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Pair of Linear Equations in Two Variables : Exercise 3.5


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Q.1      Which of the following pairs of linear equations has unique solution, no solution, or finitely many solutions ? In case there is a unique solution, find it by using cross multiplication method.
           (i) x – 3y – 3 = 0            3x – 9y – 2 = 0
           (ii) 2x + y = 5                 3x + 2y = 8
           (iii) 3x – 5y = 20             6x – 10y = 40
           (iv) x – 3y – 7 = 0           3x – 3y – 15 = 0
Sol.       (i) The given system of equations is
              x – 3y – 3 = 0
              and, 3x – 9y – 2 = 0
              These are of the form {a_1}x + {b_1}y + {c_1} = 0
              and {a_2}x + {b_2}y + {c_2} = 0
              where {a_1} = 1,{b_1} = - 3,{c_1} = - 3 and {a_2} = 3,{b_2} = - 9,{c_2} = - 2
              We have, {{{a_1}} \over {{a_2}}} = {1 \over 3},{{{b_1}} \over {{b_2}}} = {{ - 3} \over { - 9}} = {1 \over 3}and{{{c_1}} \over {{c_2}}} = {{ - 3} \over { - 2}} = {3 \over 2}
              Clearly, {{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} \ne {{{c_1}} \over {{c_2}}}
             So, the given system of equations has no solution i.e., it is inconsistent.

            (ii) The given system of equations may be written as
            2x + y – 5 = 0 and 3x + 2y – 8 = 0
            These are of the form {a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0
            where {a_1} = 2,{b_1} = 1,{c_1} = - 5\,and\,{a_2} = 3,{b_2} = 2\,\,and\,\,{c_2} = - 8
           We have, {{{a_1}} \over {{a_2}}} = {2 \over 3},{{{b_1}} \over {{b_2}}} = {1 \over 2}
           Clearly, {{{a_1}} \over {{a_2}}} \ne {{{b_1}} \over {{b_2}}}
           So, the given system of equations has a unique solution.
           To find the solution, we use the cross - multiplication method. By cross - multiplication, we have

40
          {x \over {1 \times - 8 - 2 \times \left( { - 5} \right)}} = {y \over { - 5 \times 3 - \left( { - 8} \right) \times 2}} = {1 \over {2 \times 2 - 3 \times 1}}
          \Rightarrow {x \over { - 8 + 10}} = {y \over { - 15 + 16}} = {1 \over {4 - 3}}
          \Rightarrow {x \over 2} = {y \over 1} = {1 \over 1}
          \Rightarrow x = 2, y = 1 

         Hence, the given system of equations has a unique solution given by x = 2, y = 1.

         (iii) The system of equations may be written as
         3x – 5y – 20 = 0 and 6x – 10y – 40 = 0
         The given equations are of the form
         {a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0
         where, {a_1} = 3,{b_1} = - 5,{c_1} = - 20\,\,and\,{a_2} = 6,{b_2} = - 10,{c_2} = - 40
         We have, {{{a_1}} \over {{a_2}}} = {3 \over 6} = {1 \over 2},{{{b_1}} \over {{b_2}}} = {{ - 5} \over { - 10}} = {1 \over 2}and{{{c_1}} \over {{c_2}}} = {{ - 20} \over { - 40}} = {1 \over 2}
         Clearly, {{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} = {{{c_1}} \over {{c_2}}}
         So, the given system of equations has infinitely many solutions.

         (iv) The given system of equations is
          x – 3y – 7 = 0 and 3x – 3y – 15 = 0
         The given equations are of the form
         {a_1}x + {b_1}y + {c_1} = 0\,and\,{a_2}x + {b_2}y + {c_2} = 0
         where {a_1} = 1,{b_1} = - 3,{c_1} = - 7\,\,and\,\,{a_2} = 3,{b_2} = - 3,{c_2} = - 15
         we have, {{{a_1}} \over {{a_2}}} = {1 \over 3},{{{b_1}} \over {{b_2}}} = {{ - 3} \over { - 3}} = 1               Clearly, {{{a_1}} \over {{a_2}}} \ne {{{b_1}} \over {{b_2}}}
         So, the given system of equations has a unique solution.To find the solution, we use cross-multiplication method.            By cross-multiplication, we have 

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           {x \over {45 - 21}} = {y \over { - 21 + 15}} = {1 \over { - 3 + 9}}
           {x \over {24}} = {y \over { - 6}} = {1 \over 6}
            \Rightarrow x = {{24} \over 6} = 4,y = {{ - 6} \over 6} = - 1
           Hence, the given system of equations has a unique solution given by
            x = 4, y = – 1


