Q. THE NUMBER OF VALUES OF k FOR WHICH THE SYSTEM OF EQUATIONS
x + y = 2
kx + y = 4
x + ky = 5
HAS ATLEAST ONE SOLUTION IS

• one value of k=3.5=7/2
  x=4/5
  y=6/5
• 12 years agoHelpfull: Yes(5) No(2)
• x + y = 2 equation (1)
  kx + y = 4 equation (2)
  x + ky = 5 equation (3)
  
  soln:-substitute the value of x from eqn 1 to eqn 2
  k(2-y)+4 equation (4)
  5-ky+y=2 from eqn 3
  ---------
  2k-5=2
  k=7/2
  
  Nowput the value of k in eqn 3 and substitute the value of x from eqn 1
  (2-y)+7/2y=5
  on solving we get
  y=6/5
  
  Now put the value of y in eqn 1 then in solving we get x=4/5
  So,x=4/5,y=6/5and k=7/2
  
• 11 years agoHelpfull: Yes(2) No(0)
• two values of x are 3.5 and 1
  
• 11 years agoHelpfull: Yes(1) No(0)
• two value of k= 3.5 and 1
  
• 11 years agoHelpfull: Yes(1) No(0)
• @garima.. I think ur solution is wrong
  x=2-y
  k(2-y)+y=4 i.e y=2k-4 when we put this in 3rd equation we get
  2k^2-6k+6=5
  
  which has 2 real values one approx .5 and one approx 3.5
• 12 years agoHelpfull: Yes(0) No(3)
• given
  x+y=2 =>x=2-y
  put the values of x in remaining 2 equations . then we get following equation
  y=(4-2k)/(1-k) . and y=3/(k-1) .
  equating values of y
  3/(k-1)=(4-2k)/(1-k)
  =>2k^2 -9k+7 =o
  =>k=(9+5)/4 , (9-5)/4
  =>k=3.5,1
  =>there are 2 values of k.
  
• 11 years agoHelpfull: Yes(0) No(0)
• given
  x+y=2 =>x=2-y
  put the values of x in remaining 2 equations . then we get following equation
  y=(4-2k)/(1-k) . and y=3/(k-1) .
  equating values of y
  3/(k-1)=(4-2k)/(1-k)
  =>2k^2 -9k+7 =o
  =>k=(9+5)/4 , (9-5)/4
  =>k=3.5,1
  =>there are 2 values of k.
  but on substituting,we get solution only when k=3.5
• 11 years agoHelpfull: Yes(0) No(0)