and girls $=7:5$ So, $x:8:x=7:5$ $\Rightarrow \frac{x+8}{x}=\frac{7}{5}$ Now, By cross multiplication, $\Rightarrow 5(\mathrm{x}+8)=7(\mathrm{x})$ $\Rightarrow 5\mathrm{x}+40=7\mathrm{x}$ $\Rightarrow 7\mathrm{x}-5\mathrm{x}=40$ $\Rightarrow 2\mathrm{x}=40$ $\Rightarrow \mathrm{x}=20$ $\therefore$ No. of girls $=20$ No. of boys $=\mathrm{x}+8$ or $20+8=28$ Total class strength $=20+28$ $=48\text{ students }$
• 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? Sol. Let Baichung's age be x years. Then, Baichung's father's age $=(x+29)$ years Baichung's grandfather's age $=[(x+29)+$ 26] years $=(x+55)$ years According to the question, we have $\mathrm{x}+(\mathrm{x}+29)+(\mathrm{x}+55)=135$ $\Rightarrow 3\mathrm{x}+84=135$ $\Rightarrow 3\mathrm{x}=135-84$ $\Rightarrow 3\mathrm{x}=51$ $\Rightarrow \mathrm{x}=\frac{51}{3}=17$ $\therefore$ Baichung's age is 17 years, his father's age is 46 years and his grandfather's age is 72 years.
• Fifteen years from now, Ravi's age will be four times his present age. What is Ravi's present age? Sol. Let Ravi's present age be x years. After 15 years, his age will be $(x+15)$ years. It is given that after 15 years, Ravi's age will be four times his present age. $\therefore 4\mathrm{x}=\mathrm{x}+15$ $\Rightarrow 4\mathrm{x}-\mathrm{x}=15$ $\Rightarrow 3\mathrm{x}=15$ $\Rightarrow \mathrm{x}=\frac{15}{3}=5$ Hence, Ravi's present age is 5 years.
• 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? Sol. Let the required number be x . Then, $\mathrm{x}\times \frac{5}{2}+\frac{2}{3}=-\frac{7}{12}$ $\Rightarrow \frac{5}{2}\mathrm{x}=-\frac{7}{12}-\frac{2}{3}\Rightarrow \frac{5}{2}\mathrm{x}=\frac{-7-8}{12}$ $\Rightarrow \frac{5}{2}\mathrm{x}=\frac{-15}{12}\Rightarrow \mathrm{x}=\frac{-15}{12}\times \frac{2}{5}=-\frac{1}{2}$ Hence, the required number be $-\frac{1}{2}$.
• Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, $₹50$ and $₹10$ respectively. The ratio of the number of these notes is $2:3:5$. The total cash with Lakshmi is $₹4,00,000$. How many notes of each denomination does she have? Sol. Let notes of denominations ₹ 100 , ₹ 50 and $₹10$ be $2x,3x$ and $5x$ respectively. Now, value of $2x₹100$ notes $=₹(2x\times 100)=₹200x$ values of $3\mathrm{x}₹50$ notes $=₹(3x\times 50)=₹150x$ and, value of $5\mathrm{x}₹10$ notes $=₹(5\mathrm{x}\times 10)=₹50\mathrm{x}$ But the total value of money is $₹4,00,000$ $\therefore 200x+150x+50x=400000$ $\Rightarrow 400\mathrm{x}=400000\Rightarrow \mathrm{x}=1000$ Thus, number of $₹100$ notes $=2\times 1000=$ 2000 number of $₹50$ notes $=3\times 1000=3000$ and number of $₹10$ notes $=5\times 1000=$ 5000.
