Pair of Linear Equations in Two Variables : Exercise 3.3



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Q.1      Solve the following pair of linear equations by the substitution method :
            (i) x + y = 14                                                         x – y = 4
            (ii) s – t = 3                                                           {s \over 3} + {t \over 2} = 6
            (iii) 3x + y = 3                                                       9x – 3y = 9
            (iv) 0.2x + 0.3y = 1.3                                            0.4x + 0.5y = 2.3
            (v) \sqrt 2 x + \sqrt 3 y = 0                                             \sqrt 3 x - \sqrt 8 y = 0
            (vi) {{3x} \over 2} - {{5y} \over 3} = - 2                                                {x \over 3} + {y \over 2} = {{13} \over 6}
Sol.       (i) The given system of equations is
              x + y = 14 ...(1)
              and, x – y = 4 ...(2)
              From (1), y = 14 – x
              Substituting y = 14 – x in (2), we get
              x – (14 – x) = 14
               \Rightarrow x – 14 + x = 4

               \Rightarrow 2x = 4 + 14
               \Rightarrow 2x = 18

               \Rightarrow x = 9
              Putting x = 9 in (1), we get
              9 + y = 14
               \Rightarrow y = 14 – 9
               \Rightarrow y = 5

              Hence, the solution of the given system of equations is
              x = 9, y = 5

             (ii) The given system of equations is
              s – t = 3 ...(1)
              {s \over 3} + {t \over 2} = 6 ...(2)
              From (1), s = 3 + t
              Substituting s = 3 + t in (2), we get
              {{3 + t} \over 3} + {t \over 2} = 6
               \Rightarrow 2(3 + t) + 3t = 36

               \Rightarrow 6 + 2t + 3t = 36
               \Rightarrow 5t = 30
               \Rightarrow t = 6

              Putting t = 6 in (1), we get
              s – 6 = 3
                \Rightarrow s = 3 + 6 = 9

               Hence, the solution of the given system of equations is
               s = 9, t = 6

               (iii) The given system of equations is
                3x – y = 3 ...(1)
                and, 9x – 3y = 9 ...(2)
                From (1), y = 3x – 3
                Substituting y = 3x – 3 in (2), we get
                9x – 3(3x – 3) = 9
                 \Rightarrow 9x – 9x + 9 = 9

                 \Rightarrow 9 = 9
               This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y . This situation has arisen because both the given equations are the same.
Therefore, Equations (1) and (2) have infinitely many solutions.

                 (iv) The given system of equations is
                 0.2x + 0.3y = 1.3
                  \Rightarrow 2x + 3y = 13...(1)

                 and, 0.4x + 0.5y = 2.3 
                  \Rightarrow 4x + 5y = 23 ...(2) 

                 From (2), 5y = 23 – 4x
                  \Rightarrow y = {{23 - 4x} \over 5}

                 Substituting y = {{23 - 4x} \over 5} in (1), we get
                 2x + 3\left( {{{23 - 4x} \over 5}} \right) = 13
                  \Rightarrow 10x + 69 – 12x = 65
                  \Rightarrow – 2x = – 4
                  \Rightarrow x = 2

                 Putting x = 2 in (1), we get
                 2 × 2 + 3y = 13
                  \Rightarrow 3y = 13 – 4 = 9

                  \Rightarrow y = {9 \over 3} = 3
                 Hence, the solution of the given system of equations is
                 x = 2, y = 3

                 (v) The given system of equations is
                  \sqrt {2x} + \sqrt {3y} = 0 ...(1)
                  and, \sqrt {3x} - \sqrt {8y} = 0 ...(2)
                  From (2), \sqrt {8y} = \sqrt {3x}
                   \Rightarrow y = {{\sqrt {3x} } \over {\sqrt 8 }}

                  Substituting y = {{\sqrt {3x} } \over {\sqrt 8 }} in (1), we get
                  \sqrt {2x} + \sqrt 3 \left( {{{\sqrt {3x} } \over {\sqrt 8 }}} \right) = 0
                   \Rightarrow \sqrt {2x} + {{3x} \over {\sqrt 8 }} = 0

