Exercise 3.3

Q.1      Solve the following pair of linear equation by the elimination method and the substituti method :
           (i) x + y = 5 and 2x – 3y = 4           (ii) 3x + 4y = 10 and 2x – 2y = 2           (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7           (iv) x2+2y3=−1 and x−y3=3Sol.       (i) By elimination method :             The given system of equation is              x + y = 5              and, 2x – 3y = 4              Multiplying (1) by 3, we get              3x + 3y = 15              Adding (2) and (3), we get              5x = 19
              ⇒x=195              Putting x=195 in (1), we get              195 + y = 5
              ⇒ y = 5 – 195=25−195
              Hence, x=195, y=65              By substitution method :              The given system of equations is              x + y = 5 ...(1)              and, 2x – 3y = 4 ...(2)              From (1), y = 5 – x              Substituting y = 5 – x in (2), we get              2x – 3 (5 – x) = 4              ⇒ 2x – 15 + 3x = 4              ⇒ 5x = 4 + 15              ⇒ 5x = 19              ⇒x=195              Putting x=195 in (1), we get              195+y=5
              ⇒ y = 5 – 195=25−195=65              Hence, x=195,y=65              By geometrical method :              Graph of x + y = 5 :              We have, x + y = 5              ⇒ y = 5 – x              When x = 0 , y = 5; When x = 5, y = 0              Thus, we have the following table :

              Plotting the points A(0, 5) and B(5, 0) and drawing a line joining them, we get the graph of the equation x + y = 5 as shown.
             Graph of 2x – 3y = 4
             We have, 2x – 3y = 4
             ⇒ 3x = 2x – 4
             ⇒ y=2x−43
             When x = 2, y=4−43=0; when x = – 1, y=−2−43=−62=−2
             Thus, we have the following table :

             Plotting the points C(2, 0) and D(–1,–2) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.

              Clearly, two lines intersect at the point P(195,65).
              Hence, x=195,y=65 is the solution of the given system.
              The method of elimination is the most efficient in this case.
              (ii) By elimination method :
              The given system of equation is
               3x + 4y = 10 ...(1)
              and, 2x – 2y = 2 ...(2)
              Multiplying (2) by 2 and adding to (1), we get
              3x + 4y + 4x – 4y = 10 +4
              ⇒ 7x = 14
              ⇒ x = 2
              Putting x = 2 in (1), we get
              3(2) + 4y = 10
              ⇒ 4y = 10 – 6 = 4
              ⇒ y = 1
              Hence, x = 2 y = 1
              By substitution method :
              The given system of equaitons is
              3x + 4y = 10 ...(1)
             and, 2x – 2y = 2
             ⇒ x – y = 1 ...(2)
             From (2), y = x – 1
             Substituting y = x – 1 in (1), we get
             3x + 4(x – 1) = 10
             ⇒ 3x + 4x – 4 = 10
             ⇒ 7x = 14
             ⇒ x = 2
             Putting x = 2 in (1), we get
             3(2) + 4y = 10
             ⇒ 4y = 10 – 6 = 4
             ⇒ y = 1
             Hence, x = 2, y = 1
             By geometrical method :
             The given system of equation is
             3x + 4y = 10 ...(1)
             and, 2x – 2y = 2 
             ⇒ x – y = 1 ...(2)
             For the graph of 3x + 4y = 10
             We have, 3x + 4y = 10
             ⇒ 4y = 10 – 3x
             ⇒ y=10−3x4
             When x = 2, y=10−64=44=1; when x = – 2, y=10+64=164=4
             Thus, we have the following table :

            Plotting the points A(2, 1) and B(–2, 4) and drawing a line joining them, we get the graph of the equation 3x + 4y = 10.
           For the graph of x – y = 1 :
           We have, x – y = 1
           ⇒ y = x – 1
           When x = 3, y = 2; When x = 0, y = – 1
           Thus, we have the following table :

           Plotting the points C(3, 2) and D(0, – 1) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.

