Formulas

You often have to solve a formula --- an equation --- for a variable. The formula may come from mathematics, but it can also come from another area (such as business, economics, or science). The following ideas are often useful in solving for a variable:


Example. The volume of a cylinder of radius r and height h is

$$V = \pi r^2h.$$

Solve for h.

There is only one term containing h. All I need to do is divide by the stuff multiplying the h:

$$\eqalign{ V & = \pi r^2h \cr \dfrac{V}{\pi r^2} & = \dfrac{\pi r^2h}{\pi r^2} \cr \dfrac{V}{\pi r^2} & = h \quad\halmos \cr}$$


Example. Solve the following equation for t:

$$k = \dfrac{1}{3}r(s + t).$$

Usually, these problems can be solved in more than one way.

Here is one approach:

$$\matrix{ k & = & \dfrac{1}{3}r(s + t) & \cr 3 \cdot k & = & 3 \cdot \dfrac{1}{3}r(s + t) & \hbox{(Clear the fraction)} \cr 3 k & = & r(s + t) & \cr \dfrac{3 k}{r} & = & \dfrac{r(s + t)}{r} & \hbox{(Divide out r)} \cr \dfrac{3 k}{r} & = & s + t & \cr \dfrac{3 k}{r} - s & = & s + t - s & \hbox{(Move s to the left)} \cr \dfrac{3 k}{r} - s & = & t & \cr}$$

Another approach would begin by distributing $\dfrac{1}{3}r$ into $s + t$ . See if you can get it to work that way.


Example. The surface area of a cylindrical can of radius r and height h is

$$A = 2 \pi rh + 2 \pi r^2.$$

($2 \pi rh$ represents the area of the side of the can, while $2 \pi r^2$ is the area of the top and the bottom.) Solve for h.

In this case, I need to get the term containing h by itself before dividing:

$$\matrix{ A & = & 2 \pi r h + 2 \pi r^2 & \cr A - 2 \pi r^2 & = & 2 \pi r h + 2 \pi r^2 - 2 \pi r^2 & \hbox{(Get the h term by itself)} \cr A - 2 \pi r^2 & = & 2 \pi r h & \cr \dfrac{A - 2 \pi r^2}{2 \pi r} & = & \dfrac{2 \pi r h}{2 \pi r} & \hbox{Divide out $2 \pi r$)} \cr \dfrac{A - 2 \pi r^2}{2 \pi r} & = & h & \cr}$$

Notice that I divided the whole left side by $2 \pi r$ . And be careful! --- in $\dfrac{A - 2 \pi r^2}{2 \pi r}$ , you can't cancel $2
   \pi r$ from the top and the bottom.

Thus, $h = \dfrac{A - 2 \pi r^2}{2 \pi
   r}$ .


Example. Solve for b:

$$4 a b = 7 b + 11 a.$$

In a problem like this where the variable occurs in more than one term, it can be useful to circle the terms containing the variable to focus your attention. Notice that there are two terms ($4 a b$ and $7 b$ ) containing b.

First, I'll get the b-terms on one side. Then I'll factor out the common factor of b and solve.

$$\matrix{ 4 a b & = & 7 b + 11 a & \cr 4 a b - 7 b & = & 7 b + 11 a - 7 b & \hbox{(Get all the b terms on the left)} \cr 4 a b - 7 b & = & 11 a & \cr b(4 a - 7) & = & 11 a & \hbox{(Factor to isolate b)} \cr \dfrac{b(4 a - 7)}{4 a - 7} & = & \dfrac{11 a}{4 a - 7} & \hbox{(Divide out $4 a - 7$)} \cr b & = & \dfrac{11 a}{4 a - 7} \quad\halmos & \cr}$$


Example. Solve for v:

$$2 a v + 6 a b - 3 b v = 0.$$

This time, it's easiest to get the v-terms by themselves by moving the single non-v-term to the right:

$$\matrix{ 2 a v + 6 a b - 3 b v & = & 0 & \cr 2 a v + 6 a b - 3 b v - 6 a b & = & 0 - 6 a b & \hbox{(Move the non-v term to the right)} \cr 2 a v - 3 b v & = & -6 a b & \cr v(2 a - 3 b) & = & -6 a b & \hbox{(Factor to isolate v)} \cr \dfrac{v(2 a - 3 b)}{2 a - 3 b} & = & \dfrac{-6 a b}{2 a - 3 b} & \hbox{(Divide out $2 a - 3 b$)} \cr v & = & \dfrac{-6 a b}{2 a - 3 b} \quad\halmos & \cr}$$


Example. Solve for x:

$$5 x - 7 b = 2 a b + a x.$$

I move the x-terms to the left and the non-x-terms to the right, then factor:

$$\eqalign{ 5 x - 7 b & = 2 a b + a x \cr 5 x - 7 b + 7 b - a x & = 2 a b + a x + 7 b - a x \cr 5 x - a x & = 2 a b + 7 b \cr (5 - a) x & = 2 a b + 7 b \cr \noalign{\vskip2pt} \dfrac{(5 - a) x}{5 - a} & = \dfrac{2 a b + 7 b}{5 - a} \cr \noalign{\vskip2pt} x & = \dfrac{2 a b + 7 b}{5 - a} \quad\halmos \cr}$$


Example. Solve for y:

$$x = \dfrac{y}{y + 7}.$$

I'll clear the denominator of the fraction, multiply out, then get all the y-terms on one side:

$$\eqalign{ x & = \dfrac{y}{y + 7} \cr \noalign{\vskip2pt} (y + 7) \cdot x & = (y + 7) \cdot \dfrac{y}{y + 7} \cr \noalign{\vskip2pt} (y + 7) x & = y \cr y x + 7 x & = y \cr y x + 7 x - y x & = y - y x \cr 7 x & = y - y x \cr}$$

Now all I have to do is factor the y out on the right and divide:

$$\eqalign{ 7 x & = y - y x \cr 7 x & = (1 - x) y \cr \noalign{\vskip2pt} \dfrac{7 x}{1 - x} & = \dfrac{(1 - x) y}{1 - x} \cr \noalign{\vskip2pt} \dfrac{7 x}{1 - x} & = y \quad\halmos \cr}$$



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