Example

Solving a Photo Display Budget Problem Using a System of Inequalities

Problem: Christy sells photographs at a street fair booth. She wants at least 25 photos on display. Each small photo costs her $4 and each large photo costs $10, and she cannot spend more than $200 on display photos.

ⓐ Set up the system. Let xx = the number of small photos and yy = the number of large photos. Translating the two constraints:

  • "At least 25 photos" → x+y25x + y \geq 25
  • "No more than $200" → 4x+10y2004x + 10y \leq 200

The system is: {x+y254x+10y200\left\{\begin{array}{l} x + y \geq 25 \\ 4x + 10y \leq 200 \end{array}\right.

ⓑ Graph the system. Graph x+y=25x + y = 25 as a solid boundary line. Testing (0,0)(0, 0): 0+0=0250 + 0 = 0 \geq 25 is false, so shade the side away from the origin. Graph 4x+10y=2004x + 10y = 200 as a solid boundary line. Testing (0,0)(0, 0): 0+0=02000 + 0 = 0 \leq 200 is true, so shade the side containing the origin. The solution is the doubly-shaded overlap region.

The point (10,20)(10, 20) does not lie in the solution region, so displaying 10 small and 20 large photos would not meet both constraints.

The point (20,10)(20, 10) does lie in the solution region, so displaying 20 small and 10 large photos satisfies both constraints.

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Updated 2026-04-21

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