ANSWER TO QUESTION 1
[tex]3(h-4) = - \frac{1}{2} (24 - 6h)[/tex]
We multiply through by the Least Common Multiple which is 2.
[tex]2 \times 3(h-4) = - 2 \times \frac{1}{2} (24 - 6h)[/tex]
[tex]6(h - 4) = - 1(24 - 6h)[/tex]
We expand brackets to obtain,
[tex]6h - 24 = - 24 + 6h[/tex]
Grouping like terms, have
[tex]6h - 6h = - 24 + 24[/tex]
[tex]\Rightarrow 0 = 0[/tex]
Whenever you solve an equation and you get the above result, you don't have to get confuse.
It simply means the question does not have a UNIQUE solution.
That any real number will number will satisfy the above equation.
Hence,
[tex]h \in R[/tex]
ANSWER TO QUESTION 2
[tex]ax + bx = - c[/tex]
We factor x to obtain;
[tex]x(a + b) = - c[/tex]
We divide both sides by (a+b)
[tex]x = - \frac{ c}{a + b} [/tex]
