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

NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.6 are provided here to help students with their Class 10 exam preparations. These solutions are designed by subject experts to explain the methods of solving linear equations using substitution, elimination, and graphical methods in a detailed and step-by-step manner. The solutions are aimed at simplifying the process of solving pair of linear equations and understanding their applications.

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

Pair of Linear Equations in Two Variables

Medium

Q.

$\begin{array}{l}\text{Solve the following pair of equations by reducing them to a pair of linear equations:}\\ \text{(i)\hspace{0.33em}}\frac{\text{1}}{\text{2x}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{3y}}\text{=2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(ii)\hspace{0.33em}}\frac{\text{2}}{\sqrt{\text{x}}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{3}}{\sqrt{\text{y}}}\text{\hspace{0.33em}=\hspace{0.33em}2}\\ \frac{\text{1}}{\text{3x}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{2y}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{13}}{\text{6}}\frac{\text{4}}{\sqrt{\text{x}}}–\frac{\text{9}}{\sqrt{\text{y}}}\text{\hspace{0.33em}=–1}\\ \text{(iii)\hspace{0.33em}}\frac{\text{4}}{\text{x}}\text{\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}14\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(iv)\hspace{0.33em}}\frac{\text{5}}{\text{x\hspace{0.33em}–\hspace{0.33em}1}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{y\hspace{0.33em}–\hspace{0.33em}2}}\text{\hspace{0.33em}=\hspace{0.33em}2}\\ \frac{\text{3}}{\text{x}}–\text{\hspace{0.33em}4y\hspace{0.33em}=\hspace{0.33em}23\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{6}}{\text{x\hspace{0.33em}–\hspace{0.33em}1}}–\frac{\text{3}}{\text{y\hspace{0.33em}–\hspace{0.33em}2}}\text{\hspace{0.33em}=\hspace{0.33em}1}\\ \text{(v)\hspace{0.33em}}\frac{\text{7x\hspace{0.33em}–\hspace{0.33em}2y}}{\text{xy}}\text{\hspace{0.33em}=\hspace{0.33em}5\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(vi)\hspace{0.33em}6x\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}6xy}\\ \frac{\text{8x\hspace{0.33em}+\hspace{0.33em}7y}}{\text{xy}}\text{\hspace{0.33em}=\hspace{0.33em}15\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2x\hspace{0.33em}+\hspace{0.33em}4y\hspace{0.33em}=\hspace{0.33em}5xy}\\ \text{(vii)\hspace{0.33em}}\frac{\text{10}}{\text{x\hspace{0.33em}+\hspace{0.33em}y}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{2}}{\text{x\hspace{0.33em}–\hspace{0.33em}y}}\text{\hspace{0.33em}=\hspace{0.33em}4\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(viii)\hspace{0.33em}\hspace{0.33em}}\frac{\text{1}}{\text{3x\hspace{0.33em}+\hspace{0.33em}y}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{1}}{\text{3x\hspace{0.33em}}–\text{\hspace{0.33em}y}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{3}}{\text{4}}\\ \frac{\text{15}}{\text{x\hspace{0.33em}+\hspace{0.33em}y}}–\frac{\text{5}}{\text{x\hspace{0.33em}–\hspace{0.33em}y}}\text{=\hspace{0.33em}–2\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\frac{\text{1}}{\text{2(3x\hspace{0.33em}+\hspace{0.33em}y)}}–\frac{\text{1}}{\text{2(3x\hspace{0.33em}–\hspace{0.33em}y)}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{–1}}{\text{8}}\\ \end{array}$

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Pair of Linear Equations in Two Variables

Medium

Q.

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.

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Exercise 3.6 focuses on solving pairs of linear equations in two variables through various methods and applying them to real-life situations. By practicing this exercise, students will get a better understanding of graphical solutions, algebraic solutions, and the relationship between the coefficients and solutions of the equations. This exercise helps students improve their problem-solving skills for Class 10 exams.

These solutions are aligned with the latest CBSE syllabus, offering clear, step-by-step guidance and detailed solutions to help students perform well in board examinations.

Q1: Solve the following pair of equations by reducing them to a pair of linear equations:

(i)

$$\frac{1}{2}x + \frac{1}{3}y = 2$$ $$\frac{2}{3}x + \frac{3}{2}y = 2$$

Solution:

To solve this system, we can first eliminate the fractions by multiplying through by the least common denominator (LCD) for each equation.

