1. **State the problem:** We need to graph and find the feasibility region for the system of inequalities: 1) $5 + 3x \leq 2 + 4x$ 2) $x - 4 \leq 4x - 1$ 3) $2 + 3(4x - 2) \geq 20$ 2. **Rewrite each inequality in simpler form:** For (1): $$5 + 3x \leq 2 + 4x$$ Subtract $3x$ and $2$ from both sides: $$5 - 2 \leq 4x - 3x$$ $$3 \leq x$$ So, $$x \geq 3$$ For (2): $$x - 4 \leq 4x - 1$$ Subtract $x$ and add $4$ to both sides: $$-4 + 4 \leq 4x - x - 1 + 4$$ $$0 \leq 3x + 3$$ Subtract $3$: $$-3 \leq 3x$$ Divide by $3$: $$-1 \leq x$$ So, $$x \geq -1$$ For (3): $$2 + 3(4x - 2) \geq 20$$ Expand: $$2 + 12x - 6 \geq 20$$ Simplify: $$12x - 4 \geq 20$$ Add $4$: $$12x \geq 24$$ Divide by $12$: $$x \geq 2$$ 3. **Combine all inequalities:** $$x \geq 3, \quad x \geq -1, \quad x \geq 2$$ The strongest condition is $$x \geq 3$$. 4. **Interpretation:** The feasibility region is all $x$ values greater than or equal to $3$. Since these inequalities only involve $x$, the region is a half-line on the number line starting at $x=3$ and extending to positive infinity. 5. **Graph description:** - Draw the vertical line $x=3$. - Shade the region to the right of this line (including the line). 6. **Final answer:** The feasibility region is $$\boxed{x \geq 3}$$.