MathematicsFind the values of a&b   for which the following pair of linear equations has infinitely many solutions. 2x+y−5=0 (a+b)x+(5a−7b)y−20=0  

Find the values of a&b   for which the following pair of linear equations has infinitely many solutions.


2x+y5=0 (a+b)x+(5a7b)y20=0  


  1. A
    a = 5 and b = -3
  2. B
    a = 5 and b = 3
  3. C
    a = -5 and b = 3
  4. D
    a = -5 and b = -3 

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    Solution:

    Given pair of linear equations 2x+y5=0   and (a+b)x+(5a7b)y20=0   has infinite number of solutions. We need to find the value of a&b  .
    A system of linear equations a 1 x+ b 1 y+ c 1 =0   and a 2 x+ b 2 y+ c 2 =0   will have infinite solutions when a 1 a 2 = b 1 b 2 = c 1 c 2  .
    Now from 2x+y5=0   and (a+b)x+(5a7b)y20=0  , we have,
    a 1 =2, b 1 =1, c 1 =5   and a 2 =a+b, b 2 =5a7b, c 2 =20  
    Now, substituting the values in the condition we have,
    2 a+b = 1 5a7b = 5 20  
    On comparing, we have,
    2 a+b = 5 20 2(20)=5(a+b) 40=5(a+b) a+b= 40 5 a+b=8...(i)  
    And,
    1 5a7b = 5 20  
    20=5(5a7b) 5a7b= 20 5  
    5a7b=4...(ii)  
    Multiply equation (i), by 7, we have,
    7a+7b=56  
    Now adding the above equation to equation (ii),
    7a+7b+5a7b=56+4 12a+0=60 12a=60 a= 60 12 a=5  
    Now, substituting the value of a   in equation (i), we have,
    5+b=8 b=85 b=3  
    Thus, the value of a   is 5 and the value of b   is 3.
    Therefore, option 2 is correct.
     
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