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Question

Difficulty: EasySystems of Linear Equations

A specialty coffee shop prepares two custom coffee bean blends using Arabica and Robusta beans. The first 10-pound blend consists of 4 pounds of Arabica beans and 6 pounds of Robusta beans and costs $52\$52. The second 10-pound blend consists of 7 pounds of Arabica beans and 3 pounds of Robusta beans and costs $61\$61. What is the cost, in dollars, of 1 pound of Arabica coffee beans?

Answer: 7 dollars

Answer

The cost of 1 pound of Arabica coffee beans is 7 dollars.
By setting up the system 4A+6R=524A + 6R = 52 and 7A+3R=617A + 3R = 61, we can simplify the first equation to 2A+3R=262A + 3R = 26. Subtracting this simplified equation from the second equation eliminates RR, giving 5A=355A = 35, which simplifies directly to A=7A = 7.

Step-by-Step Solution

1
Define variables and set up the system of linear equations.
Let AA be the price per pound of Arabica beans and RR be the price per pound of Robusta beans.
Equation 1: 4A+6R=524A + 6R = 52
Equation 2: 7A+3R=617A + 3R = 61
Translating the quantitative relationship given in the problem statement into algebraic equations.
2
Simplify Equation 1 and eliminate variable RR by subtraction.
Dividing Equation 1 by 2 gives 2A+3R=262A + 3R = 26. Subtracting this from Equation 2 yields (7A+3R)−(2A+3R)=61−26(7A + 3R) - (2A + 3R) = 61 - 26, which reduces to 5A=355A = 35.
Matching coefficients of RR allows for straightforward elimination of RR.
3
Solve for variable AA.
A=7A = 7
Dividing 35 by 5 yields the price per pound of Arabica beans.

Key Concept

Solving a system of 2x2 linear equations using substitution or elimination.
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Topics

Systems of Linear Equations

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