Now substitute \( q = \frac{49}{29} \) back into \( p = \frac{2q + 8}{3} \):

["SEO Article: How to Substitute ( q = \frac{49}{29} ) Back into ( p = \frac{2q + 8}{3} ): A Step-by-Step Guide", "In mathematical problem-solving, substitution is a fundamental technique used to simplify equations and solve for unknowns. One common task is substituting a previously defined value back into the original equation to find a corresponding numeric result. Today, we explore how to substitute ( q = \frac{49}{29} ) back into the expression ( p = \frac{2q + 8}{3} ), demonstrating the process clearly for students, learners, and math enthusiasts.", "---", "### Understanding the Substitution", "When given ( q = \frac{49}{29} ), we want to plug this exact value into the equation:", "[\np = \frac{2q + 8}{3}\n]", "This step turns the abstract variable ( q ) into a specific fraction, allowing us to compute an exact or simplified result for ( p ).", "---", "### Step-by-Step Calculation", "1. Start with the original expression:", "[\np = \frac{2q + 8}{3}\n]", "2. Substitute ( q = \frac{49}{29} ):", "[\np = \frac{2\left(\frac{49}{29}\right) + 8}{3}\n]", "3. Multiply ( 2 \ imes \frac{49}{29} ):", "[\n2 \cdot \frac{49}{29} = \frac{98}{29}\n]", "4. Rewrite the equation with this result:", "[\np = \frac{\frac{98}{29} + 8}{3}\n]", "5. Express ( 8 ) with denominator 29 for easy addition:", "[\n8 = \frac{8 \cdot 29}{29} = \frac{232}{29}\n]", "6. Add the two terms in the numerator:", "[\n\frac{98}{29} + \frac{232}{29} = \frac{98 + 232}{29} = \frac{330}{29}\n]", "7. Now divide by 3:", "[\np = \frac{\frac{330}{29}}{3} = \frac{330}{29} \div 3 = \frac{330}{29} \cdot \frac{1}{3} = \frac{330}{87}\n]", "8. Simplify the fraction:", "Check if ( \frac{330}{87} ) can be reduced.\n- Both numbers are divisible by 3:\n ( 330 \div 3 = 110 )\n ( 87 \div 3 = 29 )\nThus:\n[\n\frac{330}{87} = \frac{110}{29}\n]", "---", "### Final Result", "After substituting ( q = \frac{49}{29} ) back into ( p = \frac{2q + 8}{3} ), the simplified value of ( p ) is:", "[\n\boxed{p = \frac{110}{29}}\n]", "---", "### Why This Substitution Matters", "Substituting known values back into equations is essential in algebra, calculus, and applied mathematics. It allows for numerical verification, further analysis, and real-world problem solving where variables are often determined experimentally or given beforehand. Mastering such substitutions builds a strong foundation for more advanced mathematical techniques and problem-solving strategies.", "---", "### Key Takeaways", "- Always substitute the exact value into the equation.\n- Perform operations step-by-step to maintain clarity.\n- Simplify carefully using fractions and common denominators.\n- Use division by 3 as multiplication by ( \frac{1}{3} ) for accurate results.", "---", "Keywords: substitute ( q = \frac{49}{29} ), ( p = \frac{2q + 8}{3} ), fractional substitution, algebra substitution, simplifying expressions, step-by-step math, solving equations, math tutorial, substitution method, ( p = \frac{110}{29} )", "---", "Optimizing the Article for SEO:", "This SEO-friendly article targets high-traffic academic keywords, offers clear value through step-by-step explanations, and encourages engagement with meta descriptions, header tags, and bullet-pointed sections. By structuring the content around common search queries like “substitute q into p equation” and including practical explanations, the article caters to students and learners seeking accurate, understandable math resources."]