Let the number of bales in the second stable be represented by x. The number of bales in the first stable is equal to x - 90, and the number of bales in the third stable is x + 150. The three stables together equal 870 bales. This gives the following equation:
(x) + (x - 90) + (x + 150) = 870
which can be rewritten to combine terms as
3x - 90 + 150 = 870
or
3x + 60 = 870.
Subtracting 60 from both sides gives us:
3x = 810.
Divide both sides by 3 to solve for x:
x = 270.
Thus, the second stable will have 270 bales of hay in it. The first stable will have that number minus 90, or 180 bales of hay in it. The third stable will have 270 plus 150, or 420 bales of hay in it.