Q.2      (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?  2x + 3y = 7 (a – b)x + (a + b)y = 3a + b – 2
            (ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1 (2k – 1)x + (k – 1)y = 2k + 1
Sol.       (i) We know that the system of equations
              {a_1}x + {b_1}y = {c_1}\,and\,{a_2}x + {b_2}y = {c_2}
              has infinite number of solutions if
              {{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} = {{{c_1}} \over {{c_2}}}
              Therefore, The given system of equations will have infinite number of solutions if
              {2 \over {a - b}} = {3 \over {a + b}} = {7 \over {3a + b - 2}}
               \Rightarrow {2 \over {a - b}} = {3 \over {a + b}}and{2 \over {a - b}} = {7 \over {3a + b - 2}}
               \Rightarrow 2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b
               \Rightarrow a = 5b and a – 9b = – 4
              Putting a = 5b in a – 9b = – 4, we get
              5b – 9b = – 4
               \Rightarrow – 4b = – 4
               \Rightarrow b = 1
              Putting b = 1 in a – 9b = – 4, we get 
              a = 5(1) = 5 
              Hence, the given system of equations will have infinitely many solutions if 
              a = 5 and b = 1

              (ii) We know that the system of equations
              {a_1}x + {b_1}y = {c_1}\,and\,{a_2}x + {b_2}y = {c_2}
               has no solution if {{{a_1}} \over {{a_2}}} = {{{b_1}} \over {{b_2}}} \ne {{{c_1}} \over {{c_2}}}
               So, the given system of equations will have no solution if
               {3 \over {2k - 1}} = {1 \over {k - 1}} \ne {1 \over {2k + 1}}
                \Rightarrow {3 \over {2k - 1}} = {1 \over {k - 1}}and{1 \over {k - 1}} \ne {1 \over {2k + 1}}
                Now, {3 \over {2k - 1}} = {1 \over {k - 1}}
                 \Rightarrow 3k – 3 = 2k – 1
                 \Rightarrow k = 2
                 Clearly, for k = 2, we have
                 {1 \over {k - 1}} \ne {1 \over {2k + 1}}
                 Hence, the given system of equations will have no solution if k = 2.


Q.3      Solve the following pair of linear equations by the substitution and cross - multiplication methods :
             8x + 5y = 9
             3x + 2y = 4
Sol.         The given system of equations is
                8x + 5y = 9...(1)
                and, 3x + 2y = 4...(2)
                By graphical method :
                For the graph of 8x + 5y = 9
                We have, 8x + 5y = 9
                 \Rightarrow 5y = 9 – 8x
                 \Rightarrow y = {{9 - 8x} \over 5}
                 \Rightarrow When x = 3, y = {{9 - 24} \over 5} = {{ - 15} \over 5} = - 3 ; when x = – 2, y = {{9 + 16} \over 5} = {{25} \over 5} = 5
                Thus, we have the following table : 

42
               Plot the points A(3, – 3) and B(–2, 5) on a graph paper. Join A and B and extend it on both sides to obtain the graph of 8x + 5y = 9 as shown.
43
              For the graph of 3x + 2y = 4
              We have,
              3x + 2y = 4
               \Rightarrow 2y = 4 – 3x
               \Rightarrow y = {{4 - 3x} \over 2}
              When x = 0, y = {{4 - 0} \over 2} = {4 \over 2} = 2;
              When x = 2, y = {{4 - 6} \over 2} = {{ - 2} \over 2} = - 1
             Thus, we have the following table :

44
              Plot the points C(0, 2) and D(2, – 1) on the same graph paper. Join C and D and extend it on both sides to obtain the graph of 3x + 2y = 4.
             Clearly, the two lines intersect at B(–2, 5).
             Hence, x = – 2, y = 5 is the solution of the given system of equations.
             By substitution method :
             Substituting y = {{9 - 8x} \over 5} in (2), we get
             3x + 2\left( {{{9 - 8x} \over 5}} \right) = 4
              \Rightarrow 15x + 18 – 16x = 20
              \Rightarrow – x = 2
              \Rightarrow x = – 2
             Putting x = – 2 in (1), we get
             8(–2) +5y = 9
              \Rightarrow 5y = 9 + 16 = 25
              \Rightarrow y = {{25} \over 5} = 5
             Hence, x = – 2, y = 5 is the solution of the given system of equations.
             By cross - multiplication method :
45
              {x \over { - 20 + 18}} = {y \over { - 27 + 32}} = {1 \over {16 - 15}}
               \Rightarrow {x \over { - 2}} = {y \over 5} = {1 \over 1}
               \Rightarrow x = {{ - 2} \over 1} = - 2\,\,and\,\,y = {5 \over 1} = 5
              Hence, x = – 2, y = 5 is the solution of the given system of equations.
              The method of cross - multiplication is more efficient.


