Let's use #n# to represent an integer (whole number). Since we need three integers, let's define them like this:
#color(blue)(n)=#1st integer
#color(red)(n+1)=#2nd integer
#color(green)(n+2)=#3rd integer
We know we can define the second and third integers as #n+1# and #n+2# due to the problem telling us that the integers are consecutive (in order)
Now we can make our equation since we know what it's going to equal:
#color(blue)(n)+color(red)(n+1)+color(green)(n+2)=48#
Now that we've set up the equation, we can solve by combining like terms:
#3n+3=48#
#3n=45# #color(blue)(" ""Subtract " 3 " from both sides")#
#n=15# #color(blue)(" "45/3=15)#
Now that we know what #n# is, we can plug it back into our original definitions:
#color(blue)(n)=15# #color(blue)(" 1st Integer")#
#color(red)(15+1)=16# #color(red)(" 2nd Integer")#
#color(green)(15+2)=17# #color(green)(" 3rd Integer")#
#color(blue)(15)+color(red)(16)+color(green)(17)=48# #" True"#
