Exercise 3.2

Question .1      Solve the following pair of linear equations by the substitution method 

            (i) x + y = 14                                                         x – y = 4
            (ii) s – t = 3                                                           s3+t2=6
            (iii) 3x + y = 3                                                       9x – 3y = 9
            (iv) 0.2x + 0.3y = 1.3                                            0.4x + 0.5y = 2.3
            (v) 2–√x+3–√y=0                                             3–√x−8–√y=0             (vi) 3x2−5y3=−2                                                x3+y2=136

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
              ⇒ x – 14 + x = 4

              ⇒ 2x = 4 + 14
              ⇒ 2x = 18

              ⇒ x = 9

              Putting x = 9 in (1), we get

              9 + y = 14
              ⇒ y = 14 – 9
              ⇒ 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)

              s3+t2=6 ...(2)

              From (1), s = 3 + t

  Substituting s = 3 + t in (2), we get
              3+t3+t2=6
              ⇒2(3+t)+3t=36
              ⇒ 6 + 2t + 3t = 36
              ⇒ 5t = 30
              ⇒ t = 6
              Putting t = 6 in (1), we get
              s – 6 = 3
               ⇒ 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
                ⇒ 9x – 9x + 9 = 9
                ⇒ 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
                 ⇒ 2x + 3y = 13...(1)
                 and, 0.4x + 0.5y = 2.3 
                 ⇒ 4x + 5y = 23 ...(2) 
                 From (2), 5y = 23 – 4x
                 ⇒ y=23−4x5
                 Substituting y=23−4x5 in (1), we get
                 2x+3(23−4x5)=13
                 ⇒ 10x + 69 – 12x = 65
                 ⇒ – 2x = – 4
                 ⇒ x = 2
                 Putting x = 2 in (1), we get
                 2 × 2 + 3y = 13
                 ⇒ 3y = 13 – 4 = 9
                 ⇒ y=93=3
                 Hence, the solution of the given system of equations is
                 x = 2, y = 3
                 (v) The given system of equations is
                  2x−−√+3y−−√=0 ...(1)
                  and, 3x−−√−8y−−√=0 ...(2)
                  From (2), 8y−−√=3x−−√
                  ⇒ y=3x√8√
                  Substituting y=3x√8√ in (1), we get
                  2x−−√+3–√(3x√8√)=0
                  ⇒ 2x−−√+3x8√=0
                  ⇒ 16x−−−√+3x=0
                  ⇒ 4x + 3x = 0
                  ⇒ 7x = 0 
                  ⇒ x = 0
                  Putting x = 0 in (1), we get
                  0 + 3–√y = 0
                  ⇒ y = 0
                  Hence, the solution of the given system of equations is
                  x = 0, y = 0
                  (vi) The given system of equations is
                  3x2−5y3=−2
                  ⇒ 9x – 10y = – 12 ...(1)
                  and, x3+y2=136
                  ⇒ 2x + 3y = 13 ...(2)
                  From (1), 10y = 9x + 12
                  ⇒ y=9x+1210
                  Substituting y=9x+1210 in (2), we get
                  2x+3(9x+1210)=13
                  ⇒ 20x + 27x + 36 = 130
                  ⇒ 47x = 130 – 36
                  ⇒ 47x = 94
                  ⇒ x=9447=2
                  Putting x = 2 in (1), we get
                  9 × 2 – 10y = – 12
                  ⇒ –10y = –12 – 18   
⇒ – 10y = – 30
                  ⇒ y=−30−10=3
                  Hence, the solution of the given system of equations is
                  x = 2, y = 3

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Question.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
                 ⇒y=11−2x3                  Substituting y=11−2x3 in (2), we get                 2x−4(11−2x3)=−24                  ⇒ 6x – 44 + 8x = – 72                 ⇒ 14x = –72 + 44
                 ⇒ 14x = –28                 ⇒ x=−2814=−2                  Putting x = –2 in (1), we get                 2(– 2) + 3y = 11                 ⇒ – 4 + 3y = 11                 ⇒ 3y = 11 + 4                 ⇒ 3y = 15                 ⇒ y=153=5                   Putting x = – 2, y = 5 in y = mx + 3, we get                  5 = m(–2) + 3
                  ⇒ –2m = 5 – 3                  ⇒ – 2m = 2                  ⇒ m = – 1

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Question.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.
Sol.      (i) Let the numbers be x and y (x > y). Then,            x – y = 26            and, x = 3y            From (2), y=x3 in (1), we get            x−x3=26
            ⇒3x−x=78             ⇒ 2x = 78             ⇒ x = 39            Putting x = 39 in (2), we get            ⇒ 39 = 3y            ⇒ y=393=13              Hence, the required numbers are 39 and 13.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. find them.
              (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
              ⇒ 2y = 180 – 18
              ⇒ 2y = 162
              ⇒ y=1622=81
              Putting y = 81 in (2), we get
              x = 81 + 18 = 99
              Thus, the angles are 99° and 81°.
(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.
              (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 
               ⇒ y=1750−3x5
               Substituting y=1750−3x5 in (1), we get
               7x+6(1750−3x5)=3800
               ⇒ 35x + 10500 – 18x = 19000
               ⇒ 17x = 19000 – 10500
               ⇒ 17x = 8500
               ⇒ x=850017=500
               Putting x = 500 in (2), we get
               3(500) + 5y = 1750
               ⇒ 5y = 1750 – 1500
               ⇒ 5y = 250
               ⇒ y=2505=50
               Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs 50.
               
(iv) A fraction becomes 911 , if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it become 56 . Find the fraction.
                (v) Let the fraction be xy .
                Then, according to the given conditions, we have
                x+2y+2=911 and x+3y+3=56
                ⇒ 11x + 22 = 9y + 18 and 6x + 18 = 5y + 15
                ⇒ 11x – 9y = – 4 ...(1)
                and, 6x – 5y = – 3....(2)
                From (2), 5y = 6x + 3
                ⇒ y=6x+35
                Substituting y=6x+35 in (1), we get
                11x−9(6x+35)=−4
                ⇒ 55x – 54x – 27 = –20
                ⇒ x = – 20 + 27
                ⇒ x = 7
                Putting x = 7 in (1), we get
                11(7) – 9y = – 4
                ⇒ – 9y = – 4 – 77
                ⇒ –9y = – 81
                ⇒ y=−81−9=9                  Hence, the given fraction is 79 .
(v) Five years hence, the age of Nuri will three times that of his son. Five years ago,Nuri's age was seven times that of his son. What are the present ages?
                 (v) Let the present age of Nurui  be x years and the present age of his son be y years.
                 Five years hence, Nuri's age = (x + 5) years
                                                Son's age = (y + 5) years
                 Five years ago, Nuri's age = (y – 5) years 
                 Son's age = (y – 5) years                 As per question, we get
                 (x + 5) = 3(y + 5)
                 ⇒ x + 5 = 3y + 15
                 ⇒ x – 3y =10
                 ⇒ x – 3y = 15 – 5
                 and, (x – 5) = 7(y – 5)
                 ⇒ x – 5 = 7y – 35
                 ⇒ x – 7y = – 30
                 ⇒ x – 7y = – 35 + 5
                 From (1), x = 3y + 10
                 Substituting x = 3y +10 in (2), we get
                 3y + 10 – 7y = – 30
                 ⇒ – 4y = – 30 – 10
                 ⇒ – 4y = – 40
                 ⇒ y = 10
                 Putting y = 10 in (1),we get
                 x – 3 × 10 = 10
                 ⇒ x = 10 + 30 = 40
                 Hence, present age of Nuri is 40 years and that his son is 10 years