Let's call the first integer #i#. Because they are consecutive integers by definition we can add #1# to get the next integer and #2# to get the third integer. Therefore are three consecutive integers are: #i#, #i + 1# and #i + 2#.
Because the three integers sun to or add up to #-33# we can write:
#i + i + 1 + i + 2 = -33#
We can now solve for #i# as follows:
#3i + 3 = -33#
#3i + 3 - color(red)(3) = -33 - color(red)(3)#
#3i + 0 = -36#
#3i = -36#
#(3i)/color(red)(3) = -36/color(red)(3)#
#(color(red)(cancel(color(black)(3)))i)/cancel(color(red)(3)) = -12#
#i = -12#
The first integer is #-12#
The second integer is #i + 1# or #-12 + 1 = -11#
The third integer is #i + 2# or #-12 + 2 = -10#
The three consecutive integers whose sum is -33 are -12, -11 and -10.
