Question

What is a viable solution to 25x + 15y 25 ≥ 200

Answer 1

To find a viable solution to the inequality \( 25x + 15y \geq 200 \), we can start by rearranging it to solve for one of the variables, such as \( y \).

### Step 1: Rearranging the Inequality
Let's isolate \( y \):

\[
25x + 15y \geq 200
\]
Subtract \( 25x \) from both sides:

\[
15y \geq 200 - 25x
\]
Now divide every term by \( 15 \):

\[
y \geq \frac{200 - 25x}{15}
\]

### Step 2: Simplifying the Right Side
Next, simplify the right side of the inequality:

\[
y \geq \frac{200}{15} - \frac{25x}{15}
\]
This simplifies to:

\[
y \geq \frac{40}{3} - \frac{5x}{3}
\]

### Step 3: Choosing Values for \( x \)
Now, we can choose a value for \( x \) and find a corresponding value for \( y \) that satisfies the inequality.

#### Example 1: Let \( x = 0 \)
Substituting \( x = 0 \):

\[
y \geq \frac{40}{3} \approx 13.33
\]

We can use a viable solution, for example \( (0, 14) \).

#### Example 2: Let \( x = 3 \)
Substituting \( x = 3 \):

\[
y \geq \frac{200 - 25(3)}{15} = \frac{200 - 75}{15} = \frac{125}{15} \approx 8.33
\]

A viable solution could be \( (3, 9) \).

### Summary of Viable Solutions
Two examples of viable solutions to the inequality \( 25x + 15y \geq 200 \) are:
1. \( (0, 14) \)
2. \( (3, 9) \)

You can find many other solutions by choosing different values of \( x \).