Solve the system of equations: \(2x + 3y = 7\) and \(4x - y = 5\).

["# Solve the System of Equations: (2x + 3y = 7) and (4x - y = 5)", "Solving a system of equations is a fundamental skill in algebra, widely used in science, engineering, economics, and everyday problem-solving. Today, we’ll walk through how to solve the system:", "[\n\begin{align}\n(1) \quad & 2x + 3y = 7 \\n(2) \quad & 4x - y = 5\n\end{align}\n]", "This article includes step-by-step instructions, explanation of methods, and real-world relevance to help you master solving systems of equations.", "---", "## Why Solve Systems of Equations?", "Systems of equations model situations where two or more relationships interact simultaneously—such as pricing and demand, motion paths, or resource allocation. Understanding how to find their solution allows accurate predictions and informed decision-making.", "---", "## Method 1: Solving by Substitution", "The substitution method works well when one equation is easily solved for one variable.", "### Step 1: Solve Equation (2) for (y)", "Start with equation (2):\n[\n4x - y = 5\n]", "Isolate (y):\n[\n- y = 5 - 4x\n]\n[\ny = 4x - 5\n]", "### Step 2: Substitute into Equation (1)", "Replace (y) in equation (1) with (4x - 5):\n[\n2x + 3(4x - 5) = 7\n]", "Expand:\n[\n2x + 12x - 15 = 7\n]", "Combine like terms:\n[\n14x - 15 = 7\n]", "Add 15 to both sides:\n[\n14x = 22\n]", "Divide by 14:\n[\nx = \frac{22}{14} = \frac{11}{7}\n]", "### Step 3: Find (y)", "Now substitute (x = \frac{11}{7}) into (y = 4x - 5):\n[\ny = 4\left(\frac{11}{7}\right) - 5 = \frac{44}{7} - \frac{35}{7} = \frac{9}{7}\n]", "---", "## Method 2: Solving by Elimination", "Another efficient approach is elimination, where you eliminate one variable by aligning and adding/subtracting equations.", "### Rewrite the System:", "[\n\begin{align}\n(1) \quad & 2x + 3y = 7 \\n(2) \quad & 4x - y = 5\n\end{align}\n]", "### Eliminate (x): Multiply Equation (2) by 2", "Multiply equation (2) by 2 to match coefficients of (x):\n[\n2(4x - y) = 2(5) \Rightarrow 8x - 2y = 10\n]", "Now write both equations with same (x) coefficients:\n[\n\begin{align}\n(1) \quad & 2x + 3y = 7 \\n(2') \quad & 8x - 2y = 10\n\end{align}\n]", "To align, subtract (4 \ imes (1)) from ( (2') ) (since (8x = 4 \ imes 2x)), but easier to multiply equation (1) by 4:\n[\n4(2x + 3y) = 4(7) \Rightarrow 8x + 12y = 28\n]", "Now subtract ( (8x - 2y = 10) ) from ( (8x + 12y = 28) ):\n[\n(8x + 12y) - (8x - 2y) = 28 - 10\n]\n[\n8x + 12y - 8x + 2y = 18\n]\n[\n14y = 18 \Rightarrow y = \frac{18}{14} = \frac{9}{7}\n]", "Now plug ( y = \frac{9}{7} ) back into equation (2):\n[\n4x - \frac{9}{7} = 5\n]\n[\n4x = 5 + \frac{9}{7} = \frac{35}{7} + \frac{9}{7} = \frac{44}{7}\n]\n[\nx = \frac{44}{7} \div 4 = \frac{44}{28} = \frac{11}{7}\n]", "Same solution:\n[\nx = \frac{11}{7}, \quad y = \frac{9}{7}\n]", "---", "## Final Answer", "[\n\boxed{x = \frac{11}{7}, \quad y = \frac{9}{7}}\n]", "---", "## Verification", "Plug the values back into both original equations:", "1. (2x + 3y = 2\left(\frac{11}{7}\right) + 3\left(\frac{9}{7}\right) = \frac{22}{7} + \frac{27}{7} = \frac{49}{7} = 7) ✅\n2. (4x - y = 4\left(\frac{11}{7}\right) - \frac{9}{7} = \frac{44}{7} - \frac{9}{7} = \frac{35}{7} = 5) ✅", "---", "## Summary", "Solving systems of equations like (2x + 3y = 7) and (4x - y = 5) reveals the intersection point ((x, y)) where both relationships are simultaneously true. Whether using substitution or elimination, accuracy hinges on careful algebraic manipulation and verification. Mastering these techniques empowers you to solve real-world problems involving multiple constraints.", "---", "Keywords: Solve system of equations, substitution method, elimination method, solve (2x + 3y = 7) and (4x - y = 5), algebra practice, linear equations, coordinate geometry."]