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How to Solve Formulas for Specified Variables

“Formula calculation” is a fancy name for solving equations with more than one variable. When solving equations like that, you will have one variable alone on the left side of the equation and all the other variables on the right side of the equation. But how do you know which variable should be alone on the left side? Usually, the problem will tell you what to solve for.

When you do calculations with a formula, you use the exact same rules you are used to from solving equations:

You change sides and change signs, and you multiply/divide all terms by the factors you want to remove.

Think About This

Problems for formula calculation are often worded like this:

Solve this equation with respect to $\dots$
Find $\dots$ with respect to $\dots$

Let’s look at some examples to give you a sense of what we mean.

I have chosen some formulas you have seen before. That way, you already have some knowledge of what they mean, and which variables are moved around.

Example 1

a^{2} + b^{2} = c^{2}

Find $a$ as a function of $b$ and $c$ .

\begin{array}{l} a^{2} + b^{2} & = c^{2} \\ a^{2} & = c^{2} - b^{2} \\ \sqrt{a^{2}} & = \pm \sqrt{c^{2} - b^{2}} \\ a & = \pm \sqrt{c^{2} - b^{2}} \end{array}

Example 2

Solve $A = l \cdot b$ with respect to $b$ .

\begin{array}{l} A & = l \cdot b \\ \frac{A}{l} & = \frac{l \cdot b}{l} \\ \frac{A}{l} & = \frac{l \cdot b}{l} \\ \frac{A}{l} & = b \\ b & = \frac{A}{l} \end{array}

Example 3

Solve $s = v \cdot t$ with respect to $v$ .

\begin{array}{l} s & = v \cdot t \\ \frac{s}{t} & = \frac{v \cdot t}{t} \\ \frac{s}{t} & = \frac{v \cdot t}{t} \\ \frac{s}{t} & = v \\ v & = \frac{s}{t} \end{array}

Example 4

How fast did you ride your bike to school, when you know that the school is $7 km$ away, and it took you $18 min$ to get there?

Here, you need to use the formula from the last example. It is important that you get the correct time unit, so we transform 18 min into hours. You get

18 min = 18 \div 60 = 0.3 h

Now, you can insert the numbers into the formula for speed (or velocity, v) from the previous example:

\begin{array}{l} s & = t \cdot v \\ 7 km & = 0.3 h \cdot v & | \div 0.3 h \\ \frac{7 km}{0.3 h} & = v \\ v & = 23.3 km/h \end{array}

That means you are riding your bike at a speed of $23.3$ km/h.

Note! Notice that you can cross-cancel words, for example units, in the same way you cross-cancel numbers and letters when you work with fractions.

Example 5

Solve $A = \frac{g \cdot h}{2}$ for $h$ .

\begin{array}{l} A & = \frac{g \cdot h}{2} \\ 2 \cdot A & = \frac{g \cdot h}{2} \cdot 2 \\ 2 A & = \frac{g \cdot h}{2} \cdot 2 \\ 2 A & = g \cdot h \\ \frac{2 A}{g} & = \frac{g \cdot h}{g} \\ \frac{2 A}{g} & = \frac{g \cdot h}{g} \\ \frac{2 A}{g} & = h \\ h & = \frac{2 A}{g} \end{array}

Example 6

The area of a circle with radius $r$ is

A = π r^{2} .

Find the formula for the radius $r$ as a function of the area $A$ .

Think of this as an equation with $r$ as the unknown variable and $A$ and $π$ as constants. Your goal is to get $r$ on its own. We begin with dividing by $π$ : $\begin{array}{l} \frac{A}{π} & = \frac{π r^{2}}{π} \\ \frac{A}{π} & = r^{2} \end{array}$

Next, you take the square root of both sides to get $r$ on its own:

\sqrt{\frac{A}{π}} = \sqrt{r^{2}} = r .

Note! You can’t have a negative radius, so you only use the positive value of the square root.

Finally, you turn the expression to get $r$ on the left hand side of the equation:

r = \sqrt{\frac{A}{π}} .

You have now found a formula for the radius $r$ as a function of the area $A$ .

Example 7

Imagine that you have a circle with an area of $86.7 {cm}^{2}$ . What is the radius of your circle?

You use the formula for $r$ that you found in Example 6 and input the area given in the assignment. That gives you

r = \sqrt{\frac{86.7 {cm}^{2}}{π}} = 5.25 cm .

The radius of a circle with an area of $86.7$ cm² is $5.25$ cm.

Note! If you prefer, you can insert the numbers first, and then rearrange the formula.

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