Wednesday, June 8, 2011

WORKSHEET 1 Pair of Linear Equations in Two variables

1.  Aftab tells his daughter, "Seven years ago, I was se    ven 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 the present age of Aftab be x years and her daughter age be y years
    Seven years ago

    Aftab’s age = (x – 7)
    Her daughter’s age =(y – 7)
    (x – 7) =7(y – 7) ……………………….. (1)
    Three years from now
    Aftab’s age = x + 3
    Her daughter age = y + 3
    x + 3 = 3(y + 3) ……………………….. (2)
    (1) becomes,
    x – 7 = 7(y – 7)
    x – 7 = 7y – 49
    x – 7y = -42 ……………………….. (3)
    (2) becomes
    x + 3 = 3(y + 3)
    = 3y + 9
    x - 3y = 6 ………………………(4)
    Algebraically the two situations can be represented as follows:
    x – 7y + 42 = 0 , x - 3y - 6 = 0 where x and y are respectively the ages of Aftab and his daughter.
    Graphic Representation :
    x – 7y = -42 (1)
    x – 3y = 6 (2)
    => x – 7y = -42
    -7y = -42 – x
    y =
    when x = 0,
    y =  = = 6
    when x = 7,
    y =  =  = 7
    when x = 14,
    y =  =  = 8
    when x = -7,
    y =  = 5
    when x = -14,
    y =  =  = 4
    x
    -14
    -7
    0
    7
    14
    y
    4
    5
    6
    7
    6
    => x – 3y = 6
    -3y = 6 – x
    y = 
    when x = 0,
    y =  = = -2
    when x = 3,
    y =  =  = -1
    when x = 6,
    y =  = = 0
    when x = -3,
    y =  =  = -3
    when x = -6,
    y =  = = -4
    x
    -6
    -3
    0
    3
    6
    y
    -4
    -3
    -2
    -1
    0

 2.  The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another but and 2 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
  • Solution:Let the cost of each bat be Rs.x
    Let the cost of each ball be Rs.y
    Algebraically
    3x + 6y = 3900 …………………(1)
    (1) Þ x + 2y = 1300
    x + 2y = 1300 …………..(2)
    X
    1000
    700
    y = 

    150

    300

 3.  The cost of 2kg of apples and 1kg 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:Cost per kg of apple = Rs. x
    Cost per kg of grapes = Rs. y
    Algebraically 2x + y = 160 ……………….(1)
    4x + 2y = 300 ……………….(2)
    from (1) y = 160 – 2x
    Graphically
    x
    50
    60
    Y = 
    60
    40

 4.  Form the pair of linear equations in the following problem, and find their solutions graphically.10 students of Class X took part in a 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:Let the number of boy be x and the number of girls be y
    x + y = 10 ………………………. (1) (given)
    y = x + 4 ……………………. (2) (given)
    x + y = 10 …..(1)
    => y = 10 – x
    When x = -1
    y = 10 – (-1)
    = 11
    When x = 0
    y = 10 – 0
    = 10
    When x = 1
    y = 10 – 1
    = 9
    when x = 2
    y = 10 – 2
    = 8
    when x = 3
    y = 10 – 3
    = 7
    x
    -1
    0
    1
    2
    3
    y = 10 - x
    11
    10
    9
    8
    7
    y = x + 4 …………………….(2)
    Let x = -1
    y = x + 4
    = -1 + 4
    = 3
    Let x = 0
    y = x + 4
    = 0 +4 = 4
    Let x = 1
    y = x + 4
    = 1 + 4
    = 5
    Let x = 2
    y = x + 4
    = 2 + 4
    = 6

    Let x = 3
    y = x + 4
    = 3 + 4
    = 7
    x
    -1
    0
    1
    2
    3
    y = x + 4
    3
    4
    5
    6
    7
    The solution thus obtained graphically is (3, 7). Number of girls = 7 and number of boys = 3.

