Exercise 3.4

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 a1x+b1y+c1=0
              and a2x+b2y+c2=0
              where a1=1,b1=−3,c1=−3 and a2=3,b2=−9,c2=−2
              We have, a1a2=13,b1b2=−3−9=13andc1c2=−3−2=32
              Clearly, a1a2=b1b2≠c1c2
             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 a1x+b1y+c1=0anda2x+b2y+c2=0
            where a1=2,b1=1,c1=−5anda2=3,b2=2andc2=−8
           We have, a1a2=23,b1b2=12
           Clearly, a1a2≠b1b2
           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

          x1×−8−2×(−5)=y−5×3−(−8)×2=12×2−3×1
         ⇒x−8+10=y−15+16=14−3
         ⇒x2=y1=11
         ⇒ 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
         a1x+b1y+c1=0anda2x+b2y+c2=0
         where, a1=3,b1=−5,c1=−20anda2=6,b2=−10,c2=−40
         We have, a1a2=36=12,b1b2=−5−10=12andc1c2=−20−40=12
         Clearly, a1a2=b1b2=c1c2
         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
         a1x+b1y+c1=0anda2x+b2y+c2=0
         where a1=1,b1=−3,c1=−7anda2=3,b2=−3,c2=−15          we have, a1a2=13,b1b2=−3−3=1                Clearly, a1a2≠b1b2          So, the given system of equations has a unique solution.To find the solution, we use cross-multiplication method.            By cross-multiplication, we have 

           x45−21=y−21+15=1−3+9
           x24=y−6=16
           ⇒ x=246=4,y=−66=−1
           Hence, the given system of equations has a unique solution given by
            x = 4, y = – 1

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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           
Sol.       (i) We know that the system of equations              a1x+b1y=c1anda2x+b2y=c2               has infinite number of solutions if              a1a2=b1b2=c1c2               Therefore, The given system of equations will have infinite number of solutions if              2a−b=3a+b=73a+b−2               ⇒ 2a−b=3a+band2a−b=73a+b−2               ⇒ 2a + 2b = 3a – 3b and 6a + 2b – 4 = 7a – 7b              ⇒ a = 5b and a – 9b = – 4              Putting a = 5b in a – 9b = – 4, we get              5b – 9b = – 4              ⇒ – 4b = – 4              ⇒ 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) 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
              (ii) We know that the system of equations
              a1x+b1y=c1anda2x+b2y=c2
               has no solution if a1a2=b1b2≠c1c2
               So, the given system of equations will have no solution if
               32k−1=1k−1≠12k+1
               ⇒ 32k−1=1k−1and1k−1≠12k+1
                Now, 32k−1=1k−1
                ⇒ 3k – 3 = 2k – 1
                ⇒ k = 2
                 Clearly, for k = 2, we have
                 1k−1≠12k+1
                 Hence, the given system of equations will have no solution if k = 2.

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Q.3      Solve the following pair of linear equations by the substitution and cross - multiplication methods :             8x + 5y = 9             3x + 2y = 4Sol.         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                ⇒ 5y = 9 – 8x
                ⇒y=9−8x5                 ⇒ When x = 3, y = 9−245=−155=−3 ; when x = – 2, y = 9+165=255=5                 Thus, we have the following table : 

               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.

              For the graph of 3x + 2y = 4
              We have,
              3x + 2y = 4
              ⇒ 2y = 4 – 3x
              ⇒ y = 4−3x2
              When x = 0, y = 4−02=42=2 ;
              When x = 2, y = 4−62=−22=−1
             Thus, we have the following table :

              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 = 9−8x5 in (2), we get
             3x+2(9−8x5)=4
             ⇒ 15x + 18 – 16x = 20
             ⇒ – x = 2
             ⇒ x = – 2
             Putting x = – 2 in (1), we get
             8(–2) +5y = 9
             ⇒ 5y = 9 + 16 = 25
             ⇒ y=255=5
             Hence, x = – 2, y = 5 is the solution of the given system of equations.
             By cross - multiplication method :

              x−20+18=y−27+32=116−15
              ⇒ x−2=y5=11
              ⇒ x=−21=−2andy=51=5
              Hence, x = – 2, y = 5 is the solution of the given system of equations.
              The method of cross - multiplication is more efficient.