Answer
$2c + 3$
Restriction: $c \ne 0$
Work Step by Step
Let's first take a look at the problem to see what we can cancel out from the numerator and denominator:
$\dfrac{6c^2 + 9c}{3c}$
We can factor out $3c$ from the numerator:
$\dfrac{3c(2c + 3)}{3c}$
We can divide the numerator and denominator by $3c$:
$2c + 3$
To find out if there are any restrictions on the variables, we need to find which values of $c$ will cause the denominator to equal zero, which would make the whole expression undefined. Let's set the denominator equal to zero and solve for $c$:
$3c = 0$
Divide each side by $3$:
$c = 0$
Restriction: $c \ne 0$