• I have a total of $₹300$ in coins of denomination ₹ 1 , ₹ 2 and $₹5$. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160 . How many coins of each denomination are with me ? Sol. Let the number of $₹5$ coins be $x$. Then, Number of $₹2$ coins $=3\mathrm{x}$ and the number of Re coins $=160-\mathrm{x}-3\mathrm{x}$ $=160-4\mathrm{x}$ Now, value of $₹5$ coins $=₹(5\times x)=₹5x$ values of $₹2$ coins $=₹(2\times 3x)=₹6x$ and, values of $₹1$ coins $=₹1\times (160-4x)$ $=₹(160-4x)$ But the total value of the money $=₹300$ $\therefore 5x+6x+(160-4x)=300$ $\Rightarrow 7\mathrm{x}=300-160\Rightarrow 7\mathrm{x}=140\Rightarrow \mathrm{x}=20$ Thus, number of $₹5$ coins $=20$ number of $₹2$ coins $=60$ and, the number of $₹1$ coins $=80$
• The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25 . The total prize money distributed is ₹ 3000 . Find the number of winners, if the total number of participants is 63 . Sol. Let x be the number of the winners. So, the number of participants who does not win is (63-x). Prize money given to x winners $=$ $=₹(100\times x)$ $=₹100\mathrm{x}$ and prize money given to ( $63-\mathrm{x}$ ) participants $=$ ₹ $25(63-x)$ But the total prize money distributed $=₹3000$ $\therefore 100\mathrm{x}+25(63-\mathrm{x})=3000$ $\Rightarrow 4\mathrm{x}+(63-\mathrm{x})=120$ [On dividing both sides by 25] $\Rightarrow 3\mathrm{x}+63=120\Rightarrow 3\mathrm{x}=120-63$ $\Rightarrow 3\mathrm{x}=57\Rightarrow \mathrm{x}=\frac{57}{3}=19$ $\therefore$ The number of winners is 19 .

EXERCISE : 2.3

Solve the following equations and check your results: (1) $3x=2x+18$ (2) $5\mathrm{t}-3=3\mathrm{t}-5$ (3) $5x+9=5+3x$ (4) $4\mathrm{z}+3=6+2\mathrm{z}$ (5) $2\mathrm{x}-1=14-\mathrm{x}$ (6) $8\mathrm{x}+4=3(\mathrm{x}-1)+7$ (7) $x=\frac{4}{5}(x+10)$ (8) $\frac{2\mathrm{x}}{3}+1=\frac{7\mathrm{x}}{15}+39$. (9) $2y+\frac{5}{3}=\frac{26}{3}-y$ (10) $3\mathrm{m}=5\mathrm{m}-\frac{8}{5}$ Sol. (1) $3x=2x+18$ $\Rightarrow 3\mathrm{x}-2\mathrm{x}=18$ [Transposing 2 x to LHS] $\Rightarrow \mathrm{x}=18$ (2) $5t-3=3t-5$ $\Rightarrow 5\mathrm{t}-3\mathrm{t}=-5+3$ [Transposing $3t$ to LHS and -3 to RHS] $\Rightarrow 2\mathrm{t}=-2$ $\Rightarrow \frac{2\mathrm{t}}{2}=\frac{-2}{2}$ [Dividing both sides by 2] $\Rightarrow \mathrm{t}=-1$ (3) $5x+9=5+3x$ $5\mathrm{x}-3\mathrm{x}=5-9$ [Transposing 3x to LHS and 9 to RHS] $\Rightarrow \frac{2\mathrm{x}}{2}=-\frac{4}{2}$ [Dividing both sides by 2] $\Rightarrow \mathrm{x}=-2$ (4) $4\mathrm{z}+3=6+2\mathrm{z}$ $\Rightarrow 4\mathrm{z}-2\mathrm{z}=6-3$ [Transposing 2 z to LHS and 3 to RHS] $\Rightarrow 2\mathrm{z}=3$ [Dividing both