7(-\frac{5}{3}) + 3(15) + c = -4 \Rightarrow -\frac{35}{3} + 45 + c = -4 \Rightarrow c = -4 + \frac{35}{3} - 45 = -\frac{112}{3}

["Title: Step-by-Step Solution to Solve 7(–\frac{5}{3}) + 3(15) + c = –4", "Solving linear equations step-by-step is a foundational math skill with wide applications in academic and real-world problem solving. Here’s a clear breakdown of how to solve the equation:\n7(–\frac{5}{3}) + 3(15) + c = –4", "### Step 1: Simplify each term\nBegin by simplifying the multiplication terms:\n- First term: ( 7 \ imes \left(–\frac{5}{3}\right) = -\frac{35}{3} )\n- Second term: ( 3 \ imes 15 = 45 ), so the equation becomes:\n [\n -\frac{35}{3} + 45 + c = –4\n ]", "### Step 2: Combine constant terms\nNow combine the constant values (45) and keep (c) on the left side:\n[\n-\frac{35}{3} + 45 + c = –4\n]\nConvert 45 to a fraction with denominator 3:\n[\n45 = \frac{135}{3}\n]\nSo:\n[\n-\frac{35}{3} + \frac{135}{3} + c = –4 \Rightarrow \frac{100}{3} + c = –4\n]", "### Step 3: Isolate variable (c)\nSubtract (\frac{100}{3}) from both sides:\n[\nc = –4 – \frac{100}{3}\n]\nConvert (-4) to a fraction with denominator 3:\n[\n-4 = -\frac{12}{3}\n]\nNow subtract:\n[\nc = -\frac{12}{3} – \frac{100}{3} = -\frac{112}{3}\n]", "### Final Answer\n[\nc = -\frac{112}{3}\n]", "---", "Summary:\nTo solve equation ( 7(–\frac{5}{3}) + 3(15) + c = –4 ), simplify each term, combine constants, and isolate (c). The calculated value is (c = -\frac{112}{3}), demonstrating clear algebraic manipulation and consistent arithmetic. This method helps build confidence in solving equations critical for STEM education and daily math challenges.", "---", "Keywords: solve linear equations, step-by-step algebra, linear equation solution, negative fractions in algebra, solve for c, math problem solving, fractional coefficients, algebraic simplification", "Meta Description: Learn how to solve (7(–\frac{5}{3}) + 3(15) + c = –4) step-by-step with full explanation, intermediate simplifications, and final answer (c = -\frac{112}{3}). Ideal for students mastering algebra."]