For the first equation, multiply through by 6 (LCM of 2 and 3):

$$6 \times \left( \frac{1}{2}x + \frac{1}{3}y \right) = 6 \times 2$$ $$3x + 2y = 12$$

For the second equation, multiply through by 6 (LCM of 3 and 2):

$$6 \times \left( \frac{2}{3}x + \frac{3}{2}y \right) = 6 \times 2$$ $$4x + 9y = 12$$

Now, solve the system of equations:

1. $3x + 2y = 12$

2. $4x + 9y = 12$

We can solve by substitution or elimination. Let's use elimination. Multiply the first equation by 3 and the second equation by 2 to align the coefficients of $x$:

$$9x + 6y = 36$$ $$8x + 18y = 24$$

Now subtract the first equation from the second:

$$(8x + 18y) - (9x + 6y) = 24 - 36$$ $$-x + 12y = -12$$ $$x = 12y + 12$$

Substitute this back into one of the original equations to find the values of $x$ and $y$.

(ii)

$$\frac{1}{3}x + \frac{1}{2}y = \frac{13}{6}$$ $$\frac{4}{3}x - \frac{9}{2}y = -1$$

Solution:

To simplify, multiply both sides of each equation by the least common denominators.

For the first equation, multiply by 6:

$$6 \times \left( \frac{1}{3}x + \frac{1}{2}y \right) = 6 \times \frac{13}{6}$$ $$2x + 3y = 13$$

For the second equation, multiply by 6:

$$6 \times \left( \frac{4}{3}x - \frac{9}{2}y \right) = 6 \times (-1)$$ $$8x - 27y = -6$$

Now we have:

1. $2x + 3y = 13$

2. $8x - 27y = -6$

Now solve this system using substitution or elimination. We can multiply the first equation by 4 and subtract to eliminate $x$.

Q2: 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.

Solution:

Let the speed of Ritu in still water be $x$ km/hr and the speed of the current be $y$ km/hr.

• Downstream, Ritu’s speed is $x + y$ km/hr, and she covers 20 km in 2 hours:

  $$\frac{20}{x + y} = 2 \quad \Rightarrow \quad x + y = 10$$

• Upstream, Ritu’s speed is $x - y$ km/hr, and she covers 4 km in 2 hours:

  $$\frac{4}{x - y} = 2 \quad \Rightarrow \quad x - y = 2$$

Now, we have the system of linear equations:

1. $x + y = 10$

2. $x - y = 2$

Add these two equations:

$$(x + y) + (x - y) = 10 + 2$$ $$2x = 12 \quad \Rightarrow \quad x = 6$$

Substitute $x = 6$ into $x + y = 10$:

$$6 + y = 10 \quad \Rightarrow \quad y = 4$$

So, the speed of Ritu in still water is 6 km/hr, and the speed of the current is 4 km/hr.

(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.

Solution:

Let the amount of work one woman can do in 1 day be $x$ and the amount of work one man can do in 1 day be $y$.

• 2 women and 5 men can finish the work in 4 days:

  $$4(2x + 5y) = 1 \quad \Rightarrow \quad 2x + 5y = \frac{1}{4}$$

• 3 women and 6 men can finish the work in 3 days:

  $$3(3x + 6y) = 1 \quad \Rightarrow \quad 3x + 6y = \frac{1}{3}$$

Now, solve the system of equations:

1. $2x + 5y = \frac{1}{4}$

2. $3x + 6y = \frac{1}{3}$

Multiply the first equation by 3 and the second equation by 2 to eliminate the fractions and solve for $x$ and $y$.

FAQs: Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.6

Q1. What is the focus of Exercise 3.6?
Answer:
Exercise 3.6 focuses on solving pairs of linear equations using methods such as substitution, elimination, and graphical representation. It emphasizes the application of these methods to real-life problems and their solutions.

Q2. How do I solve pair of linear equations graphically?
Answer:
To solve the pair of linear equations graphically:

1. Plot both equations on a Cartesian plane.

2. Identify the point of intersection of the two lines, which represents the solution of the system.

Q3. What are the key methods used to solve linear equations in this exercise?
Answer:
The substitution method, elimination method, and graphical method are the main approaches used to solve linear equations in this exercise.

Q4. What is the significance of graphical solutions in this exercise?
Answer:
Graphical solutions help visualize the relationship between two linear equations and identify the point of intersection, which represents the solution. This method also helps in understanding the nature of the system, whether it has a unique solution, no solution, or infinite solutions.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step explanations for solving linear equations using multiple methods. They help students understand the underlying concepts of linear equations and apply them confidently in both algebraic and graphical forms. Practicing these solutions ensures better exam preparation and improved problem-solving abilities.