sides by 2] $\Rightarrow \frac{2\mathrm{z}}{2}=\frac{3}{2}$ $\Rightarrow \mathrm{z}=\frac{3}{2}$ (5) $2x-1=14-x$ $\Rightarrow 2\mathrm{x}+\mathrm{x}=14+1$ [Transposing $-x$ to LHS and -1 to RHS] $\Rightarrow 3\mathrm{x}=15$ [Dividing both sides by 3 ] $\Rightarrow \frac{3\mathrm{x}}{3}=\frac{15}{3}$ $\Rightarrow \mathrm{x}=5$ (6) $8x+4=3(x-1)+7$ $\Rightarrow 8\mathrm{x}+4=3\mathrm{x}-3+7$ $\Rightarrow 8\mathrm{x}-3\mathrm{x}=4-4$ [Transposing 3x to LHS and 4 to RHS] $\Rightarrow 5\mathrm{x}=0$ [Dividing both sides by 5] $\Rightarrow \frac{5\mathrm{x}}{5}=\frac{0}{5}$ $\Rightarrow \mathrm{x}=0$ (7) $\mathrm{x}=\frac{4}{5}(\mathrm{x}+10)$ [Multiplying both sides by 5] $\Rightarrow 5\mathrm{x}=4(\mathrm{x}+10)$ [On expanding brackets] $\Rightarrow 5\mathrm{x}=4\mathrm{x}+40$ [Transposing 4 x to LHS] $\Rightarrow 5\mathrm{x}-4\mathrm{x}=40$ $\Rightarrow \mathrm{x}=40$ (8) $\frac{2\mathrm{x}}{3}+1=\frac{7\mathrm{x}}{15}+3$ Multiplying both sides by the LCM of 3 and 15 i.e. 15 , we get $10\mathrm{x}+15=7\mathrm{x}+45$ [Transposing 7x to LHS and 15 to RHS] $\Rightarrow 10\mathrm{x}-7\mathrm{x}=45-15$ $\Rightarrow 3\mathrm{x}=30$ [Dividing both sides by 3] $\Rightarrow \frac{3\mathrm{x}}{3}=\frac{30}{3}$ $\Rightarrow \mathrm{x}=10$ (9) $2\mathrm{y}+\frac{5}{3}=\frac{26}{3}-\mathrm{y}$ [Transposing - y to LHS and 5/3 to RHS] $\Rightarrow 2y+y=\frac{26}{3}-\frac{5}{3}$ $\Rightarrow 3\mathrm{y}=\frac{21}{3}\Rightarrow 3\mathrm{y}=7$ [Dividing both sides by 3 ] $\Rightarrow \frac{3\mathrm{y}}{3}=\frac{7}{3}$ $\Rightarrow y=\frac{7}{3}$ (10) $3m=5m-\frac{8}{5}$ [Transposing 5m to LHS] $\Rightarrow 3\mathrm{m}-5\mathrm{m}=-\frac{8}{5}$ $\Rightarrow -2\mathrm{m}=-\frac{8}{5}$ [Dividing both sides by 2] $\Rightarrow \frac{-2\mathrm{m}}{2}=-\frac{8}{5}\times \frac{1}{2}$ $\Rightarrow -\mathrm{m}=-\frac{4}{5}$ $\Rightarrow \mathrm{m}=\frac{4}{5}$

EXERCISE : 2.4

• Amina thinks of a number and subtracts $5/2$ 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? Sol. Let Amina thinks of a number x . It is given that $8\left(x-\frac{5}{2}\right)=3x$ $\Rightarrow 8\mathrm{x}-20=3\mathrm{x}\Rightarrow 8\mathrm{x}-3\mathrm{x}=20$ $\Rightarrow 5\mathrm{x}=20\Rightarrow \mathrm{x}=\frac{20}{5}=4$ Hence, the number thought is 4 .
• 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? Sol. Let the numbers be x and 5 x . It is given that $2(\mathrm{x}+21)=5\mathrm{x}+21$ $\Rightarrow 2x+42=5x+21$ $\Rightarrow 2\mathrm{x}-5\mathrm{x}=21-42$ $\Rightarrow -3\mathrm{x}=-21$ $\Rightarrow \frac{-3\mathrm{x}}{-3}=\frac{-21}{-3}$ $\Rightarrow \mathrm{x}=7$ Hence, the required numbers are 7 and 35.
