# NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.6) Exercise 3.6

NCERT, or the National Council of Educational Research, is self-governing. In 1961 it was established by the Indian government itself. Its main function is to help the Central government and the state governments reach the goal of improving the quality of education in the country. The NCERT is supposed to conduct, coordinate, and promote research in fields that are related to school education. They are supposed to create and publish all the written text required for school education, like textbooks, newsletters, journals, educational kits, multimedia digital material, etc.

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High school students frequently exhibit a strong dislike for and dread of Mathematics.

It can be concluded that the majority of students experience this fear due to an unclear understanding of the foundational concepts in Mathematics. This concern can be easily overcome if adequate time is spent explaining the foundations to the students and encouraging regular practice. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can assist in understanding the foundational concepts.

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The main objective of CBSE is to enhance skill development through the integration of inputs that are job-oriented and job-linked, as well as assessment and evaluation procedures. In order for students from CBSE-affiliated schools to acquire Class 10 examination certification, the CBSE is responsible for setting test requirements and conducting exams at the end of Class 10.

The board exams for Class 10 are thought to be one of the most significant tests a student will ever take. Based on their performance in the Class 10 board exams, students get to choose the subject in which they want to continue their education.Therefore, it is crucial for students in Class 10 to take their board exams seriously.

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## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables (Ex 3.6) Exercise 3.6

Extramarks’ mathematics professionals strive to provide students with tools that can be used in a variety of ways.The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 is one of these resources. Students can quickly and easily access various resources, including the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6, by simply visiting the website.They do not need to search elsewhere on the internet for information to assist them in their board examination preparation. Reliable information can be found on the Extramarks website in the form of its resources, such as the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6. Due to the fact that these were created by Extramarks’ subject-matter experts, the students can completely rely on them for their preparations without any doubt. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 can be downloaded in PDF format for offline access.

Exercise 3.6 Class 10 Maths Is a part of Chapter 3 for Class 10 Mathematics. Chapter 3 in Class 10 Mathematics is based on – Pair of Linear Equations. The 3.6 Maths Class 10 is just one part of Chapter 3. Two-variable linear equations are defined for a line that can be drawn on a graph. Points are the answers to linear equations. A straight line will result from the two-variable linear equation described in this chapter. With the help of the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6, students can better understand the concept of graph plotting and drawing straight lines with linear equations in two variables.

Students can access the PDF for Class 10 Chapter 3 Mathematics Solutions on Extramarks to aid in their learning of the questions given in this particular chapter. Every question given in these exercises has solutions on the Extramarks website by experts in Mathematics.The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 have been updated to reflect the most recent CBSE Class 10 Maths Chapter 3 syllabus.

### Linear Equation –

A linear equation is a straight-line equation. The formula is written as axe + by + c = 0, where a, b, and c are real numbers (a0 and b0), and x and y are the two variables.

In this case, the coefficients are a and b, and the equation’s constant is c.

Pair of Two Linear Equations –

The term “pair of linear equations in two variables” refers to two linear equations with the same two variables.

• a1x + b1y + c1 = 0
• a2x + b2y + c2 = 0

### Importance of a Pair of Linear Equations in Two Variables

Linear equations have numerous uses in both Mathematics and everyday life. Equality with variables and the equal symbol (=) is known as an algebraic equation. An equation of degree one is a linear equation. Word problems are a common way for mathematicians to apply their knowledge, and solving them typically involves the use of Linear Equations.

The uses of Linear Equations in daily life are countless. To use Algebra to solve real-world issues, students have to translate the given scenario into mathematical statements that make it crystal clear how the variables relate to the information presented. The procedures for converting a situation into a mathematical statement are as follows:

• Students have to first an Algebraic Expression that accurately explains the situation by turning the problem statement into a mathematical statement.
• Then they have to determine the unknowns in the question and give them variables (a quantity whose value might vary depending on the mathematical situation).
• Then the students need to cite the information, phrases, and keywords after carefully reading the problem several times. And then sequentially arrange the facts they have collected.
• Now the students need to use the Algebraic Expression and the information in the problem statement to frame an equation and then solve it using methodical equation solving approaches.

