Question:** A space habitat sustainable agriculture technology developer is optimizing nutrient solutions. If the cost of nutrient X is $p$ dollars per liter and nutrient Y is $q$ dollars per liter, and the developer finds that $7p + 5q = 35$ and $3p - 2q = 8$, determine the cost per liter of nutrients X and Y.

["Optimizing Nutrient Solutions: Solving for the Cost of Nutrients X and Y", "In the rapidly evolving field of sustainable space agriculture, precise nutrient management is critical for growing crops in extraterrestrial habitats. Developing cost-effective and efficient nutrient solutions not only ensures healthier plant production but also supports long-term mission sustainability. One essential step in this process involves determining the exact cost per liter of key nutrients. Recent research highlights a mathematical breakthrough that combines two economic constraints to reveal precise pricing: $7p + 5q = 35$ and $3p - 2q = 8$, where $p$ is the cost of nutrient X per liter and $q$ is the cost of nutrient Y per liter.", "Solving this system of equations enables nutrient developers and aerospace engineers to optimize budgets while maintaining high-quality growth conditions in space habitats. Let’s explore how these equations are solved and what they reveal about sustainable nutrient sourcing.", "### The Equations: Background and Context", "We are given:", "1. $7p + 5q = 35$\n2. $3p - 2q = 8$", "These linear equations represent real-world constraints—budget allocations, supply availability, and cost-efficiency considerations—critical for designing closed-loop life support systems aboard spacecraft and space stations.", "### Step 1: Solving the System of Equations", "We use the elimination method to solve for $p$ and $q$.", "Equation A:\n$7p + 5q = 35$", "Equation B:\n$3p - 2q = 8$", "#### Step 1: Eliminate q", "Multiply Equation A by 2 and Equation B by 5 to align coefficients for $q$:", "- $2(7p + 5q) = 2(35)$ → $14p + 10q = 70$\n- $5(3p - 2q) = 5(8)$ → $15p - 10q = 40$", "Now add the two equations:", "$$\n(14p + 10q) + (15p - 10q) = 70 + 40\n$$\n$$\n29p = 110\n$$\n$$\np = \frac{110}{29} \approx 3.79 \ ext{ dollars per liter}\n$$", "#### Step 2: Substitute $p$ into one equation to find $q$", "Use Equation B: $3p - 2q = 8$", "$$\n3\left(\frac{110}{29}\right) - 2q = 8\n$$\n$$\n\frac{330}{29} - 2q = 8\n$$\n$$\n2q = \frac{330}{29} - 8 = \frac{330 - 232}{29} = \frac{98}{29}\n$$\n$$\nq = \frac{49}{29} \approx 1.69 \ ext{ dollars per liter}\n$$", "### Step 2: Interpreting the Results", "The solution reveals that:", "- The cost of nutrient X is approximately $3.79 per liter\n- The cost of nutrient Y is approximately $1.69 per liter", "These values optimize affordability while maintaining balance in essential nutrient mixtures, vital for supporting photosynthesis, root development, and overall plant health in confined space environments.", "### Real-World Implications in Space Agriculture", "Nutrient solutions in space must be precisely formulated due to limited storage and transportation capacity. By solving such equations, developers not only determine costs but also forecast resource needs in orbital farms. This mathematical precision supports sustainable agriculture technologies, reducing waste and maximizing output in environments where every gram counts.", "In spacecraft and space habitats, optimizing nutrient solutions using systems like $7p + 5q = 35$ and $3p - 2q = 8$ enables better financial planning and ensures reliable food production—key pillars in humanity’s journey toward long-duration space exploration.", "---", "Conclusion", "Accurate cost modeling through linear systems is an essential tool in space agriculture innovation. The specific values $p = \frac{110}{29}$ dollars and $q = \frac{49}{29}$ dollars per liter of nutrients X and Y exemplify how data-driven decision-making enhances sustainable living beyond Earth. As space missions grow longer and more complex, optimizing such nutritional inputs will remain fundamental to thriving extraterrestrial communities."]