Question: Solve the system of inequalities: $$3x \geq 1$$ and $$6x^2 \leq 5x - 1$$. Find the values of $x$ that satisfy both inequalities simultaneously.
1. **State the problem:** We need to find the values of $x$ that satisfy both inequalities $$3x \geq 1$$ and $$6x^2 \leq 5x - 1$$ at the same time.
2. **Solve the first inequality:** $$3x \geq 1$$
Divide both sides by 3:
$$\cancel{3}x \geq \frac{1}{\cancel{3}}$$
which simplifies to:
$$x \geq \frac{1}{3}$$
3. **Solve the second inequality:** $$6x^2 \leq 5x - 1$$
Bring all terms to one side:
$$6x^2 - 5x + 1 \leq 0$$
4. **Factor the quadratic:**
Find factors of $6 \times 1 = 6$ that sum to $-5$: $-2$ and $-3$.
Rewrite:
$$6x^2 - 2x - 3x + 1 \leq 0$$
Group terms:
$$(6x^2 - 2x) - (3x - 1) \leq 0$$
Factor each group:
$$2x(3x - 1) - 1(3x - 1) \leq 0$$
Factor out common term:
$$(3x - 1)(2x - 1) \leq 0$$
5. **Find critical points:**
Set each factor to zero:
$$3x - 1 = 0 \Rightarrow x = \frac{1}{3}$$
$$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$
6. **Determine intervals where inequality holds:**
Test intervals determined by critical points $\frac{1}{3}$ and $\frac{1}{2}$:
- For $x < \frac{1}{3}$, pick $x=0$: $(3(0)-1)(2(0)-1) = (-1)(-1) = 1 > 0$ (does not satisfy)
- For $\frac{1}{3} < x < \frac{1}{2}$, pick $x=0.4$: $(3(0.4)-1)(2(0.4)-1) = (1.2-1)(0.8-1) = (0.2)(-0.2) = -0.04 \leq 0$ (satisfies)
- For $x > \frac{1}{2}$, pick $x=1$: $(3(1)-1)(2(1)-1) = (3-1)(2-1) = (2)(1) = 2 > 0$ (does not satisfy)
7. **Combine both inequalities:**
From step 2: $x \geq \frac{1}{3}$
From step 6: $\frac{1}{3} \leq x \leq \frac{1}{2}$
The solution to the system is the intersection:
$$\boxed{\frac{1}{3} \leq x \leq \frac{1}{2}}$$