Open in App 0 The given pair of linear equations are10x + 5y – (k – 5) = 0 ……(i)20x + 10y – k = 0 ……(ii)Which is of the form `a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0`, where`a_1 = 10, b_1 = 5, c_1 = -(k – 5), a_2 = 20, b_2 = 10 and c_2 = -k` For the given pair of linear equations to have infinitely many solutions, we must have `(a_1)/(a_2) = (b_1)/(b_2) =(c_1)/(c_2)``⇒ 10/20 = 5/10 = (−(k−5))/(−k)``⇒ 1/2 = (k−5)/k`⇒ 2k – 10 = k ⇒ k = 10 Hence, k = 10. Page 2The given pair of linear equations are2x + 3y – 9 = 0 ……(i)6x + (k – 2)y – (3k – 2) = 0 ……(ii)Which is of the form `a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0`, where`a_1 = 2, b_1 = 3, c_1 = -9, a_2 = 6, b_2 = k – 2 and c_2 = -(3k – 2)`For the given pair of linear equations to have infinitely many solutions, we must have`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ 2/6 = 3/(k−2) ≠ (−9)/(−(3k−2))``⇒ 2/6 = 3/(k−2) , 3/(k−2) ≠ (−9)/(−(3k−2))``⇒ k = 11, 3/(k−2) ≠ 9/((3k−2))`⇒ k = 11, 3(3k – 2) ≠ 9(k – 2)⇒ k = 11, 1 ≠ 3 (true) Hence, k = 11.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and graphically.
Let the cost of 1 bat be Rs. x and cost of I ball be Rs.y Case I. Cost of 3 bats = 3x Cost of 6 balls = 6y According to question, 3x + 6y = 3900 Case II. Cost of I bat = x Cost of 3 more balls = 3y According to question, x + 3y = 1300 So, algebraically representation be 3x + 6y = 3900 x + 3y = 1300 Graphical representation : We have, 3x + 6y = 3900 ⇒ 3(x + 2y) = 3900 ⇒ x + 2y = 1300 ⇒ a = 1300 - 2y Thus, we have following table :  We have, x + 3y = 1300 ⇒ x = 1300 - 3y Thus, we have following table : When we plot the graph of equations, we find that both the lines intersect at the point (1300. 0). Therefore, a = 1300, y = 0 is the solution of the given system of equations. |
For what value of k the following pair of linear equations has infinitely many solutions 10x 5y
Pos Terkait
Copyright © 2024 ketajaman Inc.