To understand and practise this better, the students are advised to take help from the NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6.

### Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

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### NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

The NCERT books are organised such that the subject content is divided in a less challenging way. The NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.6 and similar tools are created by the experts while keeping in mind the CBSE’s published syllabus. As the students continue to advance to the next standard, the chapters get harder. This is based on the increasing calibre of students as they move towards the next standard.

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### NCERT Solutions for Class 10 Maths  Chapter 3 Exercises

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• When practised as a profession, Mathematics can bring great career opportunities.
• In a world that is continuously changing, it is crucial to understand the world.
• It gives the students a chance to interact with the outside world.

Q.1

$\begin{array}{l}\text{Solve the following pair of equations by reducing them to a pair of linear equations:}\\ \text{(i) }\frac{\text{1}}{\text{2x}}\text{ + }\frac{\text{1}}{\text{3y}}\text{=2 (ii) }\frac{\text{2}}{\sqrt{\text{x}}}\text{ + }\frac{\text{3}}{\sqrt{\text{y}}}\text{ = 2}\\ \text{ }\frac{\text{1}}{\text{3x}}\text{ + }\frac{\text{1}}{\text{2y}}\text{ = }\frac{\text{13}}{\text{6}}\text{ }\frac{\text{4}}{\sqrt{\text{x}}}\text{ }–\text{ }\frac{\text{9}}{\sqrt{\text{y}}}\text{ =–1}\\ \text{(iii) }\frac{\text{4}}{\text{x}}\text{ + 3y = 14 (iv) }\frac{\text{5}}{\text{x – 1}}\text{ + }\frac{\text{1}}{\text{y – 2}}\text{ = 2}\\ \text{ }\frac{\text{3}}{\text{x}}\text{ }–\text{ 4y = 23 }\frac{\text{6}}{\text{x – 1}}\text{ }–\text{ }\frac{\text{3}}{\text{y – 2}}\text{ = 1}\\ \text{(v) }\frac{\text{7x – 2y}}{\text{xy}}\text{ = 5 (vi) 6x + 3y = 6xy}\\ \text{ }\frac{\text{8x + 7y}}{\text{xy}}\text{ = 15 2x + 4y = 5xy}\\ \text{(vii) }\frac{\text{10}}{\text{x + y}}\text{ + }\frac{\text{2}}{\text{x – y}}\text{ = 4 (viii) }\frac{\text{1}}{\text{3x + y}}\text{ + }\frac{\text{1}}{\text{3x }–\text{ y}}\text{ = }\frac{\text{3}}{\text{4}}\\ \text{ }\frac{\text{15}}{\text{x + y}}\text{ }–\text{ }\frac{\text{5}}{\text{x – y}}\text{= –2 }\frac{\text{1}}{\text{2(3x + y)}}\text{ }–\text{ }\frac{\text{1}}{\text{2(3x – y)}}\text{ = }\frac{\text{–1}}{\text{8}}\\ \end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\\ \text{Let}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\\ \mathrm{then}\text{given equations can be written as:}\\ \frac{\mathrm{p}}{2}+\frac{\mathrm{q}}{3}=2\text{}⇒3\mathrm{p}+2\mathrm{q}-12=0\text{}...\text{(1)}\\ \frac{\mathrm{p}}{3}+\frac{\mathrm{q}}{2}=\frac{13}{6}\text{}⇒2\mathrm{p}+3\mathrm{q}-13=0\text{}...\text{(2)}\\ \text{Using cross-multiplication method, we get}\\ \frac{\mathrm{p}}{-26-\left(-36\right)}=\frac{\mathrm{q}}{-24-\left(-39\right)}=\frac{1}{9-4}\\ \text{}⇒\frac{\mathrm{p}}{10}=\frac{\mathrm{q}}{15}=\frac{1}{5}\\ \text{}⇒\frac{\mathrm{p}}{10}=\frac{1}{5}\text{and}\frac{\mathrm{q}}{15}=\frac{1}{5}\\ \text{}⇒\mathrm{p}=2\text{and q=3}\\ \text{}⇒\frac{1}{\mathrm{x}}=2\text{and}\frac{1}{\mathrm{y}}=3\\ \text{}⇒\mathrm{x}=\frac{1}{2}\text{and y=}\frac{1}{3}\\ \left(\mathrm{ii}\right)\text{}\\ \mathrm{Given}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{that}\\ \frac{2}{\sqrt{\mathrm{x}}}+\frac{3}{\sqrt{\mathrm{y}}}=2\\ \mathrm{and}\\ \frac{4}{\sqrt{\mathrm{x}}}-\frac{9}{\sqrt{\mathrm{y}}}=-1\\ \mathrm{Let}\text{}\frac{1}{\sqrt{\mathrm{x}}}=\mathrm{p}\text{and}\frac{1}{\sqrt{\mathrm{y}}}=\mathrm{q}\text{, then we get}\\ 2\mathrm{p}+3\mathrm{q}=2\text{}...\left(1\right)\\ 4\mathrm{p}-9\mathrm{q}=-1\text{}...\text{(2)}\\ \text{Multiplying equation (1) by 3, we get}\\ \text{6p + 9q = 6}\\ \text{Adding this to equation (2), we get}\\ \text{10p = 5}\\ ⇒\text{p=}\frac{1}{2}\\ \text{Putting value of p in equation (1), we get}\\ \text{2}×\frac{1}{2}+3\mathrm{q}=2\\ ⇒\text{}3\mathrm{q}=1\\ ⇒\text{}\mathrm{q}=\frac{1}{3}\end{array}$

