Question 202755
S=2B+Ph
Solve for h<br>
When you see the phrase: "solve for <some-variable-name>" it means "use Algebra to manipulate the equation until the variable is by itself on one side of the equation.<br>
In you problem it means: Manipulate S=2B+Ph until h is by itself. In other words we want the equation to look like:
h = [some-expression]
or
[some-expression] = h
Since the h is already on the right side we will aim for:
[some-expression] = h<br>
To get the h by itself we need to make the 2B and the P disappear, <b>using proper Algebra</b><br>
We'll start by subtracting 2B from both sides:
S - 2B = 2B + Ph - 2B
The 2B's on the right cancel giving:
S - 2B = Ph
We've made the "2B" disappear" from the right side! Now we will do the same with the "P". We can't, however, subtract "P" because "P" and "Ph" are <b>not</b> like terms! We can divide both sides by "P":
{{{(S - 2B)/P = (Ph)/P}}}
On the right side the P's cancel:
{{{(S - 2B)/P = h}}}
Guess what? We're done! We have "h" by itself so we have "solved for h".