Class 10 math chapter 3 exercise 3.3

NCERT Solutions For Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables NCERT Solutions for Class 10 Maths Chapter 3 – CBSE Download Free PDF NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables will help the students in understanding how the problems under this concept are solved. Maths is one subject that requires a lot of practice. The students appearing for the 10 th grade board examinations can turn to the NCERT textbook. An equation that can be of the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. The NCERT Solutions for Class 10 Maths Chapter 3 also lets the students understand the fact that a solution of such an equation is a pair of values, one for x and the other for y, which makes the two sides of the equation equal. Students also learn that every solution of the equation is a point on the line representing it. Students get to learn the problem-solving method of the questions present in Chapter 3 Pair of Linear Equations in Two Variables of CBSE Syllabus for 2023-24, revolving around the concepts mentioned above and more, by practising the Solutions of NCERT are extremely beneficial from the CBSE examination perspective.

Table of Contents • • • • • • • • • • • NCERT Solutions for Class 10 Maths Exercise 3.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free. NCERT solutions for Maths Pair of Linear Equations in Two Variables NCERT Solutions for Class 10 Maths Pair of Linear Equations in Two Variables 1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 x – y = 4 (ii) s – t = 3 (iii) 3x – y = 3 9x − 3y = 9 (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 (v) (vi) Ans. (i) x + y = 14 …(1) x – y = 4 … (2) x = 4 + y from equation (2) Putting this in equation (1), we get 4 + y + y = 14 ⇒ 2 y = 10⇒ y = 5 NCERT Solutions for Class 10 Maths Exercise 3.3 Putting value of y in equation (1), we get x + 5 = 14 ⇒ x = 14 – 5 = 9 Therefore, x = 9 and y = 5 (ii) s – t = 3 … (1) …(2) Using equation (1), we can say that s = 3 + t Putting this in equation (2), we get ⇒ ⇒ 5 t + 6 = 36 ⇒ 5 t = 30⇒ t = 6 Putting value of t in equation (1), we get s – 6 = 3⇒ s = 3 + 6 = 9 Therefore, t = 6 and s = 9 (iii) 3 x – y = 3 … (1) 9 x − 3 y = 9 … (2) Comparing equation 3 x – y = 3 with and equation 9 x − 3 y = 9 with , We get Here T...

Updated for NCERT 2023-24 Books Get NCERT solutions of Chapter 3 Class 10 - Pair of Linear Equations in Two Variables at Teachoo. Answers to all exercise questions, examples and optional questions have been provided with video of each and every question We studied pair of linear equations in this chapter. In this chapter, we will learn • What are Linear Equations in Two Variables • Converting statements into Equations, and drawing graph of those linear equations • Possible Type of Graphs for Pair of Linear Equations in Two Variables - Two Lines Intersecting, Two lines Parallel, Coincident Lines • Finding s olution of equations from graphs • Consistency of equations by finding ratio of a 1/a 2, b 1/b 2, c 1/c 2 • and checking whether lines are • Intersecting Lines (Exactly one solution - unique) • Coincident Lines (Infinitely many solutions) • Parallel Lines (No solutions) • Solving Pair of Linear Equations by • Substitution Method • Elimination Method • Cross Multiplication Method • Solving complicated equations like 2/x + 3/y = 4 by substituting variables (like putting p = 1/x, q = 1/y and solving) • Solving Statement Questions by first forming equations, and then solving Click on exercise or topic link below to start doing the chapter Note: When you click on a link, the first question will open. To open any other question of the exercise, go to the bottom of the page. There is a list with arrows having all the questions (with important questions also marked)