$\begin{array}{l}\mathrm{Now},\\ \mathrm{p}=\frac{1}{\sqrt{\mathrm{x}}}=\frac{1}{2}⇒\sqrt{\mathrm{x}}=2⇒\mathrm{x}=4\\ \mathrm{q}=\frac{1}{\sqrt{\mathrm{y}}}=\frac{1}{3}⇒\sqrt{\mathrm{y}}=3⇒\mathrm{y}=9\\ \mathrm{Hence},\text{}\mathrm{x}=4\text{and}\mathrm{y}=9.\\ \\ \text{(iii)}\\ \text{Given that}\\ \frac{4}{\mathrm{x}}+3\mathrm{y}=14\\ \frac{3}{\mathrm{x}}-4\mathrm{y}=23\\ \text{Substituting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{}\mathrm{in}\text{}\mathrm{above}\text{}\mathrm{equations},\text{we get}\\ \text{4p}+3\mathrm{y}-14=0\text{}...\text{(1)}\\ 3\mathrm{p}-4\mathrm{y}-23=0\text{}...\text{(2)}\\ \text{By cross-multiplication method, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-69-56}=\frac{\mathrm{y}}{-42-\left(-92\right)}=\frac{1}{-16-9}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-125}=\frac{\mathrm{y}}{50}=\frac{1}{-25}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{p}=5\text{and y}=-2\\ ⇒\mathrm{p}=\frac{1}{\mathrm{x}}=5⇒\mathrm{x}=\frac{1}{5}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=-2\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\\ \mathrm{Given}\text{ }\mathrm{that}\\ \frac{5}{\mathrm{x}-1}+\frac{1}{\mathrm{y}-2}=2\\ \mathrm{and}\\ \frac{6}{\mathrm{x}-1}-\frac{3}{\mathrm{y}-2}=1\\ \text{Putting}\frac{1}{\mathrm{x}-1}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}-2}=\mathrm{q}\text{, we get}\\ \text{5p}+\mathrm{q}=2\text{}...\text{(1)}\\ \text{6p}-3\mathrm{q}=1\text{}...\text{(2)}\\ \text{multiplying equation (1) by 3, we get}\\ \text{15p}+3\mathrm{q}=6\text{}...\text{(3)}\\ \text{Adding (2) and (3), we get}\\ \text{21}\mathrm{p}=7⇒\mathrm{p}=\frac{1}{3}\\ \text{Putting value of p in equation (1), we get}\\ \text{5}×\frac{1}{3}+\mathrm{q}=2⇒\mathrm{q}=2-\frac{5}{3}=\frac{1}{3}\\ \mathrm{Now},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}=\frac{1}{\mathrm{x}-1}=\frac{1}{3}⇒\mathrm{x}-1=3⇒\mathrm{x}=4\\ \text{and}\mathrm{q}=\frac{1}{\mathrm{y}-2}=\frac{1}{3}⇒\mathrm{y}-2=3⇒\mathrm{y}=5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{x}=4\text{and y}=5.\end{array}$ 

