Algebraic maninpulation of formula to isolate a variable I need to rearrange the following equation to get Rt on its own:
$$ADC = (\dfrac{R_t}{R_t + R_3} - \dfrac{R_2}{R_2 + R_1}) \times 10 \times 1023 $$
With a lot of help I've come up with:
$$(\dfrac{R_t}{R_t + R_3}) = (\dfrac{ADC}{10230} + \dfrac{R_2}{R_1 + R_2})$$
$$1 + \dfrac{R_3}{R_t} = \dfrac{1}{\dfrac{ADC}{10230} + \dfrac{R_2}{R_1 + R2}}$$
$$1 + \dfrac{R_3}{R_t} = \dfrac{10230(R_1+R_2)}{ADC(R_1+R_2) + 10230 R_2}$$
$$\dfrac{R_3}{R_t} = \dfrac{10230(R_1+R_2)}{ADC(R_1+R_2) + 10230 R_2}-\dfrac{ADC(R_1+R_2) + 10230 R_2}{ADC(R_1+R_2) + 10230 R_2}$$
$$\dfrac{R_3}{R_t}=\dfrac{10230(R_1+R_2)-ADC(R_1+R_2)-10230R_2}{ADC(R_1+R_2) + 10230 R_2}$$
$$R_t[10230(R_1 + R_2) - ADC(R_1 + R_2) - 10230R_2] = R_3[ADC(R_1 + R_2) + 10230R_2]$$
$$R_t = \dfrac{R_3[ADC(R_1 + R_2) + 10230R_2]}{10230(R_1 + R_2) - ADC(R_1 + R_2) - 10230R_2}$$
$$R_t = \dfrac{R_3[ADC(R_1 + R_2) + 10230R_2]}{10230 R_1 - ADC(R_1 + R_2)}$$
 A: Without having read your own reasoning in detail, I'll offer the following, which is less ad hoc, closer to a standard "classroom" approach.
$$ADC = (\dfrac{R_t}{R_t + R_3} - \dfrac{R_2}{R_2 + R_1}) \times 10 \times 1023$$
I'm going to go ahead and abbreviate $ADC$ by $a$, $\dfrac{R_2}{R_2 + R_1}$ by $b$ and $10\cdot1023$ by $c$, so we can see more clearly the shape of the equation.
$$a = (\dfrac{R_t}{R_t + R_3} - b)c=\dfrac{cR_t}{R_t + R_3}-bc$$
The key step that you're missing is to multiply both sides by $R_t + R_3$ so that your variable $R_t$ is no longer trapped under a fraction bar.
$$\begin{align}
a(R_t+R_3)&=cR_t-bc(R_t+R_3) \\
aR_t+aR_3&=cR_t-bcR_t-bcR_3 \\
(a+bc-c)R_t&=-aR_3-bcR_3
\end{align}$$
A: You start off ok:
$(\dfrac{R_t}{R_t + R_3}) = (\dfrac{ADC}{10230} + \dfrac{R_2}{R_1 + R_2})$ is correct.
$1 + \dfrac{R_3}{R_t} = \dfrac{1}{\dfrac{ADC}{10230} + \dfrac{R_2}{R_1 + R_2}}$ is a good idea because it gives you only one instance of $R_t$.
You can't  do what you tried to do next because $\frac 1 {a+b}$ is not equal to $\frac 1 a + \frac 1 b$ or $a+ b$.
$1 + \dfrac{R_3}{R_t} = \dfrac{10230(R_1+R_2)}{ADC(R_1+R_2) + 10230 R_2}$ is the next line and I will leave the rest to you.
A: No.
You can't cancel like that.
$1 + \dfrac{R_3}{R_t} = \dfrac{10230(R_1+R_2)}{ADC(R_1+R_2) + 10230 R_2}$
$\dfrac{R_3}{R_t} = \dfrac{10230(R_1+R_2)}{ADC(R_1+R_2) + 10230 R_2}-\dfrac{ADC(R_1+R_2) + 10230 R_2}{ADC(R_1+R_2) + 10230 R_2}$
$\dfrac{R_3}{R_t}=\dfrac{10230(R_1+R_2)-ADC(R_1+R_2)-10230R_2}{ADC(R_1+R_2) + 10230 R_2}$
Can you do the rest?
A: $$R_t = \dfrac{R_3[ADC(R_1 + R_2) + 10230R_2]}{10230(R_1 + R_2) - ADC(R_1 + R_2) - 10230R_2}$$
$$R_t = \dfrac{R_3[ADC(R_1 + R_2) + 10230R_2]}{10230 R_1 +10230 R_2) - ADC(R_1 + R_2) - 10230R_2}$$
$$R_t = \dfrac{R_3[ADC(R_1 + R_2) + 10230R_2]}{10230 R_1 - ADC(R_1 + R_2)}$$
