Question

graph the solution of $\frac{1}{3}sqrt{x + 5}<1$.

Explanation:

Step1: Isolate the square - root term

Multiply both sides of the inequality $\frac{1}{3}\sqrt{x + 5}<1$ by 3 to get $\sqrt{x + 5}<3$.

Step2: Consider the domain of the square - root

For $\sqrt{x + 5}$ to be a real number, we must have $x+5\geq0$, which gives $x\geq - 5$.

Step3: Solve the inequality

Square both sides of $\sqrt{x + 5}<3$ (since both sides are non - negative). We get $x + 5<9$. Subtract 5 from both sides: $x<4$.
Combining with the domain condition $x\geq - 5$, the solution of the inequality is $-5\leq x<4$.
On the number line, we have a closed circle at $x = - 5$ (because $x$ can equal $-5$) and an open circle at $x = 4$ (because $x$ cannot equal 4), and the line segment between them.

Answer:

We use the number - line representation with a closed circle at $x=-5$ and an open circle at $x = 4$, and the line segment connecting them.