\[ 7\left(\frac{2q + 8}{3}\right) + 5q = 35 \]

["Solving the Equation: 7 × ((2q + 8)/3) + 5q = 35 – Step-by-Step Guide", "Understanding how to solve equations is essential for mastering algebra and building strong problem-solving skills. One common challenge many learners face involves equations with fractions and variables in multiple places, such as:", "[\n7\left(\frac{2q + 8}{3}\right) + 5q = 35\n]", "In this article, we’ll walk through solving this equation step-by-step, providing clear explanations and practical tips that will boost your algebra skills. Whether you’re a student or just brushing up on math fundamentals, this guide will help you tackle similar problems with confidence.", "---", "### Why This Equation Matters", "Equation solving forms the foundation for topics in math ranging from basic algebra to advanced calculus. The structure of this problem—combining fractions, parentheses, and linear terms—mirrors real-world scenarios where relationships between variables must be clarified and computed accurately. Mastering such equations helps develop logical reasoning and precision, key skills in STEM fields.", "---", "### Step-by-Step Solution", "Let’s solve:", "[\n7\left(\frac{2q + 8}{3}\right) + 5q = 35\n]", "#### Step 1: Eliminate the Parentheses by Distributing", "Multiply 7 across the term inside the parentheses:", "[\n7 \cdot \frac{2q + 8}{3} + 5q = 35\n\Rightarrow \frac{7(2q + 8)}{3} + 5q = 35\n]", "Simplify the numerator:", "[\n\frac{14q + 56}{3} + 5q = 35\n]", "#### Step 2: Eliminate the Denominator", "Multiply every term in the equation by 3 to remove the fraction:", "[\n3 \cdot \left(\frac{14q + 56}{3} + 5q\right) = 3 \cdot 35\n\Rightarrow 14q + 56 + 15q = 105\n]", "#### Step 3: Combine Like Terms", "Add the q-terms:", "[\n(14q + 15q) + 56 = 105\n\Rightarrow 29q + 56 = 105\n]", "#### Step 4: Isolate the Variable", "Subtract 56 from both sides:", "[\n29q = 105 - 56\n29q = 49\n]", "Then, divide by 29:", "[\nq = \frac{49}{29}\n]", "---", "### Final Answer", "[\n\boxed{q = \frac{49}{29}}\n]", "---", "### Verification: Plugging Back to Check", "To confirm correctness, substitute ( q = \frac{49}{29} ) into the original equation:", "Left-hand side:", "[\n7\left(\frac{2 \cdot \frac{49}{29} + 8}{3}\right) + 5 \cdot \frac{49}{29} \n= 7\left(\frac{\frac{98}{29} + \frac{232}{29}}{3}\right) + \frac{245}{29}\n= 7\left(\frac{330}{29 \cdot 3}\right) + \frac{245}{29}\n= 7\left(\frac{110}{29}\right) + \frac{245}{29}\n= \frac{770}{29} + \frac{245}{29}\n= \frac{1015}{29}\n= 35\n]", "Which matches the right-hand side, confirming the solution is correct.", "---", "### Summary & Tips", "- Distribute carefully: Always multiply every term inside parentheses, especially fractions.\n- Clear denominators early: Multiplying through by the common denominator simplifies the equation and removes complications from fractions.\n- Combine terms methodically: Once fractions are removed, collect like terms to isolate the variable.\n- Check your work: Substitute back to validate that your solution satisfies the original equation.", "---", "### Related Keywords for SEO", "- Solving linear equations with fractions\n- How to solve equations with parentheses and variables \nAlgebra #MathProblems #EquationSolution #HighSchoolMath #LearningAlgebra #StepByStepMath", "---", "### Further Practice", "Try solving similar equations:", "[\na\left(\frac{bq + c}{d}\right) + mq = n\n]\nUsing this framework will strengthen your algebraic fluency.", "---", "Mastering equations like ( 7\left(\frac{2q + 8}{3}\right) + 5q = 35 ) unlocks deeper mathematical understanding— Königs! Keep practicing, and soon these expressions will feel intuitive."]