let

k = the number of keyboards, k is integer, k>0

m = the number of mice, m is integer, m>0

c = the number of cameras, c is integer, c>0

They have a budget of $1500 to spend on $30 keyboards, $20 mice, and $50 cameras:

30k + 20m + 50c = 1500 divide both sides by 10

3k + 2m + 5c = 150

the number of cursors should be equal to that of keyboards and twice the number of cameras:

m = k + 2c

plug m = k + 2c into 3k + 2m + 5c = 150

3k + 2(k + 2c) + 5c = 150

5k + 9c = 150

k = (1/5)(150 - 9c)

since k needs to be integer we have the three following solutions:

1) c = 5 cameras

k = (1/5)(150 - 9*5)

k = 21 keyboards

m = 21 + 2*5

m = 31 mice

He should buy 5 cameras, 21 keyboards and 31 mice.

2) c = 10 cameras

k = (1/5)(150 - 9*10)

k = 12 keyboards

m = 12 + 2*10

m = 32 mice

He should buy 10 cameras, 12 keyboards and 32 mice.

3) c = 15 cameras

k = (1/5)(150 - 9*15)

k = 3 keyboards

m = 3 + 2*15

m = 33 mice

He should buy 15 cameras, 3 keyboards and 33 mice.