 5.  Form the pair of linear equations in the following problem, and find their solutions graphically. 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.
  • Solution:Let the cost of one pencil be Rs. x
    Let the cost of one pen be Rs. y
    5x + 7y = 50
    y =  …………… (1)
    When x = 3
    y = 
     = 
    = 5
    When x = 10
    y = 
     = 
    = 0
    When x = -4
    y = 
     =  =10
    x
    -4
    3
    10
    Y
    10
    5
    0
    7x + 5y = 46
    y = 
    When x = 0
    y = 
     = 
    = 9.2
    When x = 8
    y = 
     = 
     = -2
    x
    0
    3
    8
    Y
    9.2
    5
    -2
    By solving graphically , cost of one pencil = Rs. 3, cost of one pen = Rs. 5.

 6.  On comparing the ratios and , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
  • Solution:(i) 5x – 4y + 8 = 0
    7x +6y – 9 = 0
    \ 
    \Lines are intersecting.
    (ii) 9x + 3y + 12 = 0
    18x + 6y + 24 = 0

    Since , the lines are coincident.
    (iii) 6x – 3y + 10 = 0
    2x – y + 9 = 0
    But 
    Since , the lines are parallel.

 7.  On comparing the ratios and , find out whether the lines representing the following pairs of linear equations are consistent , or incoincident.
(i) 3x + 2y = 5
2x – 3y = 17
(ii) 2x – 3y = 8
4x – 6y = 9

(iii) 
9x – 10y = 14
(iv) 5x – 3y = 11
-10x + 6y = -22
(v) 
  • Solution:(i) 3x + 2y = 5
    2x – 3y = 17
    Since , equations are consistent.

    (ii) 2x – 3y = 8
    4x – 6y = 9
    Here 
    \ The equations are inconsistent.
    (iii) 
    9x – 10y = 14
    \ The equations are consistent.
    (iv) 5x – 3y = 11
    -10x + 6y = -22
     = 
     = 
     = 
    Since , the equations are consistent. 
    (v) 
    2x + 3y = 12
    a1 = , a2 = 2, c1 =- 8
    a1 = 2, b2 = 3, c2 = -12
     =  = 
    =
    Since 
    The equations are consistent.

 8.  Which of the following pairs of linear equations are consistent/inconsistent ?If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0
(iv)2x – 2y – 2 = 0, 4x – 4y – 5 =0
  • Solution:x + y = 5, 2x + 2y = 10
    Equations are consistent .
    We have to draw the graphs of both the given equations
    x + y = 5
    x
    0
    1
    y = 5 - x
    5
    4
    2x + 2y = 10
    x
    0
    1
    y = 
    5
    4
    Hence the lines represented by both the equations are coincident. Thus both the equations have infinitely many solutions.
    (ii) x – y = 8, 3x – 3y = 16
     = 
     = 
    \The equation are inconsistent.
    2x + y – 6 = 0
    4x – 2y – 4 = 0
     =
    Thus as , the equations are consistent.
    2x + y – 6 = 0
    when x = 1 , y = 6 – 2x
    = 4
    x = 0 , y = 6
    x = -1 , y = 8
    x
    2
    0
    -1
    y = 6 – 2x
    2
    6
    8

    4x – 2y – 4 = 0
    4x – 2y = 4
    y = 
    = 2x – 2
    4x – 2y – 4 = 0
    x
    2
    0
    -1
    y = 2x – 2
    2
    -2
    -4

    When x = 1, y = 2 – 2 = 0
    x = 0, y = -2

    x = -1, y = -2 -2 = -4
    4x – 4y – 5 = 0
    4x – 4y = 5
    (iv) 2x – 2y – 2 = 0
    4x – 4y - 5= 0
    Equations are inconsistent

 9.  Half perimeter of a rectangular garden , whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
  • Solution:Let l be the length and b be the breadth
    l = b+ 4
    Þ l - b = 4 …………..(1)
    Perimeter = 2(l + b)
     ´ 2(l + b) = 36(given) ……………….(2)
    l – b = 4 ………… (1)
    l + b = 36 ………..(2)
    – (2) => 2l = 40
    l = = 20 m
    Substitute l = 20 in (1)
    l - b = 4
    20 – b = 4
    b = 20 – 4
    = 16.

 10.  Given the linear equation 2x +3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines (ii) parallel lines (iii) coincident lines