suppose y=-10 and x=12,
x+y = 2 (Satifies)
2x+3y = 24-30 = -6 (does not satify)

How on earth is it C?

We are given two main conditions involving variables $x$ and $y$ :
1. Sum Constraint: The sum of $x$ and $y$ is at least 2 but no more than 5 .
- Mathematically: $\(2 \leq x+y \leq 5\)$
2. Linear Combination Constraint: The total of twice $x$ and three times $y$ is non-negative and no more than 40.
- Mathematically: $\(0 \leq 2 x+3 y \leq 40\)$

Our goal is to find all possible values of $y$ that satisfy both conditions simultaneously. The answer choices provide ranges for $y$, and we need to determine which range correctly encompasses all possible $y$ values under the given constraints.

Step 1: Express $x$ in Terms of $y$ from the Sum Constraint
From the first condition:

$$
\(2 \leq x+y \leq 5\)
$$


We can express $x$ as:

$$
\(\begin{aligned}
& x \geq 2-y \\
& x \leq 5-y
\end{aligned}\)
$$


This gives us a range for $x$ based on $y$ :

$$
\(2-y \leq x \leq 5-y\)
$$


Step 2: Incorporate the Linear Combination Constraint
The second condition is:

$$
\(0 \leq 2 x+3 y \leq 40\)
$$


We can use the expressions for $x$ from Step 1 to rewrite this inequality in terms of $y$ alone.

Lower Bound: $\(2 x+3 y \geq 0\)$
Substitute the lower bound of $x$ :

$$
\(\begin{gathered}
2(2-y)+3 y \geq 0 \\
4-2 y+3 y \geq 0 \\
4+y \geq 0 \\
y \geq-4
\end{gathered}\)
$$

Upper Bound: $\(2 x+3 y \leq 40\)$
Substitute the upper bound of $x$ :

$$
\(\begin{gathered}
2(5-y)+3 y \leq 40 \\
10-2 y+3 y \leq 40 \\
10+y \leq 40 \\
y \leq 30
\end{gathered}\)
$$


So far, from the linear combination constraint, we have:

$$
\(-4 \leq y \leq 30\)
$$


However, we must also consider the sum constraint's implications on $y$.
Step 3: Analyze the Sum Constraint's Impact on $y$
From the sum constraint $\(2 \leq x+y \leq 5\)$, and knowing $\(x \geq 0\)$ (since $\(2 x+3 y \geq 0\)$ and $y$ can be negative, but $x$ must compensate to keep $\(2 x+3 y \geq 0\)$ ), let's explore the possible range for $y$.

But let's think differently: since $x$ is expressed in terms of $y$, and $x$ must be real and possibly positive, let's find the range where $x$ is valid.

From $\(x \geq 2-y\)$ and $\(x \leq 5-y\)$, for $x$ to have a valid range:

$$
\(2-y \leq 5-y\)
$$


This simplifies to $\(2 \leq 5\)$, which is always true, so the sum constraint doesn't directly limit $y$ beyond the linear combination constraint.

But wait, perhaps the sum constraint imposes additional limits when combined with the linear combination.

Let's consider the worst-case scenarios to find the bounds on $y$.

Finding the Minimum $y$ :
To find the smallest possible $y$, we need $x$ to be as large as possible to satisfy $\(2 x+3 y \geq 0\)$.
From the sum constraint, the maximum $x$ can be is $5-y$.
Substitute into the lower bound of the linear combination:

$$
\(2(5-y)+3 y \geq 0\)
$$

$$
\(\begin{gathered}
10-2 y+3 y \geq 0 \\
10+y \geq 0 \\
y \geq-10
\end{gathered}\)
$$


This is a more restrictive lower bound than before ( $y \geq-4$ ).

Finding the Maximum $y$ :
To find the largest possible $y$, we need $x$ to be as small as possible to satisfy $\(2 x+3 y \leq 40\)$.
From the sum constraint, the minimum $x$ can be is $\(2-y\)$.
Substitute into the upper bound of the linear combination:

$$
\(\begin{gathered}
2(2-y)+3 y \leq 40 \\
4-2 y+3 y \leq 40 \\
4+y \leq 40 \\
y \leq 36
\end{gathered}\)
$$


This is a more restrictive upper bound than before ( $y \leq 30$ ).
Step 4: Verify the Bounds
Now, we have:

$$
\(-10 \leq y \leq 36\)
$$


But we must ensure that within this range, there exists an $x$ such that both original conditions are satisfied.

Let's test the extremes:

Let's test the extremes:
1. When $y=-10$ :
- From $\(x+y \geq 2: x \geq 12\)$
- From $\(x+y \leq 5: x \leq 15\)$
- Choose $\(x=12\)$ :
- $\(2(12)+3(-10)=24-30=-6\)$ which violates $\(2 x+3 y \geq 0\)$.
- Choose $x=15$ :
- $\(2(15)+3(-10)=30-30=0\)$, which satisfies $\(0 \leq 2 x+3 y \leq 40\)$.

So, $y=-10$ is possible with $x=15$.
2. When $\(y=36\)$ :
- From $\(x+y \geq 2: x \geq-34\)$
- From $\(x+y \leq 5: x \leq-31\)$
- Choose $\(x=-31\)$ :
- $\(2(-31)+3(36)=-62+108=46\)$, which violates $\(2 x+3 y \leq 40\)$.
- Choose $\(x=-32\)$ :
- $\(2(-32)+3(36)=-64+108=44\)$, still violates.
- Continue until $\(x=-34\)$ :
- $\(2(-34)+3(36)=-68+108=40\)$, which satisfies the upper bound.

So, $y=36$ is possible with $x=-34$.
Step 5: Check Intermediate Values
Let's check a value in the middle, say $y=0$ :
- $\(x+0 \geq 2 \Rightarrow x \geq 2\)$
- $\(x+0 \leq 5 \Rightarrow x \leq 5\)$
- $\(2 x+3(0) \geq 0 \Rightarrow x \geq 0\)$ (already satisfied)
- $\(2 x \leq 40 \Rightarrow x \leq 20\)$ (already satisfied by $\(x \leq 5\)$ )

Any $x$ between 2 and 5 works, e.g., $x=3$ :
- $\(2(3)+3(0)=6\)$, which is within $\([0,40]\)$.

Step 6: Compare with Answer Choices
The derived range for $y$ is:

$$
\(-10 \leq y \leq 36\)
$$


Looking at the options:
- A. $\(-4 \leq y \leq 36 \rightarrow\)$ Incorrect lower bound.
- B. $\(-10 \leq y \leq 30 \rightarrow\)$ Incorrect upper bound.
- C. $\(-10 \leq y \leq 36 \rightarrow\)$ Matches our derived range.
- D. $\(0 \leq y \leq 40 \rightarrow\)$ Incorrect bounds.
- E. $\(2 \leq y \leq 5 \rightarrow\)$ Too restrictive.

Final Answer

The correct range for $y$ is given by option $\(\mathbf{C}\)$.