$\begin{array}{l}\left(\mathrm{v}\right)\\ \mathrm{We}\text{have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{7\mathrm{x}-2\mathrm{y}}{\mathrm{xy}}=5\text{}⇒\text{}\frac{7}{\mathrm{y}}-\frac{2}{\mathrm{x}}=5\text{}...\text{(1)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{8\mathrm{x}+7\mathrm{y}}{\mathrm{xy}}=15\text{}⇒\text{}\frac{8}{\mathrm{y}}+\frac{7}{\mathrm{x}}=15\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in equation (1) and (2), we get}\\ \text{}-2\mathrm{p}+7\mathrm{q}-5=0\text{}...\text{(3)}\\ \text{7p}+8\mathrm{q}-15=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we get}\\ \frac{\mathrm{p}}{-105-\left(-40\right)}=\frac{\mathrm{q}}{-35-30}=\frac{1}{-16-49}\\ ⇒\frac{\mathrm{p}}{-65}=\frac{\mathrm{q}}{-65}=\frac{1}{-65}\\ ⇒\frac{\mathrm{p}}{-65}=\frac{1}{-65}\text{and}\frac{\mathrm{q}}{-65}=\frac{1}{-65}\\ ⇒\mathrm{p}=1\text{and}\mathrm{q}=1\\ ⇒\mathrm{p}=\frac{1}{\mathrm{x}}=1\text{and q}=\frac{1}{\mathrm{y}}=1\\ ⇒\mathrm{x}=1\text{and}\mathrm{y}=1\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\\ \text{We have}\\ 6\mathrm{x}+3\mathrm{y}=6\mathrm{xy}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{6}{\mathrm{y}}+\frac{3}{\mathrm{x}}=6\text{}...\text{(1)}\\ \text{and}\\ 2\mathrm{x}+4\mathrm{y}=5\mathrm{xy}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{\mathrm{y}}+\frac{4}{\mathrm{x}}=5\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{3p}+6\mathrm{q}-6=0\text{}...\text{(3)}\\ \text{4p}+2\mathrm{q}-5=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-30-\left(-12\right)}=\frac{\mathrm{q}}{-24-\left(-15\right)}=\frac{1}{6-24}\\ ⇒\frac{\mathrm{p}}{-18}=\frac{\mathrm{q}}{-9}=\frac{1}{-18}\\ ⇒\frac{\mathrm{p}}{-18}=\frac{1}{-18}\text{and}\frac{\mathrm{q}}{-9}=\frac{1}{-18}\\ ⇒\mathrm{p}=1\text{and q}=\frac{1}{2}\\ ⇒\mathrm{p}=\frac{1}{\mathrm{x}}=1\text{and q}=\frac{1}{\mathrm{y}}=\frac{1}{2}\\ ⇒\mathrm{x}=1\text{and}\mathrm{y}=2\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\text{vii}\right)\text{We have}\\ \text{ }\frac{10}{\mathrm{x}+\mathrm{y}}+\frac{2}{\mathrm{x}-\mathrm{y}}=4\\ \mathrm{and}\text{ }\frac{15}{\mathrm{x}+\mathrm{y}}-\frac{5}{\mathrm{x}-\mathrm{y}}=-2\\ \text{Putting}\frac{1}{\mathrm{x}+\mathrm{y}}=\mathrm{p}\text{and}\frac{1}{\mathrm{x}-\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 10p}+2\mathrm{q}-4=0\text{}...\text{(3)}\\ \text{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 15p}-5\mathrm{q}+2=0\text{}...\text{(4)}\\ \text{By cross-multiplication method, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{4-20}=\frac{\mathrm{q}}{-60-20}=\frac{1}{-50-30}\\ ⇒\frac{\mathrm{p}}{-16}=\frac{\mathrm{q}}{-80}=\frac{1}{-80}\\ ⇒\frac{\mathrm{p}}{-16}=\frac{1}{-80}\text{and}\frac{\mathrm{q}}{-80}=\frac{1}{-80}\\ ⇒\mathrm{p}=\frac{1}{5}\text{and q}=1\\ ⇒\mathrm{p}=\frac{1}{\mathrm{x}+\mathrm{y}}=\frac{1}{5}\text{and q}=\frac{1}{\mathrm{x}-\mathrm{y}}=1\\ \text{Thus we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(5)}\\ \text{and x}-\mathrm{y}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(6)}\\ \text{Adding equations (5) and (6), we get}\\ \text{2x}=6⇒\mathrm{x}=3\end{array}$