• 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? Sol. Let the unit's digit be x . Then, ten's digit $=(9-x)$ Number $=10\times (9-x)+x$ $=90-10\mathrm{x}+\mathrm{x}=90-9\mathrm{x}$ Number with digits interchanged $=10\mathrm{x}+(9-\mathrm{x})=9\mathrm{x}+9$ $\therefore (90-9x)+27=9x+9$ $\Rightarrow 117-9x=9x+9$ $\Rightarrow -9x-9x=9-117$ $\Rightarrow -18x=-108$ $\Rightarrow \mathrm{x}=\frac{-108}{-18}=6$ $\therefore$ Unit's digit $=6$ and ten's digit $=3$ Hence, the number $=36$.
• One of the two digits of a two-digit number is three times the other digit. If you interchange the digits of this twodigit number and add the resulting number to the original number, you get 88. What is the original number? Sol. Let one's digit be $x$. Then, ten's digit $=3x$. $\therefore$ Number $=10\times 3\mathrm{x}+\mathrm{x}=30\mathrm{x}+\mathrm{x}=31\mathrm{x}...(1)$ Number obtained by reversing the digits $=10\mathrm{x}+3\mathrm{x}=13\mathrm{x}$ It is given that the sum of the original number and reversed number is 88 . $\therefore 31\mathrm{x}+13\mathrm{x}=88$ $\Rightarrow 44\mathrm{x}=88\Rightarrow \mathrm{x}=\frac{88}{44}=2$ Putting the value of $x$ in (1), we get Number $=31\times 2=62$.
• 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 his mother's present age. What are their present ages? Sol. Let the present age of Shobo be x years Then, Shobo's mother's age $=6x$ years After 5 years : Shobo's age $=(x+5)$ years $\therefore (\mathrm{x}+5)=\frac{1}{3}\times (6\mathrm{x})$ $\Rightarrow 3\times (\mathrm{x}+5)=3\times \frac{1}{3}\times (6\mathrm{x})$ $\Rightarrow 3x+15=6x\Rightarrow 3x-6x=-15$ $\Rightarrow -3\mathrm{x}=-15\Rightarrow \mathrm{x}=\frac{-15}{-3}=5$ $\therefore$ Present age of Shobo $=5$ years Present age of his mother $=30$ years
• 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 $₹100$ per metre it will cost the village panchayat $₹75000$ to fence the plot. What are the dimensions of the plot? Sol. Let the length and breadth of the Then, Perimeter $=2(11\mathrm{x}+4\mathrm{x})$ metres $=30\mathrm{x}$ metres Cost of fencing the plot @ ₹ 100 per meter $=₹(100\times 30x)=₹3000x$ But it is given as ₹75000. $\therefore 3000\mathrm{x}=75000\Rightarrow \mathrm{x}=\frac{75000}{3000}=25$ $\therefore$ Length $=11\times 25\mathrm{m}=275$ metres and breadth $=4\times 25\mathrm{m}=100$ metres.
• Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 2 metres of the trouser material he buys 3 metres of the shirt material. He sells the materials at $12\%$ and $10\%$ profit respectively. His total sale is ₹ 36,660 . How much trouser material did he buy? Sol. Let material bought for trousers and shirts be $2x$ metres and $3x$ metres respectively. Cost price of 2 x metres of trouser material @ ₹ 90 per metre $=₹(90\times 2\mathrm{x})=₹180\mathrm{x}$ Cost price of 3 x metres of shirt material @ $₹50$ per metre $=₹(50\times 3\mathrm{x})=₹150\mathrm{x}$ Profit in case of shirt material $=12\%$ $\therefore$ Its S.P. $=₹\left(150\mathrm{x}\times \frac{112}{100}\right)=₹168\mathrm{x}$ Profit in case of trouser material $=10\%$ $\therefore$ Its S.P. $=₹\left(180\mathrm{x}\times \frac{110}{100}\right)=198\mathrm{x}$ $\therefore$ Total selling price $=₹(168\mathrm{x}+198\mathrm{x})=$ ₹366x But the total selling price is given as ₹ 36660 . $\therefore 366\mathrm{x}=36660\Rightarrow \mathrm{x}=\frac{36660}{366}=100.16$ Thus, the trouser material purchased $=$ $2\times 100.16\mathrm{m}$ $=200.32\mathrm{m}\approx 200$ metres.
