Inequalities Simultaneous 05383A

1. **Problem c:** Solve the system of inequalities: $$ (2x + 1)^2 - 9 \leq 0 \quad \text{and} \quad 2x < 6 $$ 2. **Step 1:** Solve the first inequality: $$ (2x + 1)^2 - 9 \leq 0 $$ This can be rewritten as: $$ ((2x + 1) - 3)((2x + 1) + 3) \leq 0 $$ $$ (2x - 2)(2x + 4) \leq 0 $$ 3. **Step 2:** Find the roots: $$ 2x - 2 = 0 \Rightarrow x = 1 $$ $$ 2x + 4 = 0 \Rightarrow x = -2 $$ 4. **Step 3:** Determine the intervals where the product is non-positive: The critical points split the real line into three intervals: $(-\infty, -2)$, $[-2, 1]$, and $(1, \infty)$. Test values in each interval: - For $x = -3$: $(2(-3) - 2)(2(-3) + 4) = (-8)(-2) = 16 > 0$ - For $x = 0$: $(2(0) - 2)(2(0) + 4) = (-2)(4) = -8 \leq 0$ - For $x = 2$: $(2(2) - 2)(2(2) + 4) = (2)(8) = 16 > 0$ So the solution for the first inequality is: $$ -2 \leq x \leq 1 $$ 5. **Step 4:** Solve the second inequality: $$ 2x < 6 \Rightarrow x < 3 $$ 6. **Step 5:** Combine both inequalities (simultaneously means intersection): $$ -2 \leq x \leq 1 \quad \text{and} \quad x < 3 $$ The intersection is: $$ -2 \leq x \leq 1 $$ --- 7. **Problem d:** Solve the system of inequalities: $$ 4x^2 < 3x \quad \text{or} \quad \frac{1 - 4x}{4} > 0 $$ 8. **Step 1:** Solve the first inequality: $$ 4x^2 < 3x $$ Rewrite as: $$ 4x^2 - 3x < 0 $$ Factor out $x$: $$ x(4x - 3) < 0 $$ 9. **Step 2:** Find critical points: $$ x = 0 \quad \text{or} \quad 4x - 3 = 0 \Rightarrow x = \frac{3}{4} $$ 10. **Step 3:** Test intervals: - For $x = -1$: $(-1)(4(-1) - 3) = (-1)(-7) = 7 > 0$ - For $x = \frac{1}{2}$: $(\frac{1}{2})(4(\frac{1}{2}) - 3) = (\frac{1}{2})(2 - 3) = (\frac{1}{2})(-1) = -\frac{1}{2} < 0$ - For $x = 1$: $(1)(4(1) - 3) = (1)(1) = 1 > 0$ So the solution is: $$ 0 < x < \frac{3}{4} $$ 11. **Step 4:** Solve the second inequality: $$ \frac{1 - 4x}{4} > 0 $$ Multiply both sides by 4 (positive, so inequality direction stays): $$ 1 - 4x > 0 $$ $$ -4x > -1 $$ Divide both sides by -4 (negative, so reverse inequality): $$ x < \frac{1}{4} $$ 12. **Step 5:** Combine with "or" means union: $$ (0 < x < \frac{3}{4}) \cup (x < \frac{1}{4}) $$ Since $x < \frac{1}{4}$ is contained in $0 < x < \frac{3}{4}$ partially, the union is: $$ x < \frac{3}{4} $$ **Final answers:** - c) $$ -2 \leq x \leq 1 $$ - d) $$ x < \frac{3}{4} $$