Question 16205
find 3 consecutive integers that are equal to 99
Let x = the first integer;
So, x+1=second integer
x+2=third integer
So we have;
x+(x+1)+(x+2)=99
3x+3=99
3x=96
x=32
So our first integer is 32, can you figure out what the next two are?
Hope you get it
=)