• 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. Sol. Let the number of deer in the herd be x. Then, Number of deer grazing in the field $=\frac{\mathrm{x}}{2}$ Number of deer playing nearby $=$ $\frac{3}{4}\times \left(x-\frac{x}{2}\right)$ $=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$ Number of deer drinking water $=9$ $\therefore \frac{x}{2}+\frac{3x}{8}+9=\mathrm{x}$ $\Rightarrow 8\times \frac{\mathrm{x}}{2}+8\times \frac{3\mathrm{x}}{8}+8\times 9=8\times \mathrm{x}$ $\Rightarrow 4\mathrm{x}+3\mathrm{x}+72=8\mathrm{x}\Rightarrow 7\mathrm{x}-8\mathrm{x}=-72$ $\Rightarrow -x=-72\Rightarrow x=72$ Hence, the number of deer in the herd is 72.
• A grandfather is ten times older than his grand daughter. He is also 54 years older than her. Find their present ages. Sol. Let the age of granddaughter be x years. Then, the age of her grandfather is 10 x years. According to the question : $10\mathrm{x}-\mathrm{x}=54$ $\Rightarrow 9\mathrm{x}=54\Rightarrow \mathrm{x}=\frac{54}{9}=6$ $\therefore$ Granddaughter's age $=6$ years and grandfather's age $=10\times 6=60$ years.
• 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. Sol. Let the present age of son be x years. Then, the age of his father Aman is 3 x . 10 years ago, their ages will be ( $\mathrm{x}-10$ ) years and $(3x-10)$ years respectively. According to the question : $3x-10=5(x-10)$ $\Rightarrow 3\mathrm{x}-10=5\mathrm{x}-50$ $\Rightarrow 5\mathrm{x}-3\mathrm{x}=-10+50$ $\Rightarrow 2\mathrm{x}=40\Rightarrow \mathrm{x}=20$ $\therefore$ Son's age $=20$ years and Aman's age $=20\times 3=60$ years.

EXERCISE : 2.5

Solve the following linear equations

• $\frac{\mathrm{x}}{2}-\frac{1}{5}=\frac{\mathrm{x}}{3}+\frac{1}{4}$
• $\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$
• $x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$
• $\frac{x-5}{3}=\frac{x-3}{5}$
• $\frac{3\mathrm{t}-2}{4}-\frac{2\mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}$
• $\mathrm{m}-\frac{\mathrm{m}-1}{2}=1-\frac{\mathrm{m}-2}{3}$ Sol. 1. $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$ The denominators on two sides are 2, 3, 4 and 5 . Their LCM is 60. Multiplying both sides by 60 , we get $60\times \left(\frac{\mathrm{x}}{2}-\frac{1}{5}\right)=60\times \left(\frac{\mathrm{x}}{3}+\frac{1}{4}\right)$ $\Rightarrow 60\times \frac{x}{2}-60\times \frac{1}{5}=60\times \frac{x}{3}+60\times \frac{1}{4}$ $\Rightarrow 30\mathrm{x}-12=20\mathrm{x}+15$ $\Rightarrow 30\mathrm{x}-20\mathrm{x}=15+12$ $\Rightarrow 10\mathrm{x}=27\Rightarrow \mathrm{x}=\frac{27}{10}$
