TOPICS
Exercise - 3.2

Question-1 :-  Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 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.
(ii) 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 :-
(i) Let the no. of girls be x and no. of boys be y.
  The algebraic representation is given by the following equations:

  x + y = 10 ......(i)
  x – y = 4  ......(ii)

  Lets represent these equations in graphically: 
  x + y = 10  .......(i)
x010
y100
x – y = 4 ......(ii)
x04
y-40
Plot the points A(0, 10), B(10, 0) and P(0, -4), Q(4, 0), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations x + y = 10 and x – y = 4, as shown in following Figure: Now, Observe that the two lines representing the two equations are intersecting at the point (7, 3). Therefore, the no. of girls and no. of boys in the class are 7 and 3 respectively. So, it is consistent.
(ii) Let the cost of pencil be x and cost of pen be y.
  The algebraic representation is given by the following equations:

  5x + 7y = 50 ......(i)
  7x + 4y = 46 ......(ii)

  Lets represent these equations in graphically: 
  5x + 7y = 50 ......(i)
x310
y50
7x + 4y = 46 ......(ii)
x38
y5-2
Plot the points A(3, 5), B(10, 0) and P(3, 5), Q(8, -2), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations 5x + 7y = 50 and 7x + 4y = 46, as shown in following Figure: Now, Observe that the two lines representing the two equations are intersecting at the point (3, 5). Therefore, the cost of one pencil and cost of one pen are 3 and 5 respectively. So, it is consistent.

Question-2 :-  On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂ find out whether lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

Solution :-
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0 

  a₁ = 5, b₁ = -4, c₁ = 8
  a₂ = 7, b₂ = 6, c₂ = -9

  Here, a₁/a₂ = 5/7,
  Also, b₁/b₂ = -4/6 and
  c₁/c₂ = 8/(-9)
  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersecting each other at one point and its have only one solution.  
 
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0 

  a₁ = 9, b₁ = 3, c₁ = 12
  a₂ = 18, b₂ = 6, c₂ = 24

  Here, a₁/a₂ = 9/18 = 1/2, 
  Also, b₁/b₂ = 3/6 = 1/2 and
  c₁/c₂ = 12/24 = 1/2

  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
  Now, these linear equations are coincident and its have many solution.   
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

  a₁ = 6, b₁ = -3, c₁ = 10
  a₂ = 2, b₂ = -1, c₂ = 9

  Here, a₁/a₂ = 6/2 = 3, 
  Also, b₁/b₂ = -3/(-1) = 3 and
  c₁/c₂ = 10/9

  Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are parallel each other and its have no solution.

Question-3 :-  On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂ find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3x/2 + 5y/3 = 7; 9x - 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4x/3 + 2y = 8; 2x + 3y = 12

Solution :-
(i) 3x + 2y = 5 ; 2x – 3y = 7 
    3x + 2y - 5 = 0 ; 2x – 3y - 7 = 0 

  a₁ = 3, b₁ = 2, c₁ = -5
  a₂ = 2, b₂ = -3, c₂ = -7

  Here, a₁/a₂ = 3/2,
  Also, b₁/b₂ = 2/(-3) and
  c₁/c₂ = -5/(-7)= 5/7
  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersecting each other at one point and its have only one solution.
  So, It is consitent.  
 
(ii) 2x – 3y = 8 ; 4x – 6y = 9 
    2x – 3y - 8 = 0 ; 4x – 6y - 9 = 0

  a₁ = 2, b₁ = -3, c₁ = -8
  a₂ = 4, b₂ = -6, c₂ = -9

  Here, a₁/a₂ = 2/4 = 1/2, 
  Also, b₁/b₂ = -3/(-6) = 1/2 and
  c₁/c₂ = -8/(-9) = 8/9

  Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are parallel to each other and its have only no possible solution.
  So, It is inconsitent.  
(iii) 3x/2 + 5y/3 = 7; 9x - 10y = 14 
    3x/2 + 5y/3 - 7 = 0; 9x - 10y - 14 = 0

  a₁ = 3/2, b₁ = 5/3, c₁ = -7
  a₂ = 9, b₂ = -10, c₂ = -14

  Here, a₁/a₂ = (3/2)/9 = 1/6, 
  Also, b₁/b₂ = (5/3)/(-10) = 1/(-6) and
  c₁/c₂ = -7/(-14) = 1/2

  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersecting each other at one point and its have only one solution.
  So, It is consitent. 
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
    5x – 3y - 11 = 0 ; – 10x + 6y + 22 = 0

  a₁ = 5, b₁ = -3, c₁ = -11
  a₂ = -10, b₂ = 6, c₂ = 22

  Here, a₁/a₂ = 5/(-10) = 1/(-2), 
  Also, b₁/b₂ = (-3)/6 = 1/(-2) and
  c₁/c₂ = -11/22 = 1/(-2)

  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
  Now, these linear equations are coincident and its have many solution.
  So, It is consitent. 
(v) 4x/3 + 2y = 8; 2x + 3y = 12
    4x/3 + 2y - 8 = 0; 2x + 3y - 12 = 0

  a₁ = 4/3, b₁ = 2, c₁ = -8
  a₂ = 2, b₂ = 3, c₂ = -12

  Here, a₁/a₂ = (4/3)/2 = 2/3, 
  Also, b₁/b₂ = 2/3 and
  c₁/c₂ = -8/(-12) = 2/3

  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
  Now, these linear equations are coincident and its have many solution.
  So, It is consitent. 
    