                   \Rightarrow \sqrt {16x} + 3x = 0
                   \Rightarrow 4x + 3x = 0
                   \Rightarrow 7x = 0 
                   \Rightarrow x = 0

                  Putting x = 0 in (1), we get
                  0 + \sqrt 3 y = 0
                   \Rightarrow y = 0

                  Hence, the solution of the given system of equations is
                  x = 0, y = 0

                  (vi) The given system of equations is
                  {{3x} \over 2} - {{5y} \over 3} = - 2
                   \Rightarrow 9x – 10y = – 12 ...(1)

                  and, {x \over 3} + {y \over 2} = {{13} \over 6}
                   \Rightarrow 2x + 3y = 13 ...(2)

                  From (1), 10y = 9x + 12
                   \Rightarrow y = {{9x + 12} \over {10}}

                  Substituting y = {{9x + 12} \over {10}} in (2), we get
                  2x + 3\left( {{{9x + 12} \over {10}}} \right) = 13
                   \Rightarrow 20x + 27x + 36 = 130
                   \Rightarrow 47x = 130 – 36
                   \Rightarrow 47x = 94

                   \Rightarrow x = {{94} \over {47}} = 2
                  Putting x = 2 in (1), we get
                  9 × 2 – 10y = – 12
                   \Rightarrow –10y = –12 – 18
                   \Rightarrow – 10y = – 30
                   \Rightarrow y = {{ - 30} \over { - 10}} = 3
                  Hence, the solution of the given system of equations is
                  x = 2, y = 3


Q.2          Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of 'm' for which y = mx + 3.
Sol.          The given system of equations is
                2x + 3y = 11
                and, 2x – 4y = – 24
                From(1), 3y = 11 – 2x
                  \Rightarrow y = {{11 - 2x} \over 3}
                 Substituting y = {{11 - 2x} \over 3} in (2), we get
                 2x - 4\left( {{{11 - 2x} \over 3}} \right) = - 24
                  \Rightarrow 6x – 44 + 8x = – 72
                  \Rightarrow 14x = –72 + 44
                  \Rightarrow 14x = –28
                  \Rightarrow x = {{ - 28} \over {14}} = - 2
                 Putting x = –2 in (1), we get
                 2(– 2) + 3y = 11
                  \Rightarrow – 4 + 3y = 11
                  \Rightarrow 3y = 11 + 4
                  \Rightarrow 3y = 15
                  \Rightarrow y = {{15} \over 3} = 5
                  Putting x = – 2, y = 5 in y = mx + 3, we get
                  5 = m(–2) + 3
                   \Rightarrow –2m = 5 – 3
                   \Rightarrow – 2m = 2
                   \Rightarrow m = – 1


Q.3      Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid Rs. 155. What are the fixed charges and the charge per km ? How much does a person have to pay for travelling a distance of 25 km ?
(v) A fraction becomes {9 \over {11}}, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it become {5 \over 6}. Find the fraction.
(vi) Five years hence, the age of Jacob will three times that of his son. Five years ago, Jacob age was seven times that of his son. What are the present ages?
Sol.      (i) Let the numbers be x and y (x > y). Then,
            x – y = 26
            and, x = 3y
            From (2), y = {x \over 3} in (1), we get
            x - {x \over 3} = 26
             \Rightarrow 3x - x = 78
             \Rightarrow 2x = 78 
             \Rightarrow x = 39
            Putting x = 39 in (2), we get
             \Rightarrow 39 = 3y
             \Rightarrow y = {{39} \over 3} = 13
             Hence, the required numbers are 39 and 13.

              (ii) Let the angles be x° and y°(x > y). Then,
              x + y = 180
              and, x = y + 18
              Substituting x = y + 18 in (1), we get
              y + 18 + y = 180
               \Rightarrow 2y = 180 – 18

               \Rightarrow 2y = 162
               \Rightarrow y = {{162} \over 2} = 81

              Putting y = 81 in (2), we get
              x = 81 + 18 = 99
              Thus, the angles are 99° and 81°.

