1. **State the problem:** Solve the simultaneous equations: $$4c + 3d = 4$$ $$2c - 5d = 15$$ 2. **Method:** We will use the elimination method to solve for $c$ and $d$. 3. **Eliminate one variable:** Multiply the second equation by 2 to align coefficients of $c$: $$2 \times (2c - 5d) = 2 \times 15$$ $$4c - 10d = 30$$ 4. **Subtract the first equation from this new equation:** $$\cancel{4c} - 10d - (\cancel{4c} + 3d) = 30 - 4$$ $$-10d - 3d = 26$$ $$-13d = 26$$ 5. **Solve for $d$:** $$d = \frac{26}{-13} = -2$$ 6. **Substitute $d = -2$ into the first equation:** $$4c + 3(-2) = 4$$ $$4c - 6 = 4$$ 7. **Solve for $c$:** $$4c = 4 + 6 = 10$$ $$c = \frac{10}{4} = \frac{5}{2}$$ **Final answer:** $$c = \frac{5}{2}, \quad d = -2$$