Let's solve these inequalities step-by-step to find the solutions.
1. Inequality 1: x + 3y ≤ 60
First, we convert this inequality into an equation to find the boundary line:
[tex]\[
x + 3y = 60
\][/tex]
Solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex], we get:
[tex]\[
x = 60 - 3y
\][/tex]
2. Inequality 2: 2x + 4y ≥ 80
Similarly, we convert this inequality into an equation to understand the boundary condition:
[tex]\[
2x + 4y = 80
\][/tex]
Solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex], we get:
[tex]\[
2x = 80 - 4y \implies x = 40 - 2y
\][/tex]
3. Inequality 3: y ≥ 0
This inequality states that [tex]\( y \)[/tex] must be non-negative.
### Combining the Solutions
Given the boundaries:
- From [tex]\( x + 3y = 60 \)[/tex]: [tex]\( x = 60 - 3y \)[/tex]
- From [tex]\( 2x + 4y = 80 \)[/tex]: [tex]\( x = 40 - 2y \)[/tex]
- From non-negativity of [tex]\( y \)[/tex]: [tex]\( y \geq 0 \)[/tex]
We need to find the overlapping solutions of these inequalities.
### Intersection of Boundaries
Let's look at the lines and inequalities:
1. Equation of the line [tex]\( x + 3y = 60 \)[/tex]
- [tex]\( x = 60 - 3y \)[/tex]
2. Equation of the line [tex]\( 2x + 4y = 80 \)[/tex]
- [tex]\( x = 40 - 2y \)[/tex]
3. Non-negativity line [tex]\( y = 0 \)[/tex]
- [tex]\( y \)[/tex] must be zero or positive.
### Verifying the Constraints
To verify combinations:
- For [tex]\( x = 60 - 3y \)[/tex]
- For [tex]\( x = 40 - 2y \)[/tex]
- For [tex]\( y \geq 0 \)[/tex]
### Solution Set
- When [tex]\( y = 0 \)[/tex]:
- From [tex]\( x = 60 - 3y \)[/tex], [tex]\( x = 60 \)[/tex].
- From [tex]\( x = 40 - 2y \)[/tex], [tex]\( x = 40 \)[/tex].
Here the solution clashes because both boundaries don't align at [tex]\( y = 0 \)[/tex].
For both inequalities to hold, let's derive common solutions within feasible points:
- For [tex]\( y \geq 0 \)[/tex] with constraints:
[tex]\[
x + 3y \leq 60 \quad \text{and} \quad 2x + 4y \geq 80
\][/tex]
### Combine and Result
Thus, the solution set satisfies:
- [tex]\((60 - 3y, y)\)[/tex]
- [tex]\((40 - 2y, y)\)[/tex]
- [tex]\((x, 0)\)[/tex]
- [tex]\((y \geq 0) \& (x + 3y \leq 60) \& (2x + 4y \geq 80)\)[/tex]
So the solutions form intersections at these constraints and boundaries.
