Solve the following: c) 4x + y = 10 for y; d) a + b = 8, a − b = 4 for a and b.

The equation is:

$$4x + y = 10$$

Since this is a single linear equation with two variables, there are infinitely many solutions. We can express one variable in terms of the other. Subtract $4x$ from both sides of the equation:

$$y = 10 - 4x$$

The solution for equation c is $\boxed{y = 10 - 4x}$.

We have the following system:

$$a + b = 8 \quad (1)$$ $$a - b = 4 \quad (2)$$

$$(a + b) + (a - b) = 8 + 4$$ $$2a = 12$$

Divide both sides by 2:

$$\frac{2a}{2} = \frac{12}{2}$$ $$a = 6$$

$$a + b = 8$$

Substitute $a = 6$:

$$6 + b = 8$$

Subtract 6 from both sides:

$$b = 8 - 6$$ $$b = 2$$

The solution to the system of equations is $a = \boxed{6}$ and $b = \boxed{2}$.