h = [S / (2πr)] - r
It is a case about two-variable linear equations and we have to solve the equation to get the variable h.
Our main plan is to isolate the variable h alone at the end of the process on one side of the equation until the variable will be equal to the value on the opposite side.
[tex]\boxed{S = 2 \pi r^2 + 2 \pi rh}[/tex]
We can set it to be like this, i.e., just swapping positions on both sides but not changing the signs.
[tex]\boxed{2 \pi r^2 + 2 \pi rh = S}[/tex]
We subtract both sides with [tex]\boxed{2 \pi r^2}[/tex].
[tex]\boxed{2 \pi r^2 + 2 \pi rh - 2 \pi r^2 = S - 2 \pi r^2}[/tex]
[tex]\boxed{2 \pi rh = S - 2 \pi r^2}[/tex]
We divide both sides with [tex]\boxed{2 \pi r}[/tex]
[tex]\boxed{\frac{2 \pi rh}{2 \pi r} = \frac{S - 2 \pi r^2}{2 \pi r}}[/tex]
[tex]\boxed{h = \frac{S}{2 \pi r} - \frac{2 \pi r^2}{2 \pi r}}[/tex]
We get the results as follows: [tex]\boxed{\boxed{h= \frac{S}{2 \pi r} - r}}[/tex]
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Let's check again from the beginning.
For example, r = 2 and h = 3, then we first calculate the value of S by using the initial equation.
[tex]\boxed{S=2 \pi (2)^2 + 2 \pi (2)(3)}[/tex]
[tex]\boxed{S = 8 \pi + 12 \pi }[/tex]
[tex]\boxed{\boxed{S= 20 \pi}}[/tex]
Then, we substitute the results of S (with r = 2) into the new equation that we have compiled with h as the subject.
[tex]\boxed{h =\frac{20 \pi}{2 \pi (2)} - 2}[/tex]
[tex]\boxed{h = 5 - 2}[/tex]
We get the value h = 3 and it means that the equation has been proven.
Notes:
The formula that we discussed above is a formula for calculating the surface area of a closed cylinder.
[tex]\boxed{S = 2 \pi r^2 + 2 \pi rh}[/tex]
Keywords: S=2πr^2+2πrh, solve for h, subtract, divide, subject, to isolate the variable h alone, two-variable linear equations
