RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables has been provided here. Students can refer to these questions before their examinations for better preparation.
Neha Tanna1 Nov, 2024
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RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3: In Chapter 3, Exercise 3.3 of RD Sharma's Class 10 Maths, students explore solving pairs of linear equations in two variables using graphical and algebraic methods. This exercise delves into three main algebraic methods: substitution, elimination, and cross-multiplication.
Each method is applied to find the point of intersection, or the solution, of the two equations, which represents the values of the variables satisfying both equations simultaneously. Exercise 3.3 also includes practical problems, enhancing students' problem-solving skills and helping them understand the real-life application of linear equations in various contexts.RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Overview
Exercise 3.3 in RD Sharma’s Class 10 Maths on Pair of Linear Equations in Two Variables is crucial for developing a strong foundation in algebra. Mastering methods like substitution, elimination, and cross-multiplication empowers students to solve real-world problems involving two variables, such as calculating rates, distances, and finances. These techniques are fundamental for higher mathematics, physics, and economics, were systems of equations frequently model complex relationships. Understanding these concepts enhances logical thinking and problem-solving skills, essential for competitive exams and advanced studies. This exercise builds confidence in handling equations, laying groundwork for more advanced algebraic and analytical concepts.RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 PDF
Below, we have provided a PDF of RD Sharma Solutions for Class 10 Maths Chapter 3, Exercise 3.3, focused on Pair of Linear Equations in Two Variables. This resource includes step-by-step solutions to each question, helping students understand methods like substitution, elimination, and cross-multiplication. These solutions simplify complex problems, offering a clear approach to mastering linear equations in two variables.RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 PDF
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables
Below is the RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 Pair of Linear Equations in Two Variables -Solve the following system of equations:
1. 11x + 15y + 23 = 0
7x – 2y – 20 = 0
Solution:
The given pair of equations are 11x +15y + 23 = 0 …………………………. (i) 7x – 2y – 20 = 0 …………………………….. (ii) From (ii), 2y = 7x – 20 ⇒ y = (7x −20)/2 ……………………………… (iii) Now, substituting y in equation (i), we get ⇒ 11x + 15((7x−20)/2) + 23 = 0 ⇒ 11x + (105x − 300)/2 + 23 = 0 ⇒ (22x + 105x – 300 + 46) = 0 ⇒ 127x – 254 = 0 ⇒ x = 2 Next, putting the value of x in equation (iii), we get, ⇒ y = (7(2) − 20)/2 ∴ y= -3 Thus, the value of x and y is found to be 2 and -3, respectively.2. 3x – 7y + 10 = 0
y – 2x – 3 = 0
Solution:
The given pair of equations are 3x – 7y + 10 = 0 …………………………. (i) y – 2x – 3 = 0 ……………………………….. (ii) From (ii), y – 2x – 3 = 0 y = 2x+3 ……………………………… (iii) Now, substituting y in equation (i), we get ⇒ 3x – 7(2x+3) + 10 = 0 ⇒ 3x – 14x – 21 + 10 = 0 ⇒ -11x = 11 ⇒ x = -1 Next, putting the value of x in equation (iii), we get ⇒ y = 2(-1) + 3 ∴ y= 1 Thus, the value of x and y is found to be -1 and 1, respectively.3. 0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Solution:
The given pair of equations are 0.4x + 0.3y = 1.7 0.7x – 0.2y = 0.8 Let’s, multiply LHS and RHS by 10 to make the coefficients an integer. 4x + 3y = 17 ……………………….. (i) 7x – 2y = 8 …………………………… (ii) From (ii), 7x – 2y = 8 x = (8 + 2y)/7……………………………… (iii) Now, substituting x in equation (i), we get ⇒ 4[(8 + 2y)/7] + 3y = 17 ⇒ 32 + 8y + 21y = (17 x 7) ⇒ 29y = 87 ⇒ y = 3 Next, putting the value of y in equation (iii), we get ⇒ x = (8 + 2(3))/ 7 ⇒ x = 14/7 ∴ x = 2 Thus, the value of x and y is found to be 2 and 3, respectively.4. x/2 + y = 0.8
7/(x+ y/2) = 10
Solution:
The given pair of equations are x/2 + y = 0.8 ⇒ x + 2y = 1.6…… (a) 7/(x + y/2) = 10 ⇒7 = 10(x + y/2) ⇒7 = 10x + 5y Let’s, multiply the LHS and RHS of equation (a) by 10 for easy calculation. So, we finally get 10x + 20y = 16 ……………………….. (i) And, 10x + 5y = 7 …………………………… (ii) Now, subtracting two equations, we get ⇒ (i) – (ii) 15y = 9 ⇒ y = 3/5 Next, putting the value of y in the equation (i), we get x = [16 − 20(3/5)]/10 ⇒ (16 – 12)/10 = 4/10 ∴ x = 2/5 Thus, the value of x and y obtained are 2/5 and 3/5, respectively.5. 7(y + 3) – 2(x + 2) = 14
4(y – 2) + 3(x – 3) = 2
Solution:
The given pair of equations are 7(y+3) – 2(x+2) = 14…………………………. (i) 4(y-2) + 3(x-3) = 2……………………………….. (ii) From (i), we get 7y + 21 – 2x – 4 = 14 7y = 14 + 4 – 21 + 2x ⇒ y = (2x – 3)/7 From (ii), we get 4y – 8 + 3x – 9 = 2 4y + 3x – 17 – 2 = 0 ⇒ 4y + 3x – 19 = 0 …………….. (iii) Now, substituting y in equation (iii), 4[(2x − 3)/7] + 3x – 19=0 8x – 12 + 21x – (19 x 17) = 0 [after taking LCM] 29x = 145 ⇒ x = 5 Now, putting the value of x and in equation (ii), 4(y-2) + 3(5-3) = 2 ⇒ 4y -8 + 6 = 2 ⇒ 4y = 4 ∴ y = 1 Thus, the value of x and y obtained are 5 and 1, respectively.6. x/7 + y/3 = 5
x/2 – y/9 = 6
Solution:
The given pair of equations are x/7 + y/3 = 5…………………………. (i) x/2 – y/9 = 6………………………………..(ii) From (i), we get x/7 + y/3 = 5 ⇒3x + 7y = (5×21) [After taking LCM] ⇒ 3x =105 – 7y ⇒ x = (105 – 7y)/3……. (iv) From (ii), we get x/2 – y/9 = 6 ⇒ 9x – 2y = 108 ……………………… (iii) [After taking LCM] Now, substituting x in equation (iii), we get 9[(105 − 7y)/3] – 2y = 108 ⇒ 945 – 63y – 6y = 324 [After taking LCM] ⇒ 945 – 324 = 69y ⇒ 69y = 621 ⇒ y = 9 Now, putting the value of y in equation (iv), x = (105 − 7(9))/3 ⇒ x = (105 − 63)/3 = 42/3 ∴ x = 14 Thus, the value of x and y obtained are 14 and 9, respectively.7. x/3 + y/4 = 11
5x/6 − y/3 = −7
Solution:
The given pair of equations are x/3 + y/4 = 11…………………………. (i) 5x/6 − y/3 = −7……………………………….. (ii) From (i), we get x/3 + y/4 = 11 ⇒4x + 3y = (11×12) [After taking LCM] ⇒ 4x =132 – 3y ⇒ x = (132 – 3y)/4……. (iv) From (ii), we get 5x/6 − y/3 = −7 ⇒ 5x – 2y = -42 ……………………… (iii) [After taking LCM] Now, substituting x in equation (iii), we get 5[(132 − 3y)/4] – 2y = -42 ⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM] ⇒ 660 + 168 = 23y ⇒ 23y = 828 ⇒ y = 36 Now, putting the value of y in equation (iv), x = (132 – 3(36))/4 ⇒ x = (132 − 108)/4 = 24/4 ∴ x = 6 Thus, the value of x and y obtained are 6 and 36, respectively.8. 4/x + 3y = 8
