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

x+2y=100
2y=100x Divide with 2
y=(100x)/2
y=50x/2
x=2 y=502/2=501=49
x+2y=2+2*49=2+98=100
x=4 y=504/2=502=48
x+2y=4+2*48=4+96=100
x=6 y=506/2=503=47
x+2y=6+2*47=6+94=100
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x=44 y=5044/2=5022=28
x+2y=44+2*28=44+56=100
x=46 y=5046/2=5023=27
x+2y=46+2*27=46+54=100
x=48 y=5048/2=5024=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 
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
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x=6, y = 47
x=4, y = 48
x=2, y = 49
with this it comes to 49 pairs....is this correct? 
50=n/2+26