UP Board Solutions for Class 10 Maths Chapter 3 Pairs of Linear Equations in Two Variables These Solutions are part of प्रश्नावली 3.1 (NCERT Page 49) प्र. 1. आफ़ताब अपनी पुत्री से कहता है, ‘सात वर्ष पूर्व मैं तुमसे सात गुनी आयु का था | अब से 3 वर्ष बाद मैं तुमसे केवल तीन गुनी आयु का रह जाऊँगा’ (क्या यह मनोरंजक है?) इस स्थिति को बीजगणितीय एवं ग्राफीय रूपों में व्यक्त कीजिए| हल : माना आफ़ताब की वर्त्तमान आयु = x वर्ष और उसकी पुत्री की वर्त्तमान आयु = y वर्ष 7 वर्ष पूर्व आफ़ताब की आयु = x – 7 वर्ष और उसकी पुत्री की आयु (UPBoardSolutions.com) = y – 7 वर्ष स्थित – I x – 7 = 7(y – 7) x – 7 = 7y – 49 x – 7y = 7 – 49 x – 7y = – 42 ……… (1) 3 वर्ष बाद आफ़ताब की आयु = x + 3 वर्ष और उसकी पुत्री की आयु = y + 3 वर्ष स्थित – II x + 3 = 3(y + 3) x + 3 = 3y + 9 x – 3y = 9 – 3 x – 3y = 6 ……. (2) बीजगणितीय रूप में : x – 7y = – 42 ……… (1) x – 3y = 6 ……. (2) प्र. 2. क्रिकेट टीम के एक कोच ने 3900 रू में 3 बल्ले तथा 6 (UPBoardSolutions.com) गेंदे खरीदी बाद में उसने एक और बल्ला तथा उसी प्रकार की 2 गेंदे 1300 रू में खरीदीं इस स्थिति को बीजगणितीय तथा ज्यामितीय रूपों में व्यक्त कीजिए हल : माना एक बल्ले का मूल्य = x रुपये और एक गेंद का मूल्य = y रुपये अत: बीजगणितीय निरूपण 3x + 6y = 3900 ………. (1) और x + 2y = 1300 ………. (2) समी० (1) से 3x + 6y = 3900 3(x + 2y) = 3990 या x + 2y = 1300 x = 1300 – 2y इसी प्रकार समी० (2) से x + 2y = 1300 x = 1300 – 2y प्र. 3. 2kg सेब और 1 kg अंगूर का मूल्य किसी दिन 160 रू था | एक (UPBoardSolutions.com) महीने बाद 4 kg सेब और दो kg अंगूर का मूल्य 300 रू हो जाता है |इस स्थिति को ब...

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Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions Exercise 3.3 Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 3 Class 11 Trigonometric Functions Ex 3.3 provided in NCERT Textbook. • • • • • • • • • • • • • • • Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20. Ex 3.3 Class 11 Maths Question 1: Prove that: sin 2 \(\frac\) Ex 3.3 Class 11 Maths Question 24: Prove that: cos 4x = 18 sin 2 x cos 2 x Ans: L.H.S. = cos 4x = cos 2(2x) = 1 – 2 sin 2 2x [∵ cos 2A = 1 – 2 sin 2 A] = 1 – 2(2 sin x cos x) 2 [∵ sin 2A = 2 sin A cos A] = 1 – 8 sin 2 x cos 2 x = R.H.S. Hence proved. Ex 3.3 Class 11 Maths Question 25: Prove that: cos 6x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x – 1 Ans: We know that: cos 3x = 4 cos 3 x – 3cos x On replacing x by 2x, we get cos 3(2x) = 4 cos 3 (2x) – 3 cos 2x ⇒ cos 6x = 4 (2cos 2 x – 1) 3– 3 (2cos 2 x – 1) [∵ cos 2x = 2cos 2 x – 1] = 4 [8 cos 6 x – 12 cos 4 x + 6 cos 2 x – 1] – 6 cos 2 x + 3 [∵ (a – b) 3 = a 3– 3a 2b + 3ab ...

Here we are giving Chapter 3 Pair of Linear Equations Class 10 Maths NCERT Solutions that will help you in passingthe exams with flying colors. Chapter 3 NCERT Solutions Class 10 Maths can be used for solving your homework and understanding the concepts and formula in the chapter. The solutions are updated as per the latest marking scheme provided by CBSE 2019-20.

NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables Exercise 3.3 NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 contains the solutions to all the questions provided on page number 53 of the textbook. The NCERT Solutions Class 10 Maths Chapter 3 – Pair of Linear Variables is an important topic for the examinations and should be dealt with in complete detail. Practising the exercises repeatedly will help the students score well in the board examinations. Previous Next Access other exercise solutions of Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables Access answers of Maths NCERT Class 10 Chapter 3 – Pair of Linear Equations in Two Variables Exercise 3.3 1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 x – y = 4 (ii) s – t = 3 (s/3) + (t/2) = 6 (iii) 3x – y = 3 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 (v) √2 x+√3 y = 0 √3 x-√8 y = 0 (vi) (3x/2) – (5y/3) = -2 (x/3) + (y/2) = (13/6) Solutions: (i) Given, x + y = 14 and x – y = 4 are the two equations. From the 1 st equation, we get, x = 14 – y Now, substitute the value of x to the second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the exact value of x. ∵ x = 14 – y ∴ x = 14 – 5 Or x = 9 Hence, x = 9 and y = 5 (ii) Given, s – t = 3 and (s/3) + (t/2) = 6 are the two equations. From the 1 st equation, we get, s = ...