1. The problem asks to solve parts a and d of a given question, but since the exact question is not provided, I will demonstrate how to solve a typical algebraic problem for part a and d. 2. For part a, let's assume it is solving a linear equation: Solve for $x$ in $2x + 3 = 7$. 3. Use the formula for solving linear equations: isolate $x$ by subtracting 3 from both sides: $$2x + 3 = 7$$ $$2x + \cancel{3} - \cancel{3} = 7 - 3$$ $$2x = 4$$ 4. Divide both sides by 2 to solve for $x$: $$\frac{2x}{\cancel{2}} = \frac{4}{\cancel{2}}$$ $$x = 2$$ 5. For part d, let's assume it is solving a quadratic equation: Solve for $x$ in $x^2 - 5x + 6 = 0$. 6. Use the quadratic formula or factorization. Here, factorization is simpler: $$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$ 7. Set each factor equal to zero: $$x - 2 = 0 \Rightarrow x = 2$$ $$x - 3 = 0 \Rightarrow x = 3$$ 8. Therefore, the solutions for part d are $x = 2$ and $x = 3$. Final answers: - Part a: $x = 2$ - Part d: $x = 2$ or $x = 3$