            Clearly, two lines intersect at the point A(2, 1).
            Hence, x = 2, y = 1
            In this case, almost all the methods are equally efficient, however, geometrical method takes more time. 
            (iii) By elimination method :
            The given system of equations is
            3x – 5y – 4 = 0
            ⇒ 3x – 5y = 4 ....(1)
            and 9x = 2y + 7
            ⇒ 9x – 2y = 7 ...(2)
            Multiplying (1) by 3, we get
            9x – 15y = 12 ...(3)
            Sibtracting (2) from (3), we get
            – 13y = 5
           ⇒ y=−513
           Putting y=−513 in (1), we get
           3x –5(−513) = 4
           ⇒ 3x+2513=4
           3x = 4 – 2513
           ⇒ 3x=52−2513
           ⇒3x=2713
           ⇒ x=913
           Hence, x=913, y=−513
           By, substitution method :
           The given system of equations is
           3x – 5y – 4 = 0
           ⇒ 3x – 5y = 4 ...(1)
           and, 9x = 2y + 7
           ⇒ 9x – 2y = 7....(2) 
           From (2), 2y = 9x – 7
           ⇒ y = 9x−72
           Substituting y = 9x−72 in (1), we get
           3x−5(9x−72)=4
           ⇒ 6x – 45x + 35 = 8
           ⇒ – 39x = 8 – 35
           ⇒ – 39x = – 27
           ⇒ x = −27−39=913
           Putting x = 913 in (2), we get
           3×913−5y=4
           ⇒5y=2713−4
           ⇒5y=27−5213=−2513
           ⇒y=−513
           Hence, x=913y=−513
           By geometrical method :
           The given system of equations is
            3x – 5y – 4 = 0
            ⇒ 3x – 5y = 4 ...(1)
            and, 9x = 2y + 7
            ⇒ 9x – 2y = 7...(2)
            For the graph of 3x – 5y = 4 :
            We have, 3x – 5y = 4
            ⇒ 5y = 3x – 4
            ⇒y=3x−45]
            When x = 3 y = 9−45=55 = 1;
            When x = – 2, y = −6−45=−105=−2
           Thus, we have the following table :

          Plotting the points A(3, 1) and B(–2,–2) and drawing a line joining them, we get the graph of the equation 3x – 5y = 4.
          For the graph of 9x – 2y = 7 :
          We have, 9x – 2y = 7
          ⇒ 2y = 9x – 7
          ⇒y=9x−72
          When x = 1, y=9−72=22=1 ;
          When x = 3, y=27−72=202=10 ;
          Thus, we have the following table :

            Plotting the points C(1, 1) and D(3, 10) on the same graph paper and drawing a line joining them, we obtain the graph of equation.
            Clearly, two lines intersect at the point P(913,−513)
            Hence, x=913,y=−513
            In this case, the method of elimination is the most efficient.
(iv) By elimination method ;
            The given system of equations is
             x2+2y3=−1
             ⇒ 3x + 4y = – 6 ...(1)
             and, x−y3=3
             ⇒ 3x – y = 9...(2)
             Multiplying (2) by 4 and adding to (1), we get
             15x = 30
             ⇒ x = 2
             Putting x = 2 in (2), we get
             3(2) – y = 9
             ⇒ –y = 9 – 6 = 3
             ⇒ y = – 3
             Hence, x = 2, y = – 3
             By substitution method :              x2+2y3=−1
            ⇒ 3x + 4y = – 6 ...(1)
             and, x−y3=3
             ⇒ 3x – y = 9 ...(2)
             From (2), y = 3x – 9
             Putting y = 3x – 9 in (1), we get              3x + 4(3x – 9) = – 6
             ⇒ 3x + 12x – 36 = – 6
             ⇒ 15x = 30
             ⇒ x = 2
             Putting x = 2 in (2), we get
             3(2) – y = 9
             ⇒ – y = 9 – 6 = 3
             ⇒ y = – 3
             Hence, x = 2, y = – 3
             By geometrical method :
             The given system of equations is
             x2+2y3=−1
             ⇒ 3x + 4y = – 6 ...(1)
             and, x−y3=3
             ⇒ 3x – y = 9  ...(2)
             For the graph of 3x + 4y = – 6 :
             We have,
             3x + 4y = – 6
             ⇒ 4y = – 6 – 3x
             ⇒ y = −6−3x4
             When x = 2, y = −6−64=−124=−3;
             when x = – 2, y = −6+64=04=0
             Thus, we have the following table :

