Inequality Systems Ac869A

1. **State the problems:** We need to solve the inequalities: $$x^2 - 4x \leq 0$$ and $$x^2 - 16 > 0$$ and find the solution set that satisfies both simultaneously. 2. **Solve the first inequality:** $$x^2 - 4x \leq 0$$ Factor the left side: $$x(x - 4) \leq 0$$ The critical points are $x=0$ and $x=4$. 3. **Determine intervals for the first inequality:** Test intervals: - For $x < 0$, say $x = -1$: $(-1)(-1-4) = (-1)(-5) = 5 > 0$ (False) - For $0 \leq x \leq 4$, say $x=2$: $(2)(2-4) = 2(-2) = -4 \leq 0$ (True) - For $x > 4$, say $x=5$: $(5)(5-4) = 5(1) = 5 > 0$ (False) So the solution for the first inequality is: $$0 \leq x \leq 4$$ 4. **Solve the second inequality:** $$x^2 - 16 > 0$$ Factor as difference of squares: $$ (x - 4)(x + 4) > 0$$ Critical points are $x = -4$ and $x = 4$. 5. **Determine intervals for the second inequality:** Test intervals: - For $x < -4$, say $x = -5$: $(-5 - 4)(-5 + 4) = (-9)(-1) = 9 > 0$ (True) - For $-4 < x < 4$, say $x=0$: $(0 - 4)(0 + 4) = (-4)(4) = -16 < 0$ (False) - For $x > 4$, say $x=5$: $(5 - 4)(5 + 4) = (1)(9) = 9 > 0$ (True) So the solution for the second inequality is: $$x < -4 \quad \text{or} \quad x > 4$$ 6. **Find the intersection of both solutions:** First inequality solution: $$[0,4]$$ Second inequality solution: $$(-\infty, -4) \cup (4, \infty)$$ The intersection is the set of $x$ values that satisfy both inequalities simultaneously. Since $$[0,4]$$ and $$(-\infty, -4) \cup (4, \infty)$$ do not overlap, the intersection is empty. 7. **Conclusion:** There is no $x$ that satisfies both inequalities at the same time. **Final answer:** $$\boxed{\text{No solution}}$$