if mod(x-4)+mod(y-4)=4, then how many integer values can the set (x,y) have?

• only 16 integral soln.
  
  |x-4|+|y-4|= 4
  x & y can have values from 0 to 8 only.
  
  for x=0 & x =8 , only 1 soln in each case
  
  for x = 1 to 7 , two solns for each values of x.
  
  total = 2 + 2*7 = 16 soln.
  
  check:
  if x = 0 , y = 4
  if x = 1 , y = 3,5
  if x = 2 , y = 2,6
  if x = 3 , y = 1,7
  if x = 4 , y = 0,8
  if x = 5 , y = 1,7
  if x = 6 , y = 2,6
  if x = 7 , y = 3,5
  if x = 8 , y = 0
  
• 9 years agoHelpfull: Yes(15) No(3)
• |x-4|+|y-4|=4
  x=0,y=4 (1)
  x=1,y=3,5 (2)
  x=2,y=2,6 (2)
  x=3,y=1,7 (2)
  x=4,y=0,8 (2)
  x=5,y=1,7 (2)
  x=6,y=2,6 (2)
  x=7,y=3,5 (2)
  x=8,y=4 (1)
  than total solution should be 16.
  
• 9 years agoHelpfull: Yes(3) No(0)
• Answer will be 14
  If we solve the equation then we will get the equation: x+y=12,Now the possible pairs are (0+12),(1+11),(2+10),(3+9),(4+8),(5+7),(6+6),(6+6),(7+5),(8+4),(9+3),(10+2),(11+1),(12+0),which is equal to 14.......pls crct me if m wrng
• 9 years agoHelpfull: Yes(2) No(0)
• 16 integral solution
  
  
  if x = 0 , y = 4
  if x = 1 , y = 3,5
  if x = 2 , y = 2,6
  if x = 3 , y = 1,7
  if x = 4 , y = 0,8
  if x = 5 , y = 1,7
  if x = 6 , y = 2,6
  if x = 7 , y = 3,5
  
  -->> if x = 8 , y = 4 ......
• 9 years agoHelpfull: Yes(2) No(1)
• here is two diffrent variable...then every value of x,y has a unique value...that means it has infinite set of solution.
• 9 years agoHelpfull: Yes(0) No(3)
• ANSWER==> 16
  there are 16 solutions,
  possible combinations are (4+0),(3+1),(2+2), (0+4)
  now, for 4+0
  X can be 8 or 0 then Y can be 4. So 2 set of values
  similarly for 2+2 there are 4 set of values
  now adding all the sets of values for all 4 possible combination
  we get 16 set of values
• 9 years agoHelpfull: Yes(0) No(4)
• ans-40
  since there will be two equations. one will be in +ve and other one will be in -ve. and since it is an integer it will take both =ve and -ve set of values.
  |x-4|+|y-4|=4 and -|x-4|-|y-4|.
  
• 9 years agoHelpfull: Yes(0) No(0)
• x-4+y-4=4
  x+y=12
  possible pairs of (x,y) are (0,12),(1,11),(2,10),(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3),(10,2),(11,1),(12,0)
  so there are 13 solutions
• 9 years agoHelpfull: Yes(0) No(0)
• ans-38, there will be two equations
• 9 years agoHelpfull: Yes(0) No(0)
• |x-4|+|y-4|=4
  case 1
  if x,y>=4
  (x-4)+(y-4)=4, x+y=12
  At that case possible pair of (x,y) is (4,8),(5,7),(6,6),(7,5),(8,4)
  
  case 2
  if x=4
  -(x-4)+(y-4)=4, y-x=4
  At that case possible pair of (x,y) is (0,4),(1,5),(2,6),(3,)
  
  case 3
  if x>=4,y
• 9 years agoHelpfull: Yes(0) No(0)