Linear inequalities for Multiple Lines Questions in English

(D) Given inequality: $\frac{14x}{x+1} - \frac{9x-30}{x-4} < 0$
Simplify the expression by taking a common denominator:
$\frac{14x(x-4) - (9x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - (9x^2 + 9x - 30x - 30)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - (9x^2 - 21x - 30)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - 9x^2 + 21x + 30}{(x+1)(x-4)} < 0$
$\frac{5x^2 - 35x + 30}{(x+1)(x-4)} < 0$
Divide by $5$:
$\frac{x^2 - 7x + 6}{(x+1)(x-4)} < 0$
Factor the numerator:
$\frac{(x-1)(x-6)}{(x+1)(x-4)} < 0$
Using the wavy curve method (sign scheme) with critical points $-1, 1, 4, 6$:
The expression is negative in the intervals $(-1, 1)$ and $(4, 6)$.
Therefore, $x \in (-1, 1) \cup (4, 6)$.

(C) $1$. Identify the boundary lines of the shaded region:
- Line passing through $(1, 0)$ and $(0, 2)$: The equation is $\frac{x}{1} + \frac{y}{2} = 1$, which simplifies to $2x + y = 2$. Since the shaded region is above this line, the inequality is $2x + y \geq 2$.
- Line passing through $(1, 0)$ and $(0, -1)$ (implied by the slope): The line passes through $(1, 0)$ and $(0, -1)$, so the equation is $y - 0 = \frac{-1 - 0}{0 - 1}(x - 1)$, which simplifies to $y = x - 1$ or $x - y = 1$. Since the shaded region is to the left of this line, the inequality is $x - y \leq 1$.
- Line passing through $(0, 4)$ and $(4, 0)$ (implied by the intersection $B(10/3, 7/3)$): The line passing through $(0, 4)$ and $(4, 0)$ has the equation $\frac{x}{4} + \frac{y}{4} = 1$, which is $x + y = 4$. However, checking the intersection $B(10/3, 7/3)$ with $x + 2y = 8$, we see $10/3 + 2(7/3) = 10/3 + 14/3 = 24/3 = 8$. Thus, the line is $x + 2y = 8$. Since the shaded region is below this line, the inequality is $x + 2y \leq 8$.
$2$. Combining these with $x \geq 0, y \geq 0$, we get the system: $2x + y \geq 2, x - y \leq 1, x + 2y \leq 8, x \geq 0, y \geq 0$.
$3$. This matches option $C$.

(B) Step $1$: Identify the boundary lines.
Line $1$: $y - x = 2$ (passes through $(0, 2)$ and $(-2, 0)$).
Line $2$: $x + y = 5$ (passes through $(0, 5)$ and $(5, 0)$).
Line $3$: $x = 0$ ($y$-axis).
Line $4$: $y = 0$ ($x$-axis).
Step $2$: Determine the feasible region.
The region $y - x \geq 2$ lies above the line $y = x + 2$.
The region $x + y \leq 5$ lies below the line $x + y = 5$.
The region $x \geq 0$ and $y \geq 0$ is the first quadrant.
Step $3$: Find the intersection points.
Intersection of $y - x = 2$ and $x + y = 5$: Adding the equations gives $2y = 7$, so $y = 3.5$. Then $x = 1.5$. Point is $(1.5, 3.5)$.
Intersection of $y - x = 2$ and $x = 0$ is $(0, 2)$.
Intersection of $x + y = 5$ and $x = 0$ is $(0, 5)$.
Step $4$: The region is a triangle with vertices $(0, 2)$, $(0, 5)$, and $(1.5, 3.5)$. Since the region is enclosed by these lines, it is bounded.

(C) $1$. Identify the boundary lines of the shaded region:
- Line passing through $(0, 2)$ and $(1, 0)$: The equation is $\frac{x}{1} + \frac{y}{2} = 1 \implies 2x + y = 2$. Since the shaded region is above the line, the inequality is $2x + y \geq 2$.
- Line passing through $(1, 0)$ and $(10/3, 7/3)$: The slope $m = \frac{7/3 - 0}{10/3 - 1} = \frac{7/3}{7/3} = 1$. The equation is $y - 0 = 1(x - 1) \implies x - y = 1$. Since the shaded region is to the left of the line, the inequality is $x - y \leq 1$.
- Line passing through $(0, 4)$ and $(10/3, 7/3)$: The slope $m = \frac{7/3 - 4}{10/3 - 0} = \frac{-5/3}{10/3} = -1/2$. The equation is $y - 4 = -1/2(x - 0) \implies 2y - 8 = -x \implies x + 2y = 8$. Since the shaded region is below the line, the inequality is $x + 2y \leq 8$.
$2$. The non-negativity constraints are $x \geq 0, y \geq 0$.
$3$. Combining these, the system is $2x + y \geq 2, x - y \leq 1, x + 2y \leq 8, x \geq 0, y \geq 0$. This matches option $C$.