The pair of linear equations 2x + 3y 7 and 4x ky 5 has a unique solution if
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Here `a_1 = 4, a_2 = k, b_1 = 2, b_2 = 2`` Show
Now for the given pair to have a unique solution `a_1/a_2 != b_1/b_2` i.e `4/2 != k/2` i.e `k != 4` Therefore, for all values of k, except 4, the given pair of equations will have a unique solution.  Book: RD Sharma - MathematicsChapter: 3. Pair of Linear Equations in Two VariablesSubject: Maths - Class 10thQ. No. 14 of CCE - Formative AssessmentListen NCERT Audio Books to boost your productivity and retention power by 2X. A given system of equations may or may not have a solution. Sometimes it can also have infinitely many solutions. All these conditions for solvability are studied in this exercise. The RD Sharma Solutions Class 10 prepared by experts at BYJU’S can help students get a strong conceptual knowledge on the subject. Also, the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5 PDF given below is available for students for further clarifications. Chapter 3 Pair Of Linear…
Exercise 3.5
RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.5 Download PDF
Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations In Two Variables Exercise 3.5In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution, find it from 1 to 4: 1. x − 3y = 3 3x − 9y = 2 Solution: The given system of equations is: x − 3y – 3 = 0 3x − 9y − 2 = 0 The above equations are of the form a1 x + b1 y − c1 = 0 a2 x + b2 y − c2 = 0 Here, a1 = 1, b1 = −3, c1 = −3 a2 = 3, b2 = −9, c2 = −2 So according to the question, we get a1 / a2 = 1/3 b1 / b2 = −3/ −9 = 1/3 and, c1 / c2 = −3/ −2 = 3/2 ⇒ a1 / a2 = b1/ b2 ≠ c1 / c2 Hence, we can conclude that the given system of equation has no solution. 2. 2x + y = 5 4x + 2y = 10 Solution:
The given system of equations is: 2x + y – 5 = 0 4x + 2y – 10 = 0 The above equations are of the form a1 x + b1 y − c1 = 0 a2 x + b2 y − c2 = 0 Here, a1 = 2, b1 = 1, c1 = −5 a2 = 4, b2 = 2, c2 = −10 So according to the question, we get a1 / a2 = 2/4 = 1/2 b1 / b2 = 1/ 2 and, c1 / c2 = −5/ −10 = 1/2 ⇒ a1 / a2 = b1/ b2 = c1 / c2 Hence, we can conclude that the given system of equation has infinity many solutions. 3. 3x – 5y = 20 6x – 10y = 40 Solution: The given system of equations is: 3x – 5y – 20 = 0 6x – 10y – 40 = 0 The above equations are of the form a1 x + b1 y − c1 = 0 a2 x + b2 y − c2 = 0 Here, a1 = 3, b1 = -5, c1 = −20 a2 = 6, b2 = -10, c2 = −40 So according to the question, we get a1 / a2 = 3/6 = 1/2 b1 / b2 = -5/ -10 = 1/2 and, c1 / c2 = -20/ −40 = 1/2 ⇒ a1 / a2 = b1/ b2 = c1 / c2 Hence, we can conclude that the given system of equation has infinity many solutions. 4. x – 2y = 8 5x – 10y = 10 Solution: The given system of equations is: x – 2y – 8 = 0 5x – 10y – 10 = 0 The above equations are of the form a1 x + b1 y − c1 = 0 a2 x + b2 y − c2 = 0 Here, a1 = 1, b1 = -2, c1 = −8 a2 = 5, b2 = -10, c2 = -10 So according to the question, we get a1 / a2 = 1/5 b1 / b2 = -2/ -10 = 1/5 and, c1 / c2 = -8/ −10 = 4/5 ⇒ a1 / a2 = b1/ b2 ≠ c1 / c2 Hence, we can conclude that the given system of equation has no solution. Find the value of k for which each of the following system of equations having infinitely many solution: (9-19) |