• $\frac{\mathrm{n}}{2}-\frac{3\mathrm{n}}{4}+\frac{5\mathrm{n}}{6}=21$ The denominator on two sides are $1,2,4$, and 6. Their LCM is 12 . Multiplying both sides by 12 , we get. $12\times \left[\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}\right]=21\times 12$ $\Rightarrow 12\times \frac{\mathrm{n}}{2}-12\times \frac{3\mathrm{n}}{4}+12\times \frac{5\mathrm{n}}{6}=21\times 12$ $\Rightarrow 6\mathrm{n}-9\mathrm{n}+10\mathrm{n}=252$ $\Rightarrow 7\mathrm{n}=252$ $\Rightarrow \mathrm{n}=\frac{252}{7}\Rightarrow \mathrm{n}=36$
• $x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$ The denominators on two sides are 2,3 and 6. Their LCM is 12. Multiplying both sides by 12 , we get. $12\times \left[x+7-\frac{8x}{3}\right]=12\left[\frac{17}{6}-\frac{5x}{2}\right]$ $\Rightarrow 12\mathrm{x}+12\times 7-12\times \frac{8\mathrm{x}}{3}$ $=12\times \frac{17}{6}-12\times \frac{5x}{2}$ $\Rightarrow 12\mathrm{x}+84-32\mathrm{x}=34-30\mathrm{x}$ $\Rightarrow 12\mathrm{x}-32\mathrm{x}+30\mathrm{x}=34-84$ $\Rightarrow 10\mathrm{x}=-50$ $\Rightarrow \mathrm{x}=-\frac{50}{10}\Rightarrow \mathrm{x}=-5$
• We have, The denominators on LHS and RHS are 3 and 5 respectively. Multiplying both sides by the LCM of 3 and 5 , that is 15 , $15\times \left(\frac{\mathrm{x}-5}{3}\right)=\left(\frac{\mathrm{x}-3}{5}\right)\times 15$ $\Rightarrow 5(\mathrm{x}-5)=3(\mathrm{x}-3)$ $\Rightarrow 5\mathrm{x}-25=3\mathrm{x}-9$ $\Rightarrow 5\mathrm{x}-3\mathrm{x}=-9+25$ $\Rightarrow 2\mathrm{x}=16\Rightarrow \mathrm{x}=\frac{16}{2}=8$
• $\frac{3\mathrm{t}-2}{4}-\frac{2\mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}$ Multiplying both sides by 12 , the LCM of 4 and 3 , we get $12\left(\frac{3t-2}{4}\right)-12\left(\frac{2t+3}{3}\right)=12\times \frac{2}{3}-12\mathrm{t}$ $\Rightarrow 3(3\mathrm{t}-2)-4(2\mathrm{t}+3)=8-12\mathrm{t}$ $\Rightarrow 9\mathrm{t}-6-8\mathrm{t}-12=8-12\mathrm{t}$ $\Rightarrow \mathrm{t}-18=8-12\mathrm{t}$ $\Rightarrow \mathrm{t}+12\mathrm{t}=8+18$ $\Rightarrow 13\mathrm{t}=26$ $\Rightarrow \mathrm{t}=\frac{26}{13}=2$
