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+y≥25x + y \geq 25
  • "No more than $200" → 4x+10y≤2004x + 10y \leq 200

The system is: {x+y≥254x+10y≤200\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≥250 + 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≤2000 + 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) does not lie in the solution region, so displaying 10 small and 20 large photos would not meet both constraints.

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

Image 0

0

1

Updated 2026-04-21

Contributors are:

Who are from:

Tags

OpenStax

Elementary Algebra @ OpenStax

Ch.5 Systems of Linear Equations - Elementary Algebra @ OpenStax

Algebra

Math

Prealgebra

Related
Learn After