How do you find three consecutive numbers whose sum is $-33$ ? \[\]

Hint: We recall the definition of consecutive integers. We assume the first integer as $x,$then the next integer is $x+1$ and then the next consecutive integer is $x+2$. We add them and equate to $-33$. We solve for $x$ and then find $x+1,x+2$.

Complete answer:
We know that consecutive numbers are numbers which follow each other in order, without gaps, from smallest to largest. The consecutive integers are integers that follow each other by a difference of 1.
So let us assume the first consecutive integer as $x$, then the second consecutive integer that follow $x$will be incremented by 1 as $x+1$ and the third consecutive integer will follow $x+1$ and will be incremented by 1 as $x+1+1=x+2$.
We are given the question that the sum of three consecutive integers is $-33$. So we add the consecutive integers in terms of $x$ and equate the sum to $-33$ to have;
\[\begin{align}
  & x+x+1+x+2=-33 \\
 & \Rightarrow x+x+x+1+2=-33 \\
\end{align}\]
We add the variable terms and constant terms in the left hand side of the above equation to have;
\[\Rightarrow 3x+3=-33\]
We take 3 common in the left hand side to have;
\[\Rightarrow 3\left( x+11 \right)=-33\]
We divide both sides of above equation by 3 to have;
\[\begin{align}
  & \Rightarrow x+1=\dfrac{-33}{3} \\
 & \Rightarrow x+1=-11 \\
\end{align}\]
We subtract both sides of above equation by 3 to have;
\[\Rightarrow x=-11-1=-12\]
So the three consecutive integers are
\[\begin{align}
  & x=-12, \\
 & x+1=-12+1=-11 \\
 & x+2=-12+2=-10 \\
\end{align}\]

Note:
We can verify our results by adding the three consecutive integers as $-12+\left( -11 \right)+\left( -10 \right)=-33$. We note that when we add, subtract, multiply the same number or divide the same non-zero number both sides then the equation remains balanced which means the equality holds. If there are variable terms on both sides of the equation we should try to bring variable term at one side and constant term at other side.