• $m-\frac{m-1}{2}=1-\frac{m-2}{3}$ The denominators on two sides are 2 and 3. Their LCM is 6 . Multiplying both sides by 6 , we get. $6\times \left(m-\frac{m-1}{2}\right)=6\left(1-\frac{m-2}{3}\right)$ $\Rightarrow 6\mathrm{m}-\frac{6(\mathrm{m}-1)}{2}=6\times 1-\frac{6(\mathrm{m}-2)}{3}$ $\Rightarrow 6\mathrm{m}-3(\mathrm{m}-1)=6-2(\mathrm{m}-2)$ $\Rightarrow 6\mathrm{m}-3\mathrm{m}+3=6-2\mathrm{m}+4$ $\Rightarrow 3\mathrm{m}+2\mathrm{m}=10-3$ $\Rightarrow 5\mathrm{m}=7$ $\Rightarrow \mathrm{m}=\frac{7}{5}$

Simplify and solve the following linear equations 7. $3(t-3)=5(2t+1)$ 8. $15(y-4)-2(y-9)+5(y+6)=0$ 9. $3(5z-7)-2(9z-11)=4(8z-13)-17$ 10. $0.25(4f-3)=0.05(10f-9)$ Sol. 7. We have, $3(t-3)=5(2t+1)$ $\Rightarrow 3\mathrm{t}-9=10\mathrm{t}+5$ $\Rightarrow 10\mathrm{t}-3\mathrm{t}=-9-5$ $\Rightarrow 7\mathrm{t}=-14$ $\Rightarrow \frac{7\mathrm{t}}{7}=-\frac{14}{7}$ $\Rightarrow \mathrm{t}=-2$ 8. We have, $15(y-4)-2(y-9)+5(y+6)=0$ $15y-60-2y+18+5y+30=0$ $\Rightarrow 18\mathrm{y}-12=0$ $\Rightarrow 18\mathrm{y}=12$ $\Rightarrow \mathrm{y}=\frac{12}{18}$ $\Rightarrow \mathrm{y}=\frac{2}{3}$ 9. We have, $3(5z-7)-2(9z-11)=4(8z-13)-17$ $\Rightarrow 15z-21-18z+22=32z-52-17$ $\Rightarrow -3\mathrm{z}+1=32\mathrm{z}-69$ $\Rightarrow -3z-32z=-69-1$ $\Rightarrow -35z=-70$ $\Rightarrow \mathrm{z}=\frac{-70}{-35}=2$. 10. We have, $0.25(4f-3)=0.05(10f-9)$ $\Rightarrow \mathrm{f}-0.75=0.5\mathrm{f}-0.45$ $\Rightarrow \mathrm{f}-0.5\mathrm{f}=-0.45+0.75$ $\Rightarrow 0.5\mathrm{f}=0.30$ $\Rightarrow \mathrm{f}=\frac{0.30}{0.5}$ $\Rightarrow \mathrm{f}=0.6$

EXERCISE : 2.6

Solve the following equations

• $\frac{8x-3}{3x}=2$
• $\frac{9x}{7-6x}=15$
• $\frac{\mathrm{z}}{\mathrm{z}+15}=\frac{4}{9}$
• $\frac{3y+4}{2-6y}=\frac{-2}{5}$
• $\frac{7y+4}{y+2}=\frac{-4}{3}$ Sol.
• $\frac{8x-3}{3x}=2$ Multiplying both sides by $3x$, we get $\Rightarrow 3\mathrm{x}\times \frac{8\mathrm{x}-3}{3\mathrm{x}}=2\times 3\mathrm{x}$ $\Rightarrow 8\mathrm{x}-3=6\mathrm{x}$ $\Rightarrow 8\mathrm{x}-6\mathrm{x}=3$ $\Rightarrow 2\mathrm{x}=3$ $\Rightarrow \mathrm{x}=3/2$ Hence, $x=3/2$ is the solution of the given equation.
• $\frac{9x}{7-6x}=15$ Multiplying both sides by $7-6\mathrm{x}$, we get $(7-6x)\times \frac{9x}{7-6x}=(7-6x)\times 15$ $\Rightarrow 9\mathrm{x}=105-90\mathrm{x}$ $\Rightarrow 9\mathrm{x}+90\mathrm{x}=105$ $\Rightarrow 99\mathrm{x}=105$ $\Rightarrow \mathrm{x}=\frac{105}{99}=\frac{35}{33}$ Hence, $x=\frac{35}{33}$ is the solution of the given equation.
