Solve for $b$ in $A = \dfrac{1}{2}h\left( {B + b} \right)$?

Hint: In this question we are asked to find the value of \[b\] in terms of \[b\], To solve the expression for \[b\], you need to isolate it on one side of the equation, for this we have to apply multiplication first with 2 on both sides and then divide with \[h\] on both sides and finally subtract on both sides with \[B\], then further simplification we will get the required value for \[b\].

Complete step by step answer:
Given equation is $A = \dfrac{1}{2}h\left( {B + b} \right)$, we have to solve the equation for variable $b$,
The equation is the formula to find the area of some three-dimensional figures, in the given equation we have four variables, so, \[B\] is the base, \[h\] is the height , \[b\] is the base of the other side and \[A\] is the area of the figure, and all the four can be treated as algebraic variables.
The equation is $A = \dfrac{1}{2}h\left( {B + b} \right)$,
Now multiply both sides with 2 we get,
$ \Rightarrow A \times 2 = \dfrac{1}{2}h\left( {B + b} \right) \times 2$,
By simplifying we get,
$ \Rightarrow 2A = h\left( {B + b} \right)$,
Now divide both sides with \[h\] we get,
$ \Rightarrow \dfrac{{2A}}{h} = \dfrac{{h\left( {B + b} \right)}}{h}$,
Now simplifying by eliminating the like terms we get,
$ \Rightarrow \dfrac{{2A}}{h} = \left( {B + b} \right)$,
Now subtract both sides with $B$ we get,
$ \Rightarrow \dfrac{{2A}}{h} - B = B + b - B$,
Now simplifying we get
$ \Rightarrow \dfrac{{2A}}{h} - B = b$,
$ \Rightarrow b = \dfrac{{2A}}{h} - B$
So, the value of \[b\]is \[\dfrac{{2A}}{h} - B\],

$\therefore $ The value of \[b\] when the equation is solved in terms of \[b\] will be equal to \[\dfrac{{2A}}{h} - B\].

Note: Remember that in linear equations and from some basic properties, that when we are asked for a variable we have to treat all other terms as constants. If we treat \[A,h,B\] as a variable then the given equation will become four variables which will have infinite solutions.