To solve this problem, we need to establish a relationship between the number of outfits (let's denote it as \( x \)) and the number of pairs of shoes (let's denote it as \( y \)) that Jake can pack without exceeding the 50-pound weight limit.
Given:
- Each outfit weighs 2 pounds: \( 2x \)
- Each pair of shoes weighs 3 pounds: \( 3y \)
- The total weight must be less than 50 pounds: \( 2x + 3y < 50 \)
This inequality can be rearranged to help visualize it in a graph:
- Rearranging the Inequality: \[ 3y < 50 - 2x \] \[ y < \frac{50 - 2x}{3} \]
This represents a linear equation and can be graphed as follows:
- Finding Intercepts:
- To find the y-intercept (when \( x = 0 \)): \[ y < \frac{50 - 2(0)}{3} = \frac{50}{3} \approx 16.67 \]
- To find the x-intercept (when \( y = 0 \)): \[ 0 < \frac{50 - 2x}{3} \implies 50 - 2x > 0 \implies x < 25 \]
So the line \( y = \frac{50 - 2x}{3} \) will intercept the axes at approximately \( (25, 0) \) and \( (0, 16.67) \).
- Graphing the Region: The solution region is below the line \( y = \frac{50 - 2x}{3} \) and above the coordinate axes \( x \geq 0 \), \( y \geq 0 \).
Now, you'll need to compare this to the provided graphs (J, K, and L) to find the one that accurately depicts this area of viable outfits and shoes within the constraints.
Since I don't have the visuals for graphs J, K, and L, review each one to identify which shows a shaded area below the line \( y = \frac{50 - 2x}{3} \) in the first quadrant (the area where both \( x \) and \( y \) are positive).
If you can provide details about each graph, I can help you identify which one accurately represents Jake's packing limitations based on the baggage policy.
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