Chapter 5 - Linear Inequalities

1. Lines: Draw $4x + 3y = 60$ (passes through (15,0), (0,20)), $y = 2x$ (passes through (0,0), (5,10)), and $x = 3$.
2. Shading:
- $4x + 3y \le 60$: Region below the line.
- $y \ge 2x$: Region above/left of the line.
- $x \ge 3$: Region to the right of $x=3$.
3. Intersection Points:
- $y=2x$ and $x=3$: Point A(3, 6).
- $4x+3y=60$ and $y=2x$: $4x+6x=60 \implies 10x=60 \implies x=6, y=12$. Point B(6, 12).
- $4x+3y=60$ and $x=3$: $12+3y=60 \implies 3y=48 \implies y=16$. Point C(3, 16).
The solution is the triangular region ABC.

1. Lines:
- $3x + 2y = 150$: Intercepts (50, 0), (0, 75).
- $x + 4y = 80$: Intercepts (80, 0), (0, 20).
- $x = 15$: Vertical line.
2. Shading: Below both lines and left of $x=15$.
3. Intersection Points:
- $x=15$ and $x+4y=80$: $15+4y=80 \implies 4y=65 \implies y=16.25$. Point (15, 16.25).
- $x+4y=80$ and y-axis: (0, 20).
- $x=15$ and x-axis: (15, 0).
- Origin: (0, 0).
Note: Intersection of $3x+2y=150$ and $x+4y=80$ is at $x=44$, which is $>15$, so it's outside the feasible region.

1. Lines:
- $x + 2y = 10$: (10,0), (0,5).
- $x + y = 1$: (1,0), (0,1).
- $x = y$: Line through origin.
2. Shading:
- $x + 2y \le 10$: Below.
- $x + y \ge 1$: Above.
- $x \le y$: Above/Left of $y=x$.
3. Intersection Points:
- $x=y$ and $x+2y=10$: $3x=10 \implies x=10/3$. Point (3.33, 3.33).
- $x=y$ and $x+y=1$: $2x=1 \implies x=0.5$. Point (0.5, 0.5).
- Y-axis intersections: (0, 1) and (0, 5).

1. Lines:
- $2x + y = 4$: (2,0), (0,4).
- $x + y = 3$: (3,0), (0,3).
- $2x - 3y = 6$: (3,0), (0,-2).
2. Shading:
- $2x + y \ge 4$: Above/Right.
- $x + y \le 3$: Below/Left.
- $2x - 3y \le 6$: Above/Left.
3. Intersection Points:
- $2x+y=4$ and $x+y=3$: Subtracting gives $x=1, y=2$. Point (1, 2).
- $x+y=3$ and $2x-3y=6$: At y=0, x=3. Point (3, 0).
- $2x+y=4$ and $2x-3y=6$: Subtracting gives $4y=-2 \implies y=-0.5, x=2.25$. Point (2.25, -0.5).

1. Lines:
- $x - 2y = 3$: (3,0), (0,-1.5).
- $3x + 4y = 12$: (4,0), (0,3).
- $y = 1$: Horizontal line.
2. Shading:
- $y \ge 1$: Above $y=1$.
- $3x + 4y \ge 12$: Above/Right.
- $x - 2y \le 3$: Above/Left.
3. Intersection Points:
- $y=1$ and $3x+4y=12$: $3x+4=12 \implies 3x=8 \implies x=2.67$. Point (2.67, 1).
- $y=1$ and $x-2y=3$: $x-2=3 \implies x=5$. Point (5, 1).
The region is unbounded, lying between the two lines above $y=1$.

1. Lines:
- $x + y = 9$: (9,0), (0,9).
- $y = x$: Line through origin.
2. Shading:
- $x + y \le 9$: Below.
- $y > x$: Above/Left of $y=x$ (dotted line).
- $x \ge 0$: Right of Y-axis.
3. Intersection Points:
- $y=x$ and $x+y=9$: $2x=9 \implies x=4.5$. Point (4.5, 4.5).
- Y-axis intercepts: (0, 0) and (0, 9).
The solution is the triangle formed by these points.