Given $\var{a}x+\var{b}=\var{c}x$, we can subtract $\var{a}x$ from both sides to collect like terms, and then divide both sides by the coefficient of $x$ to get $x$ by itself.

There is often more than one way to solve an equation, one strategy used above in the first step was to get all the $x\text{'s}$ one the side with the most $x\text{'s}$, that way you end up with a positive number of $x\text{'s}$. This is not necessary, we could have put all the $x$'s on the left-hand side but notice in this question that we then would have had to move the $\var{b}$ onto the right-hand side, so it would have required more work, but nevertheless that method would result in the same result for $x$.

Given $\var{l}(\var{m}w-\var{n})=\var{p}w+\var{q}$, we can expand the brackets, get all the $w\text{'s}$ on the left-hand side, collect like terms, get all the numbers on the right-hand side, collect like terms and then divide both sides by the coefficient of $w$ to get $w$ by itself.

Given $\displaystyle{\frac{\var{d}y}{y-\var{f}}}=\var{g}$, we can multiply both sides by $(y-\var{f})$ to get rid of the fraction, get all the $y\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $y$ to get $y$ by itself.

Given $\displaystyle{\frac{z+\var{h}}{z+\var{j}}}=\var{k}$, we can multiply both sides by $(z+\var{j})$ to get rid of the fraction, get all the $z\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $z$ to get $z$ by itself.

Given $\displaystyle{\frac{x+\var{add}}{\var{denom1}}+\frac{x}{\var{denom2}}=\var{right}}$, we can multiply both sides by $\var{denom1}$ and by $\var{denom2}$ to get rid of the fractions, get all the $x\text{'s}$ on one side and the numbers on the other side, and then divide both sides by the coefficient of $x$ to get $x$ by itself.