$\begin{array}{l}\mathrm{Subtracting}\text{equation (6) from equation (5), we get}\\ \text{}\mathrm{y}=2\\ \text{Hence,}\mathrm{x}=3\text{and}\mathrm{y}=2.\\ \\ \left(\text{viii}\right)\text{}\\ \text{We have}\\ \text{}\frac{1}{3\mathrm{x}+\mathrm{y}}+\frac{1}{3\mathrm{x}-\mathrm{y}}=\frac{3}{4}\\ \text{and}\frac{1}{2\left(3\mathrm{x}+\mathrm{y}\right)}-\frac{1}{2\left(3\mathrm{x}+\mathrm{y}\right)}=\frac{-1}{8}\\ \text{Putting}\frac{1}{3\mathrm{x}+\mathrm{y}}=\mathrm{p}\text{and}\frac{1}{3\mathrm{x}-\mathrm{y}}=\mathrm{q}\text{in above equations, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}+\mathrm{q}=\frac{3}{4}\text{\hspace{0.17em}}...\text{(3)}\\ \mathrm{and}\text{}\frac{\text{p}}{2}-\frac{\mathrm{q}}{2}=\frac{-1}{8}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}-\mathrm{q}=\frac{-1}{4}\text{}...\text{(4)}\\ \text{Adding equations (3) and (4), we get}\\ \text{2p}=\frac{3}{4}-\frac{1}{4}⇒\mathrm{p}=\frac{1}{4}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\mathrm{Now},\\ \text{}\mathrm{p}=\frac{1}{3\mathrm{x}+\mathrm{y}}=\frac{1}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+\mathrm{y}=4\text{}...\text{(5)}\\ \mathrm{Also},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{q}=\frac{1}{3\mathrm{x}-\mathrm{y}}=\frac{1}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-\mathrm{y}=2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(6)}\\ \text{Adding equations (5) and (6), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}6x}=\text{6}⇒\mathrm{x}=1\\ \mathrm{Substitute}\text{}\mathrm{x}=1\text{in equation (5), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3+\mathrm{y}=4⇒\mathrm{y}=1\\ \mathrm{Hence},\text{}\mathrm{x}=1\text{and}\mathrm{y}=1.\end{array}$

Q.2 Formulate the following problem as a pair of linear equations and hence find their solution:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Ans.
(i)
Let speed of Ritu in still water be x km/h and the speed of stream be y km/h.
Speed of Ritu while rowing upstream = (x– y) km/h
Speed of Ritu while rowing downstream = (x + y) km/h
According to question,
2(x + y) = 20
or x + y = 10 …(1)
Also,
2(x – y) = 4
or x – y = 2 …(2)
Adding equations (1) and (2), we get
2x = 12
or x = 6
Putting this in equation (1), we get
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of current is 4 km/h.

