Do not use a calculator in answering this question. Showing your working, find the complex numbers $z$ and $w$ which satisfy the simultaneous equations $2iz + w = 5 + 7i,$ $(4 - i)\,\overline{w} = z - 3i.$

1. Express $z$ in terms of $\overline{w}$ from the second equation

$(4 - i)\overline{w} = z - 3i$
$z = (4 - i)\overline{w} + 3i$

2. Substitute the expression for $z$ into the first equation and simplify

$2i[(4 - i)\overline{w} + 3i] + w = 5 + 7i$
$(8i - 2i^2)\overline{w} + 6i^2 + w = 5 + 7i$
$(2 + 8i)\overline{w} - 6 + w = 5 + 7i$
$w + (2 + 8i)\overline{w} = 11 + 7i$

3. Substitute $w = a + bi$ into the simplified equation and expand

Let $w = a + bi$ and its conjugate $\overline{w} = a - bi$, where $a, b \in \mathbb{R}$.

$(a + bi) + (2 + 8i)(a - bi) = 11 + 7i$
$a + bi + 2a - 2bi + 8ai - 8bi^2 = 11 + 7i$
$a + bi + 2a - 2bi + 8ai + 8b = 11 + 7i$
$(3a + 8b) + (8a - b)i = 11 + 7i$

4. Equate the real and imaginary parts to form a system of linear equations

Comparing the real and imaginary parts:
$3a + 8b = 11 \quad \text{--- (1)}$
$8a - b = 7 \quad \text{--- (2)}$

5. Solve the system of equations to find $w$

From (2), we get $b = 8a - 7$.
Substitute into (1):
$3a + 8(8a - 7) = 11$
$3a + 64a - 56 = 11$
$67a = 67$
$a = 1$

Substitute $a = 1$ back into the expression for $b$:
$b = 8(1) - 7 = 1$

Therefore, $w = 1 + i$.

6. Substitute $\overline{w}$ back into the expression for $z$ to solve for $z$

$z = (4 - i)(1 - i) + 3i$
$z = 4 - 4i - i + i^2 + 3i$
$z = 4 - 5i - 1 + 3i$
$z = 3 - 2i$