Q.4      Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel  charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes {1 \over 3} when 1 is subtracted from the numerator and it becomes {1 \over 4} when 8 is added to its denominator. find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test ?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Sol.       (i) Let the fixed charges be Rs. x and charge per day be Rs. y.
             Therefore, By the given conditions,
              x + 20y = 1000 ...(1)
              and, x + 26y = 1180 ...(2)
              Subtracting (1) from (2), we get
              6y = 180
               \Rightarrow y = 30
              From (1), x + 20 × 30 = 1000
               \Rightarrow x = 1000 – 600 = 400
              Therefore, Fixed charge = Rs. 400
              and cost of food per day = Rs. 30

           (ii) Let the fraction be {x \over y}.
             According to the given conditions :
             {{x - 1} \over y} = {1 \over 3}
              \Rightarrow 3x – 3 = y
              \Rightarrow 3x – y – 3 = 0...(1)
             and, {x \over {y + 8}} = {1 \over 4}
              \Rightarrow 4x = y + 8
              \Rightarrow 4x – y – 8 = 0
             Solving (1) and (2) by cross - multiplication method.

46
              {x \over {8 - 3}} = {y \over { - 12 + 24}} = {1 \over { - 3 + 4}}
               \Rightarrow {x \over 5} = {y \over {12}} = {1 \over 1}
               \Rightarrow x = 5, y = 12
              Hence, the required fraction is {5 \over {12}}.

             (iii) Let Yash answer x questions correctly and y questions incorrectly. According to the given conditions :
              3x – y = 40 ...(1)
              and, 4x – 2y = 50 ...(2)
              Multiplying (1) by 2 and subtracting (2) from the result so obtained, we get
              2x = 30
               \Rightarrow x = 15
              From (1), 3 × 15 – y = 40
               \Rightarrow – y = 40 – 45
               \Rightarrow – y = – 5
               \Rightarrow y = 5
              Therefore, Total number of questions in the test are 15 + 5 = 20.

              (iv) Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of Y be y km/hr.
              Case I : When two cars move in the same direction.

47
            Let these cars meet at point Q. Then,
            Distance travelled by car X = AQ
            Distance travelled by car Y = BQ
            It is given that two cars meet in 5 hours.
            Therefore, Distance travelled by car A in 5 hours = 5x km
             \Rightarrow AQ = 5x
            Distance travelled by' car Y in 5 hours = 5y km
             \Rightarrow BQ = 5y
            Clearly, AQ – BQ = AB
             \Rightarrow 5x – 5y = 100 [Since, AB = 100 km]
             \Rightarrow x – y = 20
            Case II : When two cars move in opposite directions.
            Let these cars meet at point P. Then,
            Distance traveled be car X = AP

            Distance traveled be car Y = BP
            In this case, two cars meet in 1 hour.
            Therefore, Distance traveled by car X in 1 hour = x km
             \Rightarrow AP = x
            Distance travelled by car Y in 1 hour = y km
 \Rightarrow BP = y  
            Clearly, AP + PB = AB
             \Rightarrow x + y = 100 ...(2) [Since, AB = 100 km]
            Solving (1) and (2), we have
            x = 60, y = 40
            Hence, speed of car X is 60 km/hr and speed of car Y 40 km/hr.

         (v) Let the length and breadth of the rectangle be x and y units respectively. Then,
           Area = xy sq. units
           If length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq. units.
           Therefore, xy – 9 = (x – 5) (y + 3)
            \Rightarrow xy – 9 = xy + 3x – 5y – 15
            \Rightarrow 3x – 5y – 6 = 0 ...(1)
           When length is increased by 3 units and breadth by units, the area is increased by 67 sq. units.
           Therefore, xy + 67 = (x + 3) (y + 2)
            \Rightarrow xy + 67 = xy + 2x + 3y + 6
            \Rightarrow 2x + 3y – 61 = 0 ...(2)
           Solving (1) and (2), we get
           {x \over {305 + 18}} = {{ - y} \over { - 183 + 12}} = {1 \over {9 + 10}}
            \Rightarrow x = {{323} \over {19}} = 17, y = {{171} \over {19}} = 9
           Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

  

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