Find the value of k for which the system of equations has no solution 3x-4y+7=0 kx+3y-5=0
The value of k for which the system of equation has no solution :
336 Views (i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. (i) Let the number of boys be x and number of girls be y. Case I. x + y = 10 ...(i) Case II. y = x + 4 ⇒ x - y = -4 ...(ii) We have, x + y = 10 ⇒ x = 10 - y Thus, we have following table : 1113 Views The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically. Let the cost of 1 kg of apples be Rs. x and of 1 kg of grapes be Rs. y. So, algebraic representation <>2x + y = 160 4x + 2y = 300 ⇒ 2x + y = 150 Graphical representation, we have 2x + y = 160 ⇒ y = 160 - 2x We have, 2x + y = 150 ⇒ y = 150 - 2x When we plot the graph of the equation we find that two lines do not intersect i.e. they are parallel. Fig. 3.3. 1699 Views 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen. Let cost of 1 pencil be Rs. ,v and cost of 1 pen be Rs. y Case I. Cost of 5 pencils = 5x Cost of 7 pens = 7y <>According to question, 5x + 7y = 50 Case II. Cost of 7 pencils = 7x Cost of 5 pens = 5y According to question, 7x + 5y = 46 Thus, we have Thus, we have following table : Fig. 3.5. When we plot the graph of the given equation, we find that both the lines intersect at the point (3, 5). So, x = 3,y = 5 is the required solution of the pair of linear equation. Hence, the cost of 1 pencil be Rs. 3 cost of 1 pen be Rs. 5. 995 Views Aftab tells his daughter, 'Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be'. (Isn’t this interesting?) Represent this situation algebraically and graphically. Let present age of Aftab be x years and present age of his daughter be y years. Case I. Seven years ago, Age of Aftab = (x - 7) years Age of his daughter = (y - 7) years According to question : (x - 7) = 7 (y - 7) ⇒ x - 7 = 7y - 49 ⇒ x - 7y = -42 Case II. Three years later, Age of Aftab = (x + 3) years Age of his daughter = (y + 3) years Accoring to questions, x + 3 = 3 (y + 3) ⇒ x + 3 = 3y + 9 ⇒ x — 3y = 6 So, algebraic expression be x - 7y = -42 ...(i) x - 3y = 6 ...(ii) Graphical representation For eq. (i), we have x - 7y = -42 ⇒ x — 7y — 42 Thus, we have following table : From eqn. (ii), we have x -3y = 6 ⇒ x = 3y + 6 Thus, we have following table When we plot the graph of equations. We find that both the lines intersect at the point (42, 12). Therefore, x = 42, y = 12 is the solution of the given system of equations. Fig. 3.1. 1935 Views 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. 2043 Views |