Question

1. graph:

$x + y \geq 2$
$4x + y \geq - 1$

Explanation:

Step1: Rewrite in slope-intercept form

For $x + y \geq 2$:
$y \geq -x + 2$
For $4x + y \geq -1$:
$y \geq -4x - 1$

Step2: Find intercepts for first line

$x + y = 2$:
x-intercept: set $y=0$, $x=2$ → $(2, 0)$
y-intercept: set $x=0$, $y=2$ → $(0, 2)$
Draw solid line (≥ includes equality).

Step3: Find intercepts for second line

$4x + y = -1$:
x-intercept: set $y=0$, $x=-\frac{1}{4}$ → $(-0.25, 0)$
y-intercept: set $x=0$, $y=-1$ → $(0, -1)$
Draw solid line (≥ includes equality).

Step4: Test points for shading

For $y \geq -x + 2$: test $(0,0)$: $0 \geq 2$? No. Shade above line.
For $y \geq -4x - 1$: test $(0,0)$: $0 \geq -1$? Yes. Shade above line.
Shade overlapping region.

Step5: Find intersection of lines

Solve

$$\begin{cases} y = -x + 2 \\ y = -4x - 1 \end{cases}$$

Set equal: $-x + 2 = -4x - 1$
$3x = -3$ → $x=-1$, $y=3$
Intersection: $(-1, 3)$

Answer:

1. Draw a solid line for $y = -x + 2$ through $(2,0)$ and $(0,2)$, shade the region above this line.
2. Draw a solid line for $y = -4x - 1$ through $(-0.25,0)$ and $(0,-1)$, shade the region above this line.
3. The overlapping shaded region (including the lines) is the solution, with the intersection point of the two lines at $(-1, 3)$.