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Pair of Linear Equations in Two Variables : Exercise - 3.6 (Mathematics NCERT Class 10th)


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Q.1       Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) {1 \over {2x}} + {1 \over {3y}} = 2

{1 \over {3x}} + {1 \over {2y}} = {{13} \over 6}
(ii) {2 \over {\sqrt x }} + {3 \over {\sqrt y }} = 2
{4 \over {\sqrt x }} - {9 \over {\sqrt y }} = - 1
(iii) {4 \over x} + 3y = 14 
{3 \over x} - 4y = 23
(iv) {5 \over {x - 1}} + {1 \over {y - 2}} = 2
       
{6 \over {x - 1}} - {3 \over {y - 2}} = 1
(v) {{7x - 2y} \over {xy}} = 5
{{8x + 7y} \over {xy}} = 15
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
(vii) {{10} \over {x + y}} + {2 \over {x - y}} = 4
{{15} \over {x + y}} - {5 \over {x - y}} = - 2
(viii) {1 \over {3x + y}} + {1 \over {3x - y}} = {3 \over 4}
{1 \over {2\left( {3x + y} \right)}} - {1 \over {2\left( {3x - y} \right)}} = {{ - 1} \over 8}

Sol.       (i) Taking {1 \over x} = u and {1 \over y} = v. The given system of equations become
              {1 \over 2}u + {1 \over 3}v = 2
               \Rightarrow 3u + 2v = 12 ...(1)

              {1 \over 3}u + {1 \over 2}v = {{13} \over 6}
               \Rightarrow 2u + 3v = 13...(2) 

              Multiplying (1) by 3 and (2), we have
              9u + 6v = 36 ...(3)
              and, 4u + 6v = 26 ...(4) 
              Subtracting (4) from (3), we get
              5u = 10
               \Rightarrow u = 2
              Putting u = 2 in (3), we get
             18 + 6v = 36
              \Rightarrow 6v = 18
              \Rightarrow v = 3
             Now, u = 2
              \Rightarrow {1 \over x} = 2
              \Rightarrow x = {1 \over 2}
             and, v = 3
              \Rightarrow {1 \over y} = 3
              \Rightarrow y = {1 \over 3}
             Hence, the solution is x = {1 \over 2}, y = {1 \over 3}.

            (ii) The given system of equations is
            {2 \over {\sqrt x }} + {3 \over {\sqrt y }} = 2 and {4 \over {\sqrt x }} - {9 \over {\sqrt y }} = - 1
            Putting u = {1 \over {\sqrt x }} and v = {1 \over {\sqrt y }} .Then , the given equations become
            2u + 3v = 2 ...(1)
            and, 4u – 9v = – 1 ...(2)
            Multiplying (1) by 3, we get
            6u + 9v = 6...(3)
            Adding (2) and (3), we get
            10u = 5
             \Rightarrow u = {5 \over {10}} = {1 \over 2}
             Putting u = {1 \over 2} in (1), we get
             2 \times {1 \over 2} + 3v = 2
              \Rightarrow 3v = 1
              \Rightarrow v = {1 \over 3}
             Now, u = {1 \over 2}
              \Rightarrow {1 \over {\sqrt x }} = {1 \over 2}
              \Rightarrow {\sqrt x } = 2
              \Rightarrow x = 4

             and, v = {1 \over 3}
              \Rightarrow {1 \over {\sqrt y }} = {1 \over 3}
              \Rightarrow {\sqrt y } = 3
              \Rightarrow y = 9
             Hence, the solution is x = 4, y = 9.

            (iii) The given system of equations is
            {4 \over x} + 3y = 14...(1) 
            and, {3 \over x} - 4y = 23...(2)
            Multiplying (1) by 4 and (2) by 3, we get
            {{16} \over x} + 12y = 56...(3)
            and, {9 \over x} - 12y = 69...(4)
            Adding (3) and (4), we get
            {{25} \over x} = 125
             \Rightarrow x = {{25} \over {125}} = {1 \over 5}
            Putting x = {1 \over 5} in (1), we get
            4 \times 5 + 3y = 14
             \Rightarrow 3y = 14 – 20
             \Rightarrow 3y = – 6
             \Rightarrow y = – 2
            Hence, the solution is x = {1 \over 5}, y = – 2.

