First equation:
You have to isolate 4h in one side of the equation and all other terms in the other side.
To "pass" the terms from one side of the equation to the other you have to perform the inverse operation of which they are doing in both sides of the equal sign.
So, for example, the term "s" is positive, the inverse operation is "-s"
[tex]\begin{gathered} ab=s+2d+3t+4h \\ ab-s=s-s+2d+3t+4h \\ ab-s=2d+3t+4h \end{gathered}[/tex]Next you have to subtract term 2d to both sides of the equation
[tex]\begin{gathered} ab-s-2d=2d-2d+3t+4h \\ ab-s-2d=3t+4h \end{gathered}[/tex]Finally subtract 3t from both sides of the equation:
[tex]\begin{gathered} ab-s-2d-3t=3t-3t+4h \\ ab-s-2d-3t=4h \end{gathered}[/tex]Second equation:
In this part you have to solve for h.
Note that "h" is being multiplied by 4, to cancel this multiplication you have to divide it by 4.
And to keep the equality valid, any operation performed in one side of the = sign must be performed in the other side, so you have ti divide both sides by 4.
[tex]\begin{gathered} ab-s-2d-3t=4h \\ \frac{ab-s-2d-3t}{4}=\frac{4h}{4} \\ \frac{1}{4}ab-\frac{1}{4}s-\frac{2}{4}d-\frac{3}{4}t=h \\ \frac{1}{4}ab-\frac{1}{4}s-\frac{1}{2}d-\frac{3}{4}t=h \end{gathered}[/tex]