Subtracting these gives $ 3a = 4 $ → $ a = \frac{4}{3} $. Substituting back, $ b = -4 $. Then, from $ 7a + 3b + c = 7 $, we find $ c = \frac{35}{3} $. Finally, $ d = -6 $. Thus,

["Understanding Linear Equations: Solving Step-by-Step for $ c $ and $ d $", "Solving linear equations is a fundamental skill in algebra, enabling us to find unknown values through logical substitution and arithmetic. In this tutorial, we’ll analyze a clear sequence of substitutions that leads from a simple equation to computed variable values, showing how each step contributes to the final result.", "---", "### The Starting Equation: $ 3a = 4 $", "The process begins with the equation:", "$$\n3a = 4\n$$", "To isolate $ a $, divide both sides by 3:", "$$\na = \frac{4}{3}\n$$", "This straightforward division gives the first key value.", "---", "### Finding $ b $ by substitution", "Next, substitute $ a = \frac{4}{3} $ into the expression for $ b $, given by:", "$$\nb = -4\n$$", "Even though $ b $ is not derived algebraically from $ a $, its known value allows direct use in the next equation.", "---", "### Solving for $ c $ using $ 7a + 3b + c = 7 $", "Now substitute $ a = \frac{4}{3} $ and $ b = -4 $ into the equation:", "$$\n7a + 3b + c = 7\n$$", "Compute $ 7a $:\n$$\n7 \cdot \frac{4}{3} = \frac{28}{3}\n$$", "Compute $ 3b $:\n$$\n3 \cdot (-4) = -12\n$$", "Add these with $ c $:\n$$\n\frac{28}{3} - 12 + c = 7\n$$", "Convert $ -12 $ to thirds: $ -12 = -\frac{36}{3} $, so:\n$$\n\frac{28}{3} - \frac{36}{3} = -\frac{8}{3}\n$$", "Thus:\n$$\n-\frac{8}{3} + c = 7\n$$", "Solve for $ c $:\n$$\nc = 7 + \frac{8}{3} = \frac{21}{3} + \frac{8}{3} = \frac{29}{3}\n$$", "Wait—here we notice a discrepancy. The problem states $ c = \frac{35}{3} $. Let’s reevaluate carefully.", "Wait: check if the original equation is correctly interpreted.", "Suppose instead the equation is:", "$$\n7a + 3b + c = 7\n$$", "Using $ a = \frac{4}{3} \Rightarrow 7a = \frac{28}{3} $,\n$ b = -4 \Rightarrow 3b = -12 $,\nSo:\n$$\n\frac{28}{3} - 12 + c = 7\n\quad \Rightarrow \quad\nc = 7 + 12 - \frac{28}{3}\n= 19 - \frac{28}{3}\n= \frac{57 - 28}{3} = \frac{29}{3}\n$$", "Still $ \frac{29}{3} $, not $ \frac{35}{3} $.", "This suggests a possible typo in the original problem statement. However, assuming the intended final value for $ c $ is $ \frac{35}{3} $, we double-check:", "$$\n7a + 3b + c = 7 \\n\Rightarrow \frac{28}{3} - 12 + c = 7 \\n\Rightarrow c = 7 - \frac{28}{3} + 12 = 19 - \frac{28}{3} = \frac{57 - 28}{3} = \frac{29}{3}\n$$", "So unless $ b = -5 $ instead of $ -4 $, we cannot reach $ \frac{35}{3} $. Try substituting $ b = -5 $:", "$$\n3b = 3(-5) = -15 \\n\Rightarrow \frac{28}{3} - 15 + c = 7 \\nc = 7 + 15 - \frac{28}{3} = 22 - \frac{28}{3} = \frac{66 - 28}{3} = \frac{38}{3}\n$$", "Still not $ \frac{35}{3} $. Try $ b = -6 $:", "$$\n3b = -18 \\n\Rightarrow \frac{28}{3} - 18 + c = 7 \\nc = 7 + 18 - \frac{28}{3} = 25 - \frac{28}{3} = \frac{75 - 28}{3} = \frac{47}{3}\n$$", "No. Try $ a = \frac{5}{3} $? But original says $ 3a = 4 \Rightarrow a = \frac{4}{3} $.", "Thus, the value $ c = \frac{35}{3} $ cannot follow from given substitutions unless the expression or data is adjusted.", "But assuming the problem meant to say:\nFrom $ 7a + 3b + c = 25 $, then:", "$$\n\frac{28}{3} - 12 + c = 25 \\n\Rightarrow c = 25 + 12 - \frac{28}{3} = 37 - \frac{28}{3} = \frac{111 - 28}{3} = \frac{83}{3}\n$$", "Still not.", "Alternatively, perhaps the equation is:", "$$\n7a + 3b + c = 35/3?\n$$", "Unlikely.", "Given this inconsistency, we instead present the accurate derivation based on the stated values, clarifying the correct result.", "---", "### Final Correct Derivation:", "We follow the logical flow as described:", "- $ 3a = 4 \Rightarrow a = \frac{4}{3} $\n- $ b = -4 $\n- $ 7a + 3b + c = 7 $\n- $ 7 \cdot \frac{4}{3} = \frac{28}{3} $\n- $ 3 \cdot (-4) = -12 = -\frac{36}{3} $\n- $ \frac{28}{3} - \frac{36}{3} = -\frac{8}{3} $\n- So: $ -\frac{8}{3} + c = 7 $ → $ c = 7 + \frac{8}{3} = \frac{21 + 8}{3} = \frac{29}{3} $", "Then, finally:", "- $ d = -6 $", "---", "### Final Answer:", "Thus,\n$$\na = \frac{4}{3},\quad b = -4,\quad c = \frac{29}{3},\quad d = -6\n$$", "This structured substitution method is essential in algebra, ensuring accuracy and clarity in solving multi-step equations common in equations, word problems, and applied mathematics.", "---", "Keywords: solve linear equations, substitute values, algebraic step-by-step, find $ c $ and $ d $, demonstrate algebra, solve for variables, step-by-step algebra, equation solution, substitution method.", "For more insights on linear equations and substitution techniques, explore advanced algebra guides or practice with varied equations."]