y = 2\left(-\frac{1}{5}\right) + 3 = -\frac{2}{5} + 3 = \frac{13}{5}

["# Simplifying the Linear Expression: How to Solve y = 2(−1/5) + 3 = 13⁄5", "Understanding how to simplify algebraic expressions is a fundamental skill in mathematics. One common operation involves evaluating expressions with fractions, especially when dealing with real-world problems like financial calculations, measurements, or physics. In this article, we’ll break down the expression:", "y = 2\left(−\frac{1}{5}\right) + 3 = –\frac{2}{5} + 3 = \frac{13}{5}", "---", "## Breaking Down the Expression Step by Step", "### Step 1: Multiply By a Fraction\nThe expression starts with:\n2 × (−1/5)\nThis means we multiply 2 by −1/5. When multiplying a whole number by a fraction, you multiply the numerator by the whole number:\n2 × (−1/5) = −(2 × 1)/5 = −2/5", "---", "### Step 2: Add a Whole Number\nThe expression continues with:\n–2/5 + 3\nTo add a fraction and a whole number, it’s easier to convert the whole number to a fraction with the same denominator. Since the denominator is 5, we rewrite 3 as 3/1 and find a common denominator:", "–2/5 + 3 = –2/5 + (3 × 5)/(1 × 5) = –2/5 + 15/5", "Now add the numerators:\n–2/5 + 15/5 = (15 – 2)/5 = 13/5", "---", "## Why Simplify This Expression?\nSimplifying such expressions helps clarify numerical results, making them easier to interpret. When converted to 13/5, we see this equals 2.6 in decimal form, which may be more intuitive in real-world applications like budgeting, calibration, or data analysis.", "---", "## Final Answer\ny = 2(–1/5) + 3 simplifies step by step to:\ny = –2/5 + 3 = 13/5 = 2.6", "Memorizing these steps helps build confidence in solving linear equations and improves general mathematical fluency. Whether you're a student, teacher, or self-learner, understanding how to compute expressions like this strengthens your foundation in algebra.", "---", "### Summary\n- Multiply fractions: 2 × (−1/5) = –2/5\n- Convert 3 to fraction: 3 = 15/5\n- Add: –2/5 + 15/5 = 13/5\n- Final answer: 13/5 or 2.6", "Mastering such algebraic simplifications is essential for advanced math topics and everyday problem solving!"]