6/x −4y = −5
Solution:
Taking 1/x = u Then, the two equations become 4u + 3y = 8…………………… (i) 6u – 4y = -5……………………. (ii) From (i), we get 4u = 8 – 3y ⇒ u = (8 − 3y)/4 …….. (iii) Substituting u in (ii), [6(8 − 3y)/4] – 4y = -5 ⇒ [3(8−3y)/2] − 4y = −5 ⇒ 24 − 9y −8y = −5 x 2 [After taking LCM] ⇒ 24 – 17y = -10 ⇒ -17y =- 34 ⇒ y = 2 Putting y=2 in (iii), we get u = (8 − 3(2))/4 ⇒ u = (8 − 6)/4 ⇒ u = 2/4 = 1/2 ⇒ x = 1/u = 2 ∴ x = 2 So, the solution of the pair of equations given is x=2 and y =2.9. x + y/2 = 4
2y + x/3 = 5
Solution:
The given pair of equations are x + y/2 = 4 ……………………. (i) 2y + x/3 = 5……………………. (ii) From (i), we get x + y/2 = 4 ⇒ 2x + y = 8 [After taking LCM] ⇒ y = 8 – 2x …..(iv) From (ii), we get x + 6y = 15 ……………… (iii) [After taking LCM] Substituting y in (iii), we get x + 6(8 – 2x) = 15 ⇒ x + 48 – 12x = 15 ⇒ -11x = 15 – 48 ⇒ -11x = -33 ⇒ x = 3 Putting x = 3 in (iv), we get y = 8 – (2×3) ∴ y = 8 – 6 = 2 Hence, the solutions of the given system of the equation are x = 3 and y = 2, respectively.10. x + 2y = 3/2
2x + y = 3/2
Solution:
The given pair of equations are x + 2y = 3/2 …………………. (i) 2x + y = 3/2…………………… (ii) Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1, respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations. Multiplying equation (i)x1 and (ii)x2 ⇒ x + 2y = 3/2 ………………………. (iii) 4x + 2y = 3 ……………………………………………………. (iv) Subtracting equation (iii) from (iv), (4x – x) + (2y-2y) = 3x = 3 – (3/2) ⇒ 3x = 3/2 ⇒ x = 1/2 Putting x = 1/2 in equation (iv), 4(1/2) + 2y = 3 ⇒ 2 + 2y = 3 ∴ y= 1/2 The solution of the system of equations is x = 1/2 and y = 1/211. √2x – √3y = 0
√3x − √8y = 0
Solution:
The given pair of equations are √2x – √3y = 0……………………….. (i) √3x − √8y = 0……………………….. (ii) From equation (i), x = √(3/2)y ……………..(iii) Substituting this value in equation (ii), we obtain √3x − √8y = 0 ⇒ √3(√(3/2)y) − √8y = 0 ⇒ (3/√2)y – √8y = 0 ⇒ 3y – 4y = 0 ⇒ y = 0 Now, substituting y in equation (iii), we obtain ⇒ x=0 Thus, the value of x and y obtained are 0 and 0, respectively.12. 3x – (y + 7)/11 + 2 = 10
2y + (x + 11)/7 = 10
Solution:
The given pair of equations are 3x – (y + 7)/11 + 2 = 10……………….. (i) 2y + (x + 11)/7 = 10…………………….. (ii) From equation (i), 33x – y – 7 + 22 = (10 x 11) [After taking LCM] ⇒ 33x – y + 15 = 110 ⇒ 33x + 15 – 110 = y ⇒ y = 33x – 95………. (iv) From equation (ii), 14 + x + 11 = (10 x 7) [After taking LCM] ⇒ 14y + x + 11 = 70 ⇒ 14y + x = 70 – 11 ⇒ 14y + x = 59 …………………….. (iii) Substituting (iv) in (iii), we get 14 (33x – 95) + x = 59 ⇒ 462x – 1330 + x = 59 ⇒ 463x = 1389 ⇒ x = 3 Putting x = 3 in (iii), we get ⇒ y = 33(3) – 95 ∴ y= 4 The solution for the given pair of equations is x = 3 and y = 4, respectively.13. 2x – (3/y) = 9
3x + (7/y) = 2 ,y ≠ 0
Solution:
The given pair of equations are 2x – (3/y) = 9……………………………. (i) 3x + (7/y) = 2…………………………… (ii) Substituting 1/y = u, the above equations become 2x – 3u = 9 ………………………..(iii) 3x + 7u = 2………………………..(iv) From (iii) 2x = 9 + 3u ⇒ x = (9+3u)/2 Substituting the value of x from above in equation (iv), we get 3[(9+3u)/2] + 7u = 2 ⇒ 27 + 9u + 14u = (2 x 2) ⇒ 27 + 23u = 4 ⇒ 23u = -23 ⇒ u = -1 So, y = 1/u = -1 And putting u = -1 in x = (9 + 3u)/2, we get ⇒ x = [9 + 3(−1)]/2 = 6/2 ∴ x = 3 The solutions of the pair of equations given are y = 3 and x = -1, respectively.14. 0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Solution:
The given pair of equations are 0.5x + 0.7y = 0.74……………………… (i) 0.3x – 0.5y = 0.5 ………………………….. (ii) Now, let’s multiply LHS and RHS by 100 for both (i) and (ii) to make integral coefficients and constants. (i) x100 ⇒ 50x +70y = 74 ……………………….. (iii) (ii) x100 ⇒ 30x + 50y = 50 …………………………… (iv) From (iii), 50x = 74 – 70y x = (74−70y)/ 50 ……………………………… (v) Now, substituting x in equation (iv), we get 30[(74−70y)/ 50] + 50y = 50 ⇒ 222 – 210y + 250y = 250 [After taking LCM] ⇒ 40y = 28 ⇒ y = 0.7 Now, by putting the value of y in the equation (v), we get ⇒ x = [74 − 70(0.7)]/ 50=0 ⇒ x =25/ 50 = 1/2 ∴ x = 0.5 Thus, the values of x and y so obtained are 0.5 and 0.7, respectively.15. 1/(7x) + 1/(6y) = 3
1/(2x) – 1/(3y) = 5
Solution:
The given pair of equations are 1/(7x) + 1/(6y) = 3………………………….. (i) 1/(2x) – 1/(3y) = 5……………………………. (ii) Multiplying (ii) by 1/2, we get 1/(4x) – 1/(6y) = 52……………………………. (iii) Now, solving equations (i) and (iii), 1/(7x) + 1/(6y) = 3………………………….. (i) 1/(4x) – 1/(6y) = 5/2……………………………. (iii) Adding (i) + (iii), we get 1/x(1/7 + 1/4 ) = 3 + 5/2 ⇒ 1/x(11/28) = 11/2 ⇒ x = 1/14 Using x =1/ 14, we get, in (i) 1/[7(1/14)] + 1/(6y) = 3 ⇒ 2 + 1/(6y)=3 ⇒ 1/(6y) = 1 ⇒ y = 1/6 The solution for the given pair of equations is x=1/14 and y=1/6, respectively.16. 1/(2x) + 1/(3y) = 2
1/(3x) + 1/(2y) = 13/6
Solution:
Let 1/x = u and 1/y = v, So, the given equations become u/2 + v/3 = 2 ………………(i) u/3 + v/2 = 13/6 ……………(ii) From (i), we get u/2 + v/3 = 2 ⇒ 3u + 2v = 12 ⇒ u = (12 – 2v)/3 ………….(iii) Using (iii) in (ii), [(12 – 2v)/3]/3 + v/2 = 13/6 ⇒ (12 – 2v)/9 + v/2 = 13/6 ⇒ 24 – 4v + 9v = (13/6) x 18 [after taking LCM] ⇒ 24 + 5v = 39 ⇒ 5v = 15 ⇒ v = 3 Substituting v in (iii), u = (12 – 2(3))/3 ⇒ u = 2 Thus, x = 1/u ⇒ x = 1/2 and y = 1/v ⇒ y = 1/3 The solution for the given pair of equations is x = 1/2 and y = 1/3, respectively.17. 15/u + 2/v = 17
1/u + 1/v = 36/5
Solution:
Let 1/x = u and 1/y = v So, the given equations become 15x + 2y = 17 ………………………….. (i) x + y = 36/5………………………. (ii) From equation (i), we get 2y = 17 – 15x =y = (17 − 15x)/ 2 …………………. (iii) Substituting (iii) in equation (ii), we get = x + (17 − 15x)/ 2 = 36/5 2x + 17 – 15x = (36 x 2)/ 5 [after taking LCM] -13x = 72/5 – 17 = -13x = -13/5 ⇒ x = 1/5 ⇒ u = 1/x = 5 Putting x = 1/5in equation (ii), we get 1/5 + y = 36/5 ⇒ y = 7 ⇒ v = 1/y = 1/7 The solutions of the pair of equations given are u = 5 and v = 1/7, respectively.18. 3/x – 1/y = −9
2/x + 3/y = 5
Solution:
Let 1/x = u and 1/y = v So, the given equations become 3u – v = -9…………………..(i) 2u + 3v = 5 ……………………….(ii) Multiplying equation (i) x 3 and (ii) x 1, we get 9u – 3v = -27 ………………………….. (iii) 2u + 3v = 5 ……………………………… (iv) Adding equations (iii) and (iv), we get 9u + 2u – 3v + 3v = -27 + 5 ⇒ 11u = -22 ⇒ u = -2 Now putting u =-2 in equation (iv), we get 2(-2) + 3v = 5 ⇒ 3v = 9 ⇒ v = 3 Hence, x = 1/u = −1/2 and, y = 1/v = 1/319. 2/x + 5/y = 1
60/x + 40/y = 19
Solution:
Let 1/x = u and 1/y = v So, the given equations become 2u + 5v = 1…………………..(i) 60u + 40v = 19 ……………………….(ii) Multiplying equation (i) x 8 and (ii) x 1, we get 16u + 40v = 8 ………………………….. (iii) 60u + 40v = 19 ……………………………… (iv) Subtracting equation (iii) from (iv), we get 60u – 16u + 40v – 40v = 19 – 8 ⇒ 44u = 11 ⇒ u = 1/4 Now putting u = 1/4 in equation (iv), we get 60(1/4) + 40v = 19 ⇒ 15 + 40v = 19 ⇒ v = 4/ 40 = 1/10 Hence, x = 1/u = 4 and y = 1/v = 1020. 1/(5x) + 1/(6y) = 12
1/(3x) – 3/(7y) = 8
Solution:
Let 1/x = u and 1/y = v So, the given equations become u/5 + v/6 = 12…………………..(i) u/3 – 3v/7 = 8……………………….(ii) Taking LCM for both equations, we get 6u + 5v = 360………. (iii) 7u – 9v = 168……….. (iv) Subtracting (iii) from (iv), 7u – 9v – (6u + 5v) = 168 – 360 ⇒ u – 14v = -192 ⇒ u = (14v – 192)………. (v) Using (v) in equation (iii), we get 6(14v – 192) + 5v = 360 ⇒ 84v -1152 + 5v = 360 ⇒ 89v = 1512 ⇒ v = 1512/89 ⇒ y = 1/v = 89/1512 Now, substituting v in equation (v), we find u. u = 14 x (1512/89) – 192 ⇒ u = 4080/89 ⇒ x = 1/u = 89/ 4080 Hence, the solution for the given system of equations is x = 89/4080 and y = 89/ 1512.21. 4/x + 3y = 14
3/x – 4y = 23
Solution:
Taking 1/x = u, the given equations become 4u + 3y = 14…………………….. (i) 3u – 4y = 23…………………….. (ii) Adding (i) and (ii), we get 4u + 3y + 3u – 4y = 14 + 23 ⇒ 7u – y = 37 ⇒ y = 7u – 37……………………… (iii) Using (iii) in (i), 4u + 3(7u – 37) = 14 ⇒ 4u + 21u – 111 = 14 ⇒ 25u = 125 ⇒ u = 5 ⇒ x = 1/u = 1/5 Putting u= 5 in (iii), we find y y = 7(5) – 37 ⇒ y = -2 Hence, the solution for the given system of equations is x = 1/5 and y = -2.22. 4/x + 5y = 7
3/x + 4y = 5
Solution:
Taking 1/x = u, the given equations become 4u + 5y = 7…………………….. (i) 3u + 4y = 5…………………….. (ii) Subtracting (ii) from (i), we get 4u + 5y – (3u + 4y) = 7 – 5 ⇒ u + y = 2 ⇒ u = 2 – y……………………… (iii) Using (iii) in (i), 4(2 – y) + 5y = 7 ⇒ 8 – 4y + 5y = 7 ⇒ y = -1 Putting y = -1 in (iii), we find u. u = 2 – (-1) ⇒ u = 3 ⇒ x = 1/u = 1/3 Hence, the solution for the given system of equations is x = 1/3 and y = -1.23. 2/x + 3/y = 13
5/x – 4/y = -2
Solution:
Let 1/x = u and 1/y = v So, the given equations become 2u + 3v = 13………………….. (i) 5u – 4v = -2 ………………………. (ii) Adding equations (i) and (ii), we get 2u + 3v + 5u – 4v = 13 – 2 ⇒ 7u – v = 11 ⇒ v = 7u – 11…….. (iii) Using (iii) in (i), we get 2u + 3(7u – 11) = 13 ⇒ 2u + 21u – 33 = 13 ⇒ 23u = 46 ⇒ u = 2 Substituting u = 2 in (iii), we find v. v = 7(2) – 11 ⇒ v = 3 Hence, x = 1/u = 1/2 and, y = 1/v = 1/324. 2/x + 3/y = 2
4/x – 9/y = -1
Solution:
Let 1/√x = u and 1/√y = v, So, the given equations become 2u + 3v = 2………………….. (i) 4u – 9v = -1 ………………………. (ii) Multiplying (ii) by 3 and Adding equations (i) and (ii)x3, we get 6u + 9v + 4u – 9v = 6 – 1 ⇒ 10u = 5 ⇒ u = 1/2 Substituting u = 1/2 in (i), we find v 2(1/2) + 3v = 2 ⇒ 3v = 2 – 1 ⇒ v = 1/3 Since, 1/√x = u we get x = 1/u 2 ⇒ x = 1/(1/2) 2 = 4 And, 1/√y = v y = 1/v 2 ⇒ y = 1/(1/3) 2 = 9 Hence, the solution is x = 4 and y = 9.25. (x + y)/xy = 2