              (iii) Let the cost of one bat and one ball be Rs. x and Rs. y respectively. Then,
               7x + 6y = 3800 .....(1)
               and, 3x + 5y = 1750 ...(2)
               From (2), 5y = 1750 – 3x 
                \Rightarrow y = {{1750 - 3x} \over 5}

               Substituting y = {{1750 - 3x} \over 5} in (1), we get
               7x + 6\left( {{{1750 - 3x} \over 5}} \right) = 3800
                \Rightarrow 35x + 10500 – 18x = 19000
                \Rightarrow 17x = 19000 – 10500
                \Rightarrow 17x = 8500
                \Rightarrow x = {{8500} \over {17}} = 500
               Putting x = 500 in (2), we get
               3(500) + 5y = 1750
                \Rightarrow 5y = 1750 – 1500
                \Rightarrow 5y = 250
                \Rightarrow y = {{250} \over 5} = 50
               Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs 50.

               (iv) Let the fixed charges of taxi be Rs. x per km and the running charges be Rs y km/hr.
                According to the given condition, we have
                x + 10y = 105 ...(1)
                x + 15y = 155 ...(2)
                From (1), x = 105 – 10y
                Substituting x = 105 – 10y in (2), we get
                105 – 10y + 15y = 155
                  \Rightarrow 105 + 5y = 155

                  \Rightarrow 5y = 155 – 105
                  \Rightarrow 5y = 50

                  \Rightarrow y = 10
                  Putting y = 10 in (1), we get
                  x + 10 × 10 = 105
                   \Rightarrow x = 105 – 100

                   \Rightarrow x = 5
                  Total charges for travelling a distance of 25 km
                   = x + 25y = Rs (5 + 25 × 10)
                   = Rs. 255
                   Hence, the fixed charge is Rs 5, the charge per km is Rs. 10 and the total charge for travelling a distance of 25 km is Rs 255.

                (v) Let the fraction be {x \over y}.
                Then, according to the given conditions, we have
                {{x + 2} \over {y + 2}} = {9 \over {11}} and {{x + 3} \over {y + 3}} = {5 \over 6}
                 \Rightarrow 11x + 22 = 9y + 18 and 6x + 18 = 5y + 15
                 \Rightarrow 11x – 9y = – 4 ...(1)
                and, 6x – 5y = – 3....(2)
                From (2), 5y = 6x + 3
                 \Rightarrow y = {{6x + 3} \over 5}
                Substituting y = {{6x + 3} \over 5} in (1), we get
                11x - 9\left( {{{6x + 3} \over 5}} \right) = - 4
                 \Rightarrow 55x – 54x – 27 = –20

                 \Rightarrow x = – 20 + 27
                 \Rightarrow x = 7

                Putting x = 7 in (1), we get
                11(7) – 9y = – 4
                 \Rightarrow – 9y = – 4 – 77

                 \Rightarrow –9y = – 81
                 \Rightarrow y = {{ - 81} \over { - 9}} = 9

                 Hence, the given fraction is {7 \over 9}.

                 (vi) Let the present age of Jacob be x years and the present age of his son be y years.
                 Five years hence, Jacob's age = (x + 5) years
                                                Son's age = (y + 5) years
                 Five years ago, Jacob's age = (y – 5) years 
                 Son's age = (y – 5) years
                 As per question, we get
                 (x + 5) = 3(y + 5)
                  \Rightarrow x + 5 = 3y + 15
                  \Rightarrow x – 3y =10
                  \Rightarrow x – 3y = 15 – 5
                 and, (x – 5) = 7(y – 5)
                  \Rightarrow x – 5 = 7y – 35
                  \Rightarrow x – 7y = – 30
                  \Rightarrow x – 7y = – 35 + 5
                 From (1), x = 3y + 10
                 Substituting x = 3y +10 in (2), we get
                 3y + 10 – 7y = – 30

                  \Rightarrow – 4y = – 30 – 10
                  \Rightarrow – 4y = – 40
                  \Rightarrow y = 10
                 Putting y = 10 in (1),we get
                 x – 3 × 10 = 10
                  \Rightarrow x = 10 + 30 = 40
                 Hence, present age of Jacob is 40 years and that his son is 10 years.



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