Question-4 :-  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 :-
(i) x + y = 5, 2x + 2y = 10 
    x + y - 5 = 0 ; 2x + 2y - 10 = 0 

  a₁ = 1, b₁ = 1, c₁ = -5
  a₂ = 2, b₂ = 2, c₂ = -10

  Here, a₁/a₂ = 1/2,
  Also, b₁/b₂ = 1/2 and
  c₁/c₂ = -5/(-10)= 1/2
  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
  Now, these linear equations are coincident and its have many solution.
  So, It is consitent.   

  Lets represent these equations in graphically: 
  x + y = 5  ...........(i) 
x05
y50
2x + 2y = 10 ...........(ii)
x05
y50
Plot the points A(0, 5), B(5, 0) and P(0, 5), Q(5, 0), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations x + y = 5 and 2x + 2y = 10, as shown in following Figure:
 
(ii) x – y = 8, 3x – 3y = 16
    x – y - 8 = 0 ; 3x – 3y - 16 = 0

  a₁ = 1, b₁ = -1, c₁ = -8
  a₂ = 3, b₂ = -3, c₂ = -16

  Here, a₁/a₂ = 1/3, 
  Also, b₁/b₂ = -1/(-3) = 1/3 and
  c₁/c₂ = -8/(-16) = 1/2

  Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are parallel to each other and its have only no possible solution.
  So, It is inconsitent.  
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 

  a₁ = 2, b₁ = 1, c₁ = -6
  a₂ = 4, b₂ = -2, c₂ = -4

  Here, a₁/a₂ = 2/4 = 1/2, 
  Also, b₁/b₂ = 1/(-2) and
  c₁/c₂ = -6/(-4) = 3/2

  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersecting each other at one point and its have only one solution.
  So, It is consitent. 

  Lets represent these equations in graphically: 
  2x + y = 6  ...........(i) 
x03
y60
4x – 2y = 4 ...........(ii)
x01
y-20
Plot the points A(0, 6), B(3, 0) and P(0, -2), Q(1, 0), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations 2x + y = 6 and 4x – 2y = 4, as shown in following Figure:
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

  a₁ = 2, b₁ = -2, c₁ = -2
  a₂ = 4, b₂ = -4, c₂ = -5

  Here, a₁/a₂ = 2/4 = 1/2, 
  Also, b₁/b₂ = (-2)/(-4) = 1/2 and
  c₁/c₂ = -2/(-5) = 2/5

  Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are parallel to each other and its have only no possible solution.
  So, It is inconsitent. 
    

Question-5 :-  Half the 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 the width of garden and length of garden be araxand y respectively.
  According to question linear equations are:

  y = x + 4
  x - y = -4 ........(i)

  y + x = 36
  x + y = 36 ........(ii)  

  Lets represent these equations in graphically: 
  x - y = -4  ...........(i) 
x0-4
y40
x + y = 36 ...........(ii)
x036
y360
Plot the points A(0, 4), B(-4, 0) and P(0, 36), Q(36, 0), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations x - y = -4 and x + y = 36, as shown in following Figure:

Question-6 :-  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

Solution :-
(i) For intersecting lines, 
  a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂

  Given : 2x + 3y - 8 = 0 ....(i)
  According to condition the second line is :
  3x + 5y - 15 = 0 ....(ii)

  a₁ = 2, b₁ = 3, c₁ = -8
  a₂ = 3, b₂ = 5, c₂ = -15

  Here, a₁/a₂ = 2/3, 
  Also, b₁/b₂ = 3/5 and
  c₁/c₂ = -8/(-15) = 8/15
 
  Hence, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂.
(ii) For parallel lines, 
  a₁/a₂ = b₁/b₂ ≠ c₁/c₂

  Given : 2x + 3y - 8 = 0 ....(i)
  According to condition the second line is :
  4x + 6y - 18 = 0 ....(ii)

  a₁ = 2, b₁ = 3, c₁ = -8
  a₂ = 4, b₂ = 6, c₂ = -18

  Here, a₁/a₂ = 2/4, 
  Also, b₁/b₂ = 3/6 and
  c₁/c₂ = -8/(-18) = 4/9
 
  Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂. 
(iii) For coincident lines, 
  a₁/a₂ = b₁/b₂ = c₁/c₂

  Given : 2x + 3y - 8 = 0 ....(i)
  According to condition the second line is :
  6x + 9y - 24 = 0 ....(ii)

  a₁ = 2, b₁ = 3, c₁ = -8
  a₂ = 6, b₂ = 9, c₂ = -24

  Here, a₁/a₂ = 2/6 = 1/3, 
  Also, b₁/b₂ = 3/9 = 1/3 and
  c₁/c₂ = -8/(-24) = 1/3
 
  Hence, a₁/a₂ = b₁/b₂ = c₁/c₂.
    

Question-7 :-  Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution :-
  x – y + 1 = 0
  x - y = -1 ........(i)

  3x + 2y – 12 = 0
  3x + 2y = 12 ........(ii)  

  Lets represent these equations in graphically: 
  x - y = -1 ........(i)
x0-1
y10
3x + 2y = 12 ........(ii)
x04
y60
Plot the points A(0, 1), B(-1, 0) and P(0, 6), Q(4, 0), corresponding to the solutions in Table. Now draw the lines AB and PQ, representing the equations x - y = -1 and 3x + 2y = 12, as shown in following Figure:
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