Solve for x
x\in (-\infty,-6]\cup [\frac{1}{3},\infty)
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\frac{2x}{3}+4\geq 0 \frac{1}{3}-x\leq 0
For the product to be ≤0, one of the values \frac{2x}{3}+4 and \frac{1}{3}-x has to be ≥0 and the other has to be ≤0. Consider the case when \frac{2x}{3}+4\geq 0 and \frac{1}{3}-x\leq 0.
x\geq \frac{1}{3}
The solution satisfying both inequalities is x\geq \frac{1}{3}.
\frac{1}{3}-x\geq 0 \frac{2x}{3}+4\leq 0
Consider the case when \frac{2x}{3}+4\leq 0 and \frac{1}{3}-x\geq 0.
x\leq -6
The solution satisfying both inequalities is x\leq -6.
x\geq \frac{1}{3}\text{; }x\leq -6
The final solution is the union of the obtained solutions.
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