(x – y)/xy = 6
Solution:
The given pair of equations are (x + y)/xy = 2 ⇒ 1/y + 1/x = 2……. (i) (x – y)/xy = 6 ⇒ 1/y – 1/x = 6………(ii) Let 1/x = u and 1/y = v, so the equations (i) and (ii) become v + u = 2……. (iii) v – u = 6……..(iv) Adding (iii) and (iv), we get 2v = 8 ⇒ v = 4 ⇒ y = 1/v = 1/4 Substituting v = 4 in (iii) to find x, 4 + u = 2 ⇒ u = -2 ⇒ x = 1/u = -1/2 Hence, the solution is x = -1/2 and y = 1/4.26. 2/x + 3/y = 9/xy
4/x + 9/y = 21/xy
Solution:
Taking LCM for both the given equations, we have (2y + 3x)/ xy = 9/xy ⇒ 3x + 2y = 9………. (i) (4y + 9x)/ xy = 21/xy ⇒ 9x + 4y = 21………(ii) Performing (ii) – (i)x2⇒ 9x + 4y – 2(3x + 2y) = 21 – (9×2) ⇒ 3x = 3 ⇒ x = 1 Using x = 1 in (i), we find y 3(1) + 2y = 9 ⇒ y = 6/2 ⇒ y = 3 Thus, the solution for the given set of equations is x = 1 and y = 3.Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3
Solving RD Sharma’s Class 10 Maths Chapter 3, Exercise 3.3 on Pair of Linear Equations in Two Variables offers multiple benefits:Foundation in Algebra : Builds a strong base for understanding algebraic methods, essential for higher mathematics.
Problem-Solving Skills : Develops critical thinking by using substitution, elimination, and cross-multiplication techniques.
Real-World Application : Helps students model and solve real-life problems involving relationships between two variables, such as financial calculations, distances, and rates.
Competitive Exam Preparation : Enhances accuracy and speed, useful for exams.
Logical Thinking : Strengthens analytical abilities, essential for advanced studies in science and engineering.
RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.3 FAQs
What is the unique solution in linear equations in two variables?
Every linear equation in one variable has a unique solution. But a pair of linear equations have two solutions i.e. one for x and the other for y, satisfying both the equations. If a given set of linear equation intersects at a point, the solution will be unique for both the equations.
What are the conditions for no solution in linear equations?
Considering the pair of linear equations by two variables u and v. Therefore a1, b1, c1, a2, b2, c2 are real numbers. If (a1/a2) = (b1/b2) ≠ (c1/c2), then this will result in no solution.
How many solutions does a linear in two variables have?
A pair of linear equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 and its solution is a pair of values (x, y) that satisfy both equations. To solve linear equations in two variables, we must have at least two equations. A linear equation in two variables has infinitely many solutions.
When a pair of linear equations have infinitely many solutions?
If the two lines have the same y-intercept and the slope, they are actually in the same exact line. In other words, when the two lines are the same line, then the system should have infinite solutions. It means that if the system of equations has an infinite number of solutions, then the system is said to be consistent.
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