Pair of Linear Equations in Two Variables

  • Question 1
    CBSEENMA10006417

    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.

    Solution

    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.

    Question 2
    CBSEENMA10006422

    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.

    Solution

    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.

    Question 3
    CBSEENMA10006425

    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.

    Solution

    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.

    Question 4
    CBSEENMA10006439

    (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.

    Solution

    (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 :

    Fig. 3.4.

    We have, x - y = -4

    ⇒    x = y - 4

    Thus we have following table :

    When we plot the graph of the given equation, we find that both the lines intersect at me point (3, 7). So.r = 3,y = 7 is the required solution of the pair of linear equation.

    Hence, the number of boys be 3 and the number of girls be 7, who took part in quiz.

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