Given that a = (-7, 20) and b = (3, -5), find the vector 6a + b.

We are given two vector equations:

$$\vec{a} - \vec{b} = \begin{pmatrix} -8 \\ 3 \end{pmatrix} \quad (1)$$ $$3\vec{a} + \vec{b} = \begin{pmatrix} -7 \\ 29 \end{pmatrix} \quad (2)$$

Adding the left-hand sides and right-hand sides of the two equations:

$$(\vec{a} - \vec{b}) + (3\vec{a} + \vec{b}) = \begin{pmatrix} -8 \\ 3 \end{pmatrix} + \begin{pmatrix} -7 \\ 29 \end{pmatrix}$$ $$4\vec{a} = \begin{pmatrix} -8 + (-7) \\ 3 + 29 \end{pmatrix}$$ $$4\vec{a} = \begin{pmatrix} -15 \\ 32 \end{pmatrix}$$

Divide both sides by 4:

$$\vec{a} = \frac{1}{4} \begin{pmatrix} -15 \\ 32 \end{pmatrix}$$ $$\vec{a} = \begin{pmatrix} \frac{-15}{4} \\ \frac{32}{4} \end{pmatrix}$$ $$\vec{a} = \begin{pmatrix} -\frac{15}{4} \\ 8 \end{pmatrix}$$

Using equation (1): $\vec{a} - \vec{b} = \begin{pmatrix} -8 \\ 3 \end{pmatrix}$
Substitute the value of $\vec{a}$:

$$\begin{pmatrix} -\frac{15}{4} \\ 8 \end{pmatrix} - \vec{b} = \begin{pmatrix} -8 \\ 3 \end{pmatrix}$$

Rearrange to solve for $\vec{b}$:

$$\vec{b} = \begin{pmatrix} -\frac{15}{4} \\ 8 \end{pmatrix} - \begin{pmatrix} -8 \\ 3 \end{pmatrix}$$ $$\vec{b} = \begin{pmatrix} -\frac{15}{4} - (-8) \\ 8 - 3 \end{pmatrix}$$ $$\vec{b} = \begin{pmatrix} -\frac{15}{4} + 8 \\ 5 \end{pmatrix}$$

To simplify the x-component, convert 8 to a fraction with denominator 4: $8 = \frac{32}{4}$.

$$\vec{b} = \begin{pmatrix} -\frac{15}{4} + \frac{32}{4} \\ 5 \end{pmatrix}$$ $$\vec{b} = \begin{pmatrix} \frac{32 - 15}{4} \\ 5 \end{pmatrix}$$ $$\vec{b} = \begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}$$

The vectors are:

$$\boxed{\vec{a} = \begin{pmatrix} -\frac{15}{4} \\ 8 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}}$$