• $\frac{\mathrm{z}}{\mathrm{z}+15}=\frac{4}{9}$ [by cross multiplication] $\Rightarrow 9\times \mathrm{z}=4(\mathrm{z}+15)$ $\Rightarrow 9\mathrm{z}=4\mathrm{z}+60$ $\Rightarrow 9\mathrm{z}-4\mathrm{z}=60$ $\Rightarrow 5\mathrm{z}=60$ $\Rightarrow \mathrm{z}=\frac{60}{5}$ $\Rightarrow \mathrm{z}=12$ Hence, $\mathrm{z}=12$ is the solution of the given equation.
• $\frac{3y+4}{2-6y}=\frac{-2}{5}$ [By cross multiplication] $\Rightarrow 5(3y+4)=-2(2-6y)$ $\Rightarrow 15y+20=-4+12y$ $\Rightarrow 15\mathrm{y}-12\mathrm{y}=-4-20$ $\Rightarrow 3y=-24$ $\Rightarrow \mathrm{y}=\frac{-24}{3}=-8$ Hence, $\mathrm{y}=-8$ is the solution of the given equation.
• $\frac{7y+4}{y+2}=-\frac{4}{3}$ $3(7y+4)=-4(y+2)$ $\Rightarrow 21y+12=-4y-8$ $\Rightarrow 21y+4y=-8-12$ $\Rightarrow 25\mathrm{y}=-20$ $\Rightarrow \mathrm{y}=-\frac{20}{25}\Rightarrow \mathrm{y}=-\frac{4}{5}$ Hence, $y=-\frac{4}{5}$ is the solution of the given equation.
• 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. Sol. Let the present ages of Hari and Harry be $5x$ years and $7x$ years respectively. After 4 years : Hari's age $=(5x+4)$ years Harry's age $=(7x+4)$ years $\therefore \frac{5x+4}{7x+4}=\frac{3}{4}$ (Given) $\Rightarrow 4(5\mathrm{x}+4)=3(7\mathrm{x}+4)$ $\Rightarrow 20\mathrm{x}+16=21\mathrm{x}+12$ $\Rightarrow 20\mathrm{x}-21\mathrm{x}=12-16$ $\Rightarrow -\mathrm{x}=-4$, i.e., $\mathrm{x}=4$ $\therefore$ Present age of Hari $=5\times 4$ years $=20$ years Present age of Harry $=7\times 4$ years $=28$ years
• 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 $\frac{3}{2}$. Find the rational number. Sol. Let the numerator of the rational number be $x$. Then, the denominator of the rational number $=x+8$. $\therefore$ The rational number $=\frac{\mathrm{x}}{\mathrm{x}+8}$ If the numerator is increased by 17 and the denominator is decreased by 1 , the number becomes $\frac{3}{2}$. $\therefore \frac{x+17}{(x+8)-1}=\frac{3}{2}$ $\Rightarrow \frac{x+17}{x+7}=\frac{3}{2}$ $\Rightarrow 2(\mathrm{x}+17)=3(\mathrm{x}+7)[0\mathrm{n}$ cross multiplying] $\Rightarrow 2\mathrm{x}+34=3\mathrm{x}+21$ $\Rightarrow 2\mathrm{x}-3\mathrm{x}=21-34$ $\Rightarrow -\mathrm{x}=-13\Rightarrow \mathrm{x}=13$ $\therefore \frac{x}{x+8}=\frac{13}{13+8}=\frac{13}{21}$ Hence, the required rational number $=\frac{13}{21}$.

5.0Benefits of Studying Class 8 Maths Chapter 2 NCERT Solutions – Linear Equations in One Variable

• Improves Problem-Solving Skills: Helps students master the techniques required to solve linear equations efficiently.
• Strengthens Conceptual Understanding: Enhances comprehension of key algebraic concepts, forming a solid foundation for future studies.
• Boosts Confidence: Regular practice with NCERT solutions builds confidence in tackling various types of linear equations.
• Prepares for Higher Classes: Provides a strong base for understanding more complex algebraic topics in higher grades.