7. if x-y + z = 19 , y + z =20 , x-z=3 , find d value of x+4y-5z 22/38/17/none

• x-y+z=19---------(1)
  y+z=20
  thus y=20-z---------(2)
  x-z=3
  thus x =3+z---------(3)
  substitute x and y in equation (1)
  you will get,
  (3+z) - (20-z) + z = 19
  3z = 36
  z = 12
  substitute z in (2) and (3)
  thus x = 15
  and y = 8
  substitute value of x,y,z in x+4y-5z
  answer is - 13
  ans- none
• 7 years agoHelpfull: Yes(27) No(3)
• x-y+z=19....(1)
  y+z=20...(2)
  x-z=3...(3)
  by adding all these 3 equations we get,
  x-y+z+y+z+x-z=19+20+3
  =>2x+z=42...... (4)
  by comparing euation (3) & (4) we get, 3x=45. So, x=15
  so z=x-3=12 and y=20-z=8
  given equation, x+4y-5z
  =15+(4*8)-(5*12)
  =47-60
  =-13 (Ans)
  
• 7 years agoHelpfull: Yes(3) No(1)
• x-y=z=19------(1)
  y+z=20---------(2)
  Add both the equations (1) and (2)
  x+2z=39-------(3)
  x-z=3----------(4)
  solve 3 and 4
  you will get x=15,y=8 and z=12
  Put it in the equation
  15+32-60=-13
  So, the ans is none
• 7 years agoHelpfull: Yes(2) No(0)
• x - y + z = 19 ---- (1)
  y + z = 20 ---- (2)
  x - z = 3 ------ (3)
  
  From (3)
  x = z + 3 --- (4)
  
  Place (4) in (1)
  
  z + 3 - y + z = 19
  2z - y = 16 --- (5)
  
  Add (1) and (5) we get
  
  z = 12
  y = 8
  x = 15
  
  Substitute x, y, z values in x + 4y - 5z
  
  15 + 4 * 8 - 5 * 12 = -13
  
  So there is no option -13 the answer is none
• 7 years agoHelpfull: Yes(2) No(0)
• None and
  X=15;y=8;z=15
  -13 answer
• 7 years agoHelpfull: Yes(1) No(0)
• none
  
  x-y+z=19
  x-20=19 => x=39
  
  x-z=3
  39-z=3 => z=36
  
  
• 7 years agoHelpfull: Yes(1) No(1)
• X-Y+Z=19 --------------(1)
  X-Z=3 ----------------(2)
  Y+Z=20 ----------------(3)
  Now put the value of (2) in (1)
  X-20=19
  =>X=39
  
  Put the value of X in (2)
  => 39-Z=3
  => Z=36
  Now,
  Put the value of X and Z in (1)
  =>39-Y+36=19
  =>Y=56
  or
  put the value of Z in (2)
  =>Y+36=20
  => Y= -16
  
  Hence X=39 Y=56 or -16 Z=36
  How can value of Y be different ?????????
  So I think Question is wrong
• 7 years agoHelpfull: Yes(1) No(3)
• from second equation y=20-z
  from 3rd equation x=3+z
  put these value of x and y in 1st equation and solve it we get z=12
  now put this value of z in 2nd and 3rd equation and solve for y and x we get y=8 and x=15
  now put these three values of x,y,z in last equation then we get
  x+4y-5z=-13
• 7 years agoHelpfull: Yes(1) No(1)
• x-y+z=19, y+z=20, x-z=3
  3 equations 3 unknowns Thus we get x=15 y=8, z=12
  Hence x+4y-5z = 15+32-60= -13
• 7 years agoHelpfull: Yes(1) No(1)
• x/6=4
  at x=10
  10/6 give remainder 4
  y/6=5
  11/6= 5
  (10+11)/6=3
• 7 years agoHelpfull: Yes(0) No(3)
• x=15 and y=8 and z=12
  so x+4y-5z=>15+32-60=-13
• 7 years agoHelpfull: Yes(0) No(0)
• Answer is -13 (None)
  On computing 1st & 3rd,
  2x-y=22
  y=2x-22
  Putting this in 2nd,
  2x-22+z=20 => 2x+z=42
  x-z=3
  3x=45
  x=15,z=12,y=8
  15+4(8)-5(12)= -13.
• 5 years agoHelpfull: Yes(0) No(0)