The sum of two consecutive multiplies of $3$ is $69$. Find them.

Let the two consecutive multiples of $3$ are $3\left(x\right)$ and $3(x+1)$.

Given sum $=69$

$3x+3(x+1)=69$

$x+(x+1)=23$

$2x+1=23$
$2x=22$

$x=11$.

So the numbers are $33$ and $36$.