(ii)

$\begin{array}{l}\text{Let the number of days taken by a woman to finish the work}\\ \mathrm{is}\text{}\mathrm{x}\text{and the number of days taken by a man to finish the}\\ \text{work}\mathrm{is}\text{}\mathrm{y}.\\ \text{Therefore,}\\ \text{Part of the work finished by a woman in 1 day =}\frac{1}{\mathrm{x}}\\ \mathrm{and}\\ \text{part of the work finished by a man in 1 day =}\frac{1}{\mathrm{y}}\\ \mathrm{Given}\text{}\mathrm{that}\text{}\\ \mathrm{the}\text{}.\\ \text{Therefore,}\\ \text{4}\left(\frac{2}{\mathrm{x}}+\frac{5}{\mathrm{y}}\text{) = 1}\\ ⇒\text{}\frac{2}{\mathrm{x}}+\frac{5}{\mathrm{y}}=\frac{1}{4}\end{array}$

$\begin{array}{l}\mathrm{It}\text{is also given that}\\ \mathrm{the}\text{}\mathrm{work}\text{}.\\ \text{Therefore,}\\ \text{3\hspace{0.17em}\hspace{0.17em}}\left(\frac{3}{\mathrm{x}}+\frac{6}{\mathrm{y}}\right)\text{= 1}\\ ⇒\text{}\frac{3}{\mathrm{x}}+\frac{6}{\mathrm{y}}=\frac{1}{3}\\ \mathrm{P}\text{utting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in the above equations, we get}\\ \text{2p + 5q =}\frac{1}{4}⇒8\mathrm{p}+20\mathrm{q}=1\\ \text{3p + 6q =}\frac{1}{3}⇒9\mathrm{p}+18\mathrm{q}=1\\ \text{By cross-multiplication method, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{-20-\left(-18\right)}=\frac{\mathrm{q}}{-9-\left(-8\right)}=\frac{1}{144-180}\\ ⇒\frac{\mathrm{p}}{-2}=\frac{\mathrm{q}}{-1}=\frac{1}{-36}\\ ⇒\frac{\mathrm{p}}{-2}=\frac{1}{-36}\text{and}\frac{\mathrm{q}}{-1}=\frac{1}{-36}\\ ⇒\mathrm{p}=\frac{1}{18}\text{and}\mathrm{q}=\frac{1}{36}\\ ⇒\mathrm{p}=\frac{1}{\mathrm{x}}=\frac{1}{18}\text{and}\mathrm{q}=\frac{1}{\mathrm{y}}=\frac{1}{36}\\ ⇒\mathrm{x}=18\text{and}\mathrm{y}=36\\ \text{Hence, number of days taken by a woman = 18 days}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}number of days taken by a man = 36 days.}\end{array}$

$\begin{array}{l}\left(\text{iii}\right)\\ \text{Let the speeds of train and bus be x km/h and y km/h respectively.}\\ \text{According to question,}\\ \text{}\frac{60}{\mathrm{x}}+\frac{240}{\mathrm{y}}=4\text{}...\text{(1)}\\ \text{and}\\ \text{}\frac{100}{\mathrm{x}}+\frac{200}{\mathrm{y}}=\frac{25}{6}\text{}...\text{(2)}\\ \text{Putting}\frac{1}{\mathrm{x}}=\mathrm{p}\text{and}\frac{1}{\mathrm{y}}=\mathrm{q}\text{in these equations, we get}\\ \text{}60\mathrm{p}+240\mathrm{q}=4\text{}...\text{(3)}\\ \text{and}100\mathrm{p}+200\mathrm{q}=\frac{25}{6}\text{}\\ ⇒\text{}600\mathrm{p}+1200\mathrm{q}=25\text{}...\text{(4)}\\ \text{Multiplying equation (3) by 10, we get}\\ \text{}600\mathrm{p}+2400\mathrm{q}=40\text{}...\text{(5)}\\ \text{Subtracting equation (4) ​from (5), we get}\\ \text{1200q = 15}⇒\mathrm{q}=\frac{1}{80}\\ \text{Substituting in equation (3), we get}\\ \text{60p = 1}⇒\mathrm{p}=\frac{1}{60}\\ \mathrm{Now},\text{}\mathrm{p}=\frac{1}{\mathrm{x}}=\frac{1}{60}\text{and q =}\frac{1}{\mathrm{y}}=\frac{1}{80}\end{array}$

$\begin{array}{l}⇒\text{x = 60 km/h and y = 80 km/h}\\ \text{Hence, speed of train = 60 km/h}\\ \text{and\hspace{0.17em}\hspace{0.17em}speed of bus = 80 km/h}\end{array}$