# How do you graph the inequality 5/(x+3) ≥ 3/x?

The final graph will be the shape of $\frac{5}{x + 3}$ with -3 <=x <=0; x>=4.5

#### Explanation:

In order to graph the whole relationship, we need to graph both the left and right individually.

Here's the left side $\left(\frac{5}{x + 3}\right)$:

graph{5/(x+3) [-20, 20, -20, 20]}

And the right $\left(\frac{3}{x}\right)$:

graph{3/x [-20, 20, -20, 20]}

So now the question is, for what values of $x$ is $\frac{5}{x + 3} \ge \frac{3}{x}$. We'll find that wherever the result of $\frac{5}{x + 3}$, in essence the $y$ value, is larger than that of $\frac{3}{x}$:

graph{(y-(5/(x+3)))(y-3/x)=0 [-20, 20, -20, 20]}

By observation, we can see that, approaching the graph from the left to the right, that the first value of $x$ where this holds is at $x = - 3$ and continues until $x = 0$.

At $x = 4.5$, it holds again and does so until $+ \infty$:

graph{(y-(5/(x+3)))(y-3/x)=0 [4, 6, 0, 1]}

And so the final graph will be the shape of $\frac{5}{x + 3}$ with -3 <=x <=0; x>=4.5