Simple algebra questions from Head First Physics

Question

I am enjoying a book called Head First Physics, however I have run into a problem after forgetting some simple high school level algebra.

Question 1: On page 316 where did the 1/2 fractions go? Why does a become 2a? I assume the 1/2 was divided at the top of the fraction to cancel them, and then multiplied at the bottom to multiply a so it became 2a...

Question 2: On page 318 how do I expand and rearrange the 3rd motion equation? I know how to expand brackets and how to make something the subject of an equation, however this one is beyond my skill set. My best guess would be to multiply 2a by the inner bracket's contents to get 2ax - 2ax_0 and then make x the subject...

Thank you

Answer 1

For the $1$st question, note that $\frac{\frac{1}{2}v^{2} - \frac{1}{2}v_{0}^{2}}{a} = \frac{\frac{1}{2}v^{2} - \frac{1}{2}v_{0}^{2}}{a}\cdot\frac{2}{2} = \frac{v^{2}-v_{0}^{2}}{2a}$

For the $2$nd question, $2a(x-x_{0})$ = $2ax - 2ax_{0}$ by the distributive property of multiplication. Then, we solve for $x$. First, we bring all the terms not containing $x$ to the left side:

$v^{2} - v_{0}^{2} + 2ax_{0} = 2ax$

Now, we divide by $2a$ on both sides to obtain our answer:

$\frac{v^{2} - v_{0}^{2}}{2a} + x_{0} = x$

Answer 2

For question $2$, we want to isolate $x$. To do this, subtract both sides by $v_0^2$ to get:

$$v^2 - v_0^2 = 2a(x-x_0)$$

and then divide both sides by $2a$:

$$\frac{v^2 - v_0^2}{2a} = x - x_0$$

from which we add both sides by $x_0$ to arrive at:

$$\frac{v^2-v_0^2}{2a}+x_0 = x$$