Solving equations 3 and 4: From 3, $c = 7 - 3d$. Substitute into 4: $2(7 - 3d) + 4d = 8 \Rightarrow 14 - 6d + 4d = 8 \Rightarrow -2d = -6 \Rightarrow d = 3$, then $c = 7 - 9 = -2$.

["Solving Linear Equations 3 and 4: A Step-by-Step Guide to Finding $d$ and $c$", "Working through systems of linear equations is a fundamental skill in algebra, especially when solving for variables that represent unknown quantities. In this tutorial, we’ll explore how to solve equations 3 and 4 systematically, demonstrating clear algebraic manipulation to find exact values of $d$ and $c$. This method is essential not only for classroom learning but also for real-world problem-solving in science, engineering, and finance.", "---", "### Equation 3: Expressing $c$ in Terms of $d$", "We begin with Equation 3, given as:\n$$\nc = 7 - 3d\n$$", "This equation defines $c$ algebraically as a function of $d$. Like all expressions, it gives $c$’s value depending on $d$, and is valuable for substitution later.", "---", "### Substitution into Equation 4", "Equation 4 is:\n$$\n2(7 - 3d) + 4d = 8\n$$", "Instead of solving for $c$ directly, we substitute the expression for $c$ into Equation 4 to eliminate $c$ and solve only for $d$.", "Substituting $c = 7 - 3d$ into Equation 4 gives:\n$$\n2(7 - 3d) + 4d = 8\n$$", "---", "### Expanding and Simplifying", "Expand the equation:\n$$\n14 - 6d + 4d = 8\n$$", "Combine like terms:\n$$\n14 - 2d = 8\n$$", "Subtract 14 from both sides:\n$$\n-2d = 8 - 14 = -6\n$$", "---", "### Solving for $d$", "Divide both sides by $-2$:\n$$\nd = \frac{-6}{-2} = 3\n$$", "So, $d = 3$. This intermediate step reduces the original two-equation system to a simple linear equation in one variable.", "---", "### Finding $c$ Using Substituted Value", "Now substitute $d = 3$ back into Equation 3 to find $c$:\n$$\nc = 7 - 3(3) = 7 - 9 = -2\n$$", "---", "### Final Answer", "The solutions are:\n- $d = 3$\n- $c = -2$", "---", "### Why This Method Matters", "Using substitution between equations is a powerful technique because:", "- It reduces complex systems to simpler ones.\n- It relies on clear algebraic steps that minimize errors.\n- It builds confidence in solving multi-step word problems and equations common in STEM fields.", "Mastering this process helps learners move smoothly from expressing variables conceptually to finding numerical solutions—key for exams, homework, and real-life applications.", "---", "### Summary of Steps:", "1. Start with Equation 3: $c = 7 - 3d$\n2. Substitute into Equation 4: $2(7 - 3d) + 4d = 8$\n3. Expand: $14 - 6d + 4d = 8$\n4. Combine: $14 - 2d = 8$\n5. Solve: $-2d = -6 \Rightarrow d = 3$\n6. Back-substitute: $c = 7 - 3(3) = -2$", "---", "Practice makes proficient — try solving your own equations by substituting expressions, just like we did above. Your algebraic fluency will grow step by step!"]