How many ordered pairs (x, y) of counting numbers (x = 1, 2, 3, · · · , and y = 1, 2, 3, · · ·) satisfy the equation x + 2y = 100?

Answer 1

x+2y=100

2y=100-x Divide with 2

y=(100-x)/2

y=50-x/2

x=2 y=50-2/2=50-1=49

x+2y=2+2*49=2+98=100

x=4 y=50-4/2=50-2=48

x+2y=4+2*48=4+96=100

x=6 y=50-6/2=50-3=47

x+2y=6+2*47=6+94=100

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x=44 y=50-44/2=50-22=28

x+2y=44+2*28=44+56=100

x=46 y=50-46/2=50-23=27

x+2y=46+2*27=46+54=100

x=48 y=50-48/2=50-24=26

x+2y=48+2*26=48+52=100

x=(2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48)

y=(49,48,47,46,45,44,43,42,41,40,39,38,37,36,35,34,33,32,31,30,29,28,27,26)

24 pairs

Answer 2

thank you for the response;

why is x is limited to 48 , and y to 26;

why can't we take x upto 98?

x=98, y = 1
x=96, y = 2
x=94, y = 3
..
..
..
..
x=6, y = 47
x=4, y = 48
x=2, y = 49

with this it comes to 49 pairs....is this correct?

Answer 3

50=n/2+26