Question

1. $x+2y < -6$

$-4x-2y < -10$

Explanation:

Step1: Isolate y in first inequality

Start with $x + 2y < -6$. Subtract $x$ from both sides, then divide by 2.
$2y < -x -6$
$y < -\frac{1}{2}x - 3$

Step2: Isolate y in second inequality

Start with $-4x - 2y < -10$. Add $4x$ to both sides, divide by $-2$ (reverse inequality sign).
$-2y < 4x -10$
$y > -2x + 5$

Step3: Identify boundary lines

First boundary: $y = -\frac{1}{2}x - 3$ (dashed, shade below).
Second boundary: $y = -2x + 5$ (dashed, shade above).

Step4: Find intersection of boundaries

Set $-\frac{1}{2}x - 3 = -2x + 5$. Solve for $x$:
$-\frac{1}{2}x + 2x = 5 + 3$
$\frac{3}{2}x = 8$
$x = \frac{16}{3} \approx 5.33$
Substitute $x$ to find $y$:
$y = -2(\frac{16}{3}) + 5 = -\frac{32}{3} + \frac{15}{3} = -\frac{17}{3} \approx -5.67$

Answer:

The solution is the region that satisfies both $y < -\frac{1}{2}x - 3$ and $y > -2x + 5$, bounded by the dashed lines $y = -\frac{1}{2}x - 3$ (shade below) and $y = -2x + 5$ (shade above), with their intersection at $(\frac{16}{3}, -\frac{17}{3})$.