Rs aggarwal class 10 solutions 2022-23

Students can download Class 10 Rs Aggarwal Chapter 2 Solution free of cost here that will help them in the preparation of their board examinations. These RS Aggarwal Class 10 Ch 2 solutions are prepared by our subject experts at Vedantu, who have years of experience in teaching and the solutions are therefore accurate and reliable. RS Aggarwal Solutions Class 10 Polynomials provide an in-depth understanding of the concepts of the chapter and teach you how to format your answers in the exam correctly. These solutions are written in accordance with the CBSE guidelines to help you score well in your examinations. The important topics of the chapter- Zeros of a polynomial and the relationship between coefficients and zeros of quadratic polynomials are extensively covered in the PDF. This solution will enable students to resolve their queries regarding the problems included in the exercises right away. By referring to these solutions, you will be able to define a strategy to approach these problems and score more in the exams. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download free PDFs of Class 10 Maths and Class 10 Science NCERT Solutions to help you to revise the complete syllabus and score more marks in your examinations. Where Can I Download RS Aggarwal Class 10 Solutions- Polynomials? You can download RS Aggarwal Class 10 Solutions- Polynomials on Vedantu. It has all the relevant study material that the...

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Chapter Name RS Aggarwal Chapter 3 Linear Equations in Two Variables Book Name RS Aggarwal Mathematics for Class 10 Other Exercises • Exercise 3B • Exercise 3C • Exercise 3D • Exercise 3E • Exercise 3F Related Study NCERT Solutions for Class 10 Maths Exercise 3A Solutions 1. Solve the system of equations graphically: 2x + 3y = 2, x – 2y = 8 Solution On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x axis and y-axis, respectively. Graph of 2x + 3y = 2 2x + 3y = 2 ⇒ 3y = (2 – 2x) ⇒ 3y = 2(1 – x) ⇒ y = 2(1 – x)/3 …(i) Putting x = 1, we get y = 0 Putting x = -2, we get y = 2 Putting x = 4, we get y = -2 Thus, we have the following table for the equation 2x + 3y = 2 x 1 -2 4 y 0 2 -2 Now, plot the points A(1, 0), B(-2, 2) and C(4, -2) on the graph paper. Join AB and AC to get the graph line BC. Extend it on both ways. Thus, the line BC is the graph of 2x + 3y = 2. Graph of x - 2y = 8 x – 2y = 8 ⇒ 2y = (x – 8) ⇒ y = (x – 8)/2 …(ii) Putting x = 2, we get y = -3 Putting x = 4, we get y = -2 Putting x = 0, we get y = -4 Thus, we have the following table for the equation x – 2y = 8. x 2 4 0 y -3 -2 -4 Now, plot the points P(0, -4) and Q(2, -3). The point C(4, -2) has already been plotted. JoinPQ and QC and extend it on both ways. Thus, line PC is the graph of x – 2y = 8. The two graph lines intersect at C(4, -2). ∴ x = 4 and y = -2 are the solutions of the given system of equations. 2. Solve the system of equations graphically: 3x + 2y = 4, 2x ...