            (iv) Let u = {1 \over {x - 1}},v = {1 \over {y - 2}}. Then, the given system of equations becomes
            5u + v = 2 ...(1)
            and, 6u – 3v = 1 ...(2)
            Multiplying (1) by 3, we get
            15u + 3v = 6 ...(3)
            Adding (2) and (3), we get
            21u = 7
             \Rightarrow  u = {1 \over 3}
            Putting u = {1 \over 3} in (1), we get
            {5 \over 3} + v = 2
             \Rightarrow v = 2 - {5 \over 3} = {{6 - 5} \over 3} = {1 \over 3}
            Now, u = {1 \over 3}
             \Rightarrow {1 \over {x - 1}} = {1 \over 3}
             \Rightarrow x – 1 = 3
             \Rightarrow x = 4
             v = {1 \over 3}
              \Rightarrow {1 \over {y - 2}} = {1 \over 3}
              \Rightarrow y – 2 = 3
              \Rightarrow y = 5
            Hence, the solution is x = 4, y = 5.

           (v) The given system of equations is
           {{7x - 2y} \over {xy}} = 5
            \Rightarrow {7 \over y} - {2 \over x} = 5
           and, {{8x + 7y} \over {xy}} = 15
            \Rightarrow {8 \over y} + {7 \over x} = 15
            Let u = {1 \over x}, v = {1 \over y}. Then, the above equations become
            7 – 2u = 5 ...(1)
            and, 8v + 7u = 15 ...(2)
            Multiplying (1) by 7 and (2), we get
            49v – 14u = 35 ...(3)
             and, 16v + 14u = 30
             Adding (3) and (4), we get
             65v = 65
              \Rightarrow v = 1
             Putting v = 1 in (1), we get
             7 – 2u = 5
              \Rightarrow – 2u = – 2
              \Rightarrow u = 1
             Now, u = 1
              \Rightarrow {1 \over x} = 1
              \Rightarrow x = 1
              and, v = 1
               \Rightarrow {1 \over y} = 1
               \Rightarrow y = 1
              Hence, the solution is x = 1, y = 1.

              (vi) The given system of equations is 6x + 3y = 6xy and 2x + 4y = 5xy, where x and y are non -zero.
              Since x \ne 0,y \ne 0, , we have xy  \ne 0.
              On dividing each one of the given equations by xy we get
              {3 \over x} + {6 \over y} = 6 and {4 \over x} + {2 \over y} = 5
              Taking {1 \over x} = u and {1 \over y} = v, the above equations become
               3u + 6v = 6...(1) 
               and, 4u + 2v = 5...(2) 
               Multiplying (2) by 3, we get
               12u + 6v = 15 ...(3)
               Subtracting (1) from (3), we get
               9u = 15 – 6 = 9
                \Rightarrow u = 1
               Putting u = 1 in (1), we get
               3×1 + 6v = 6
                \Rightarrow 6v = 6 – 3 = 3
                \Rightarrow v = {1 \over 2}
                Now, u = 1
                 \Rightarrow {1 \over x} = 1
                 \Rightarrow x = 1
                and, v = {1 \over 2}
                 \Rightarrow {1 \over y} = {1 \over 2}
                 \Rightarrow y = 2
                Hence, the given system of equations has one solution x = 1, y = 2.

                (vii) The given system of equations is
                {{10} \over {x + y}} + {2 \over {x - y}} = 4
                {{15} \over {x + y}} - {5 \over {x - y}} = - 2
                Putting u = {1 \over {x + y}} and v = {1 \over {x - y}}. Then, the given equations become
                10u +2v = 4
                 \Rightarrow 5u + v = 2 ...(1)
                and, 15u – 5v = – 2...(2)
                Multiplying (1) by 5, we get
                25u + 5v = 10 ...(3)
                Adding (2) and (3), we get
                40u = 8
                 \Rightarrow u = {1 \over 5}
                Putting u = {1 \over 5} in (1), we get
                5\left( {{1 \over 5}} \right) + v = 2
                 \Rightarrow v = 2 – 1 = 1
                 Now, u = {1 \over 5}
                 \Rightarrow x + y = 5 ...(4)
                v = 1
                  \Rightarrow {1 \over {x - y}} = 1
                 \Rightarrow x – y = 1
                Adding (4) and (5), we get
                2x = 6
                 \Rightarrow x = 3
                When x = 3, then from (4), we get
                 3 + y = 5
                  \Rightarrow y = 5 – 3 = 2
                 Hence, the given system of equations has one solution
                 x = 3, y = 2

