### Instruction: Solve
\[\frac{x}{x + 3} \ge 0.\]Enter your answer using interval notation.### Response: To solve this inequality, I need to find the values of x that make the fraction non-negative.
A fraction is non-negative if its numerator and denominator have the same sign (positive or positive zero) or if the numerator is zero.
So I can split the problem into two cases: either x = 0 or x and x + 3 have the same sign.
If x = 0, then the fraction is zero, which is non-negative.
So x = 0 is one possible solution.
If x and x + 3 have the same sign, then I need to avoid the values of x that make x + 3 zero, since that would make the fraction undefined.
So x + 3 = 0 implies x = -3, which is a critical value that I need to exclude from the solution.
To determine the sign of x and x + 3, I can use a sign chart.
I plot the critical values x = 0 and x = -3 on a number line and test the sign of the fraction in each interval.
For x < -3, I can pick any value, say x = -4.
Then the fraction is \[\frac{-4}{-4 + 3} = \frac{-4}{-1} = 4,\] which is positive.
So the fraction is positive for x < -3.
For -3 < x < 0, I can pick any value, say x = -1.
Then the fraction is \[\frac{-1}{-1 + 3} = \frac{-1}{2},\] which is negative.
So the fraction is negative for -3 < x < 0.
For x > 0, I can pick any value, say x = 1.
Then the fraction is \[\frac{1}{1 + 3} = \frac{1}{4},\] which is positive.
So the fraction is positive for x > 0.
Therefore, the values of x that make the fraction non-negative are x = 0 or x < -3 or x > 0.
In interval notation, this is \[(-\infty, -3) \cup \{0\} \cup (0, \infty).\]