Question

solve the following inequality.
$3x \leq \frac{9x + 3}{3}$
all real numbers
no solutions

Explanation:

Step1: Simplify the right-hand side

$\frac{9x + 3}{3} = 3x + 1$

Step2: Rewrite the original inequality

$3x \leq 3x + 1$

Step3: Isolate the variable term

Subtract $3x$ from both sides: $3x - 3x \leq 3x + 1 - 3x$
$0 \leq 1$

Answer:

All real numbers