Example

Solving a Recipe Mixture Application Using a System of Equations

Systems of linear equations can determine the exact quantities of different ingredients required in a recipe to satisfy both a weight requirement and a budget constraint. For instance, if a large batch of chili requires a combined total of 2020 pounds of beans and ground beef, with an average target cost of $3\$3 per pound, this scenario can be modeled algebraically. If beans cost $1\$1 per pound and ground beef costs $5\$5 per pound, one can assign variables such as bb for the pounds of beans and gg for the pounds of ground beef. The total weight constraint translates to the equation b+g=20b + g = 20. The cost constraint is expressed by equating the individual costs to the total mixture value: 1b+5g=20(3)1b + 5g = 20(3), which simplifies to b+5g=60b + 5g = 60. To solve this system by elimination, the weight equation can be multiplied by 1-1 (resulting in bg=20-b - g = -20) and added to the cost equation. This eliminates bb and yields 4g=404g = 40, meaning g=10g = 10. Substituting 1010 back into the first equation (b+10=20b + 10 = 20) reveals that b=10b = 10. Thus, exactly 1010 pounds of beans and 1010 pounds of ground beef must be purchased to perfectly meet the recipe and budget requirements.

0

1

Updated 2026-05-25

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.4 Systems of Linear Equations - Intermediate Algebra @ OpenStax

Algebra

Related
Learn After