               (viii) Taking u = {1 \over {3x + y}} and v = {1 \over {3x - y}}. The give system of equations becomes
                u + v  = {3 \over 4}...(1)
                {1 \over 2}u - {1 \over 2}v = {{ - 1} \over 8}
                 \Rightarrow u – v = {{ - 1} \over 4}...(2)
                Adding (1) and (2), we get
                 2u = {3 \over 4} - {1 \over 4}
                  \Rightarrow 2u = {2 \over 4} = {1 \over 2}
                  \Rightarrow u = {1 \over 4}
                 Putting u = {1 \over 4} in (1), we get
                 {1 \over 4} + v = {3 \over 4}
                  \Rightarrow v = {3 \over 4} - {1 \over 4} = {2 \over 4} = {1 \over 2}
                 Now, u = {1 \over 4}
                  \Rightarrow {1 \over {3x + y}} = {1 \over 4}
                  \Rightarrow 3x + y = 4...(3) 
                 and, v = {1 \over 2}
                  \Rightarrow 3x – y = 2...(4)
                  \Rightarrow {1 \over {3x - y}} = {1 \over 2}
                 Adding (3) and (4), we get
                 Putting x = 1 in (3), we get
                 3x + y = 4 
                  \Rightarrow y = 4 – 3 = 1
                 Hence, the solution is x = 1, y = 1.


Q.2      Formulate the following problems as a pair if equations, and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken be 1 man alone.
(iii) Roohi travels 300 km to her home partly be train and partly be bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes linger. find the speed of the train and the bus separately.
Sol.       (i) Let the speed of the boat in still water be x km/hr. and the speed of the stream by y km/hr. Then,
               Speed upstream = (x – y) km/hr
               Speed downstream = (x + y) km/hr
               Time taken to cover 20 km downstream = 2 hours
                \Rightarrow {{20} \over {x + y}} = 2
                \Rightarrow x +y = 10...(1)
               Time taken to cover 4 km upstream = 2 hours
                 \Rightarrow {4 \over {x - y}} = 2
                 \Rightarrow x – y = 2...(2)
                Adding (1) and (2), we get
                2x = 12
                 \Rightarrow x = 6
                Putting x = 6 in (1), we get
                6 + y = 10
                 \Rightarrow y = 10 – 6 = 4
                Hence, the speed of boat in still water = 6km/hr
                and, speed of stream = 4 km/hr

                (ii) Let 1 woman can finish the embroidery in x days and 1 man can finish the embroidery in y days.
                Then,
                 1 woman's 1 day's work = {1 \over x}
                 1 man's 1 day's work = {1 \over y}
                 Therefore, {2 \over x} + {5 \over y} = {1 \over 4} and {3 \over x} + {6 \over y} = {1 \over 3}
                 Putting {1 \over x} = u and {1 \over y} = v, these equations become
                 2u + 5v = {1 \over 4}...(1)
                 and, 3u + 6v = {1 \over 3}...(2)
                 Multiplying (1) by 3 and (2) by 2 and subtracting, we get
                 3v = {1 \over {12}}
                  \Rightarrow v = {1 \over {36}}
                 Putting v = {1 \over {36}} in (1), we get
                 2u = \left( {{1 \over 4} - {5 \over {36}}} \right) = {{9 - 5} \over {36}}
                  = {4 \over {36}} = {1 \over 9}
                  \Rightarrow u = {1 \over {18}}
                 Now, u = {1 \over {18}}
                  \Rightarrow {1 \over x} = {1 \over {18}}
                  \Rightarrow x = 18
                  and,
                  v = {1 \over {36}}
                   \Rightarrow {1 \over y} = {1 \over {36}}
                   \Rightarrow y = 36
                  Thus, I woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.

                   (iii) Let the speed of train be c km/hr and speed of bus be y km/hr.
                    Then, {{60} \over x} + {{240} \over y} = 4 ...(1)
                    and, {{100} \over x} + {{200} \over y} = 4 + {{10} \over {60}} = {{25} \over 6} ...(2)
                    Let {1 \over x} = u and {1 \over y} = v, than eqns. (1) and (2) becomes
                    60u +240v = 4 ...(3)
                    100u + 200v = {{25} \over 6} ... (4)
                    Multiplying eqn. (3) by 5 and eqn. (4) by 6, we get                               

001

                   \Rightarrow u = {{ - 5} \over { - 300}} = {1 \over {60}}
                  From eqn. (3)
                  60 × {1 \over {60}} + 240 × v = 4 

                   \Rightarrow v = {3 \over {240}} = {1 \over {80}}
                  Since, x = {1 \over u} = 60 and y = {1 \over v} = 80
                  Therefore, Speed of train is 60 km/h and speed of bus is 80 km/h.



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