            Plotting the points A(2,–3) and B(–2, 0) and drawing a line joining them, we get the graph of the equation 3x +4y = – 6.
           For the graph of 3x – y = 9 :
           We have, 3x – y = 9
           ⇒ y = 3x – 9
           When x = 3, y = 6
           ⇒ y = 3x – 9
           When x = 3, y = 9 – 9 = 0;
           When x = 4, y = 12 – 9 = 3
          Thus, we have the following table :

           Plotting the points C(3, 0) and D(4, 3) on the same graph paper and drawing a line joining them, we obtain the graph of the equation.
              Clearly, the two lines intersect at A(2, –3).
              Hence, x = 2, y = – 3
              In this case, almost all the methods are equally efficient, however, geometrical method takes more time.

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Q.2      Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 if we only add 1 to the denominator. What is the fraction ?

Sol.       (i) Let x be the numerator and y be the denominator of the fraction. So, the fraction is xy.
              By given conditions : x+1y−1=1
              ⇒x+1=y−1
              ⇒ x – y = – 2...(1)
              and, xy+1=12
              ⇒ 2x = y + 1 ...(2)
              ⇒ 2x – y = 1
              Subtracting (2) from (1), we get
              (x – y) – (2x – y) = – 2 – 1
              ⇒ x – y – 2x + y = – 3
              ⇒ – x = – 3
               ⇒ x = 3
              Substituting x = 3 in (1), we get
              3 – y = –2
              ⇒ y = 5
              Hence, the required fraction is 35.
(ii) Five year ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
             (ii) Let the present age of Nuri = x years
             and, the present age of Sonu = y years
             Five years ago, Nuri's age = (x – 5) years
             Sonu's age = (y – 5) years
             As per conditions, x – 5 = 3(y – 5)
             ⇒ x – 5 = 3y – 15
             ⇒ x – 3y = – 15 + 5
             ⇒ x – 3y = – 10 ...(1)
             Ten years later, Nuri's age = (x + 10) years
             Sonu's age = (y + 10)
             As per condition, x + 10 = 2(y + 10)
             ⇒ x + 10 = 2y + 20
             ⇒ x – 2y = 20 – 10
             ⇒ x – 2y = 10 ...(2)
             Subtracting (2) from (1), we get
             – y = – 20
             ⇒ y = 20
             Putting y = 20 in (2), we get
             x – 2(20) = 10
             ⇒ x = 10 + 40 = 50
             Therefore, Nuri's present age = 50 years
             and Sonu's present age = 20 years
(iii) The sum of the digits of a two - digit number 9. Also, nine times this number is twice the number obtained by reversingthe order of the digits. Find the number.
             (iii) Let the digits in the units's place and ten's place be x and y respectively.
             Therefore, Number = 10y + x
              If the digits age reversed, the new number = 10x + y
              As per conditions : x + y = 9 ....(1)
              and, 9(10y + x) = 2(10x + y)
              ⇒ 90y + 9x = 20x +2y
              ⇒ 20x – 9x + 2y – 90y = 0
              ⇒ 11x – 88 y = 0 ...(2)
              From (1), y = 9 – x
              Putting y = 9 – x in (2), we get
              11x –88(9 – x) = 0
              ⇒ 11x – 88 × 9 + 88x = 0
              ⇒ 99x = 88 × 9
              ⇒ x = 88×999=8
              Putting x = 8 in (1), we get
              8 + y = 9
              ⇒ y = 1
              Hence, the number = 10y + x = 10 × 1 + 8 = 18