Solving a Formula For a Specific Variable

Solving a Formula For a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In the first two examples from the previous lesson, we used the formula \(d=rt\). This formula gives the value of \(d\), distance, when you substitute in the values of \(r\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\), the rate and time. But in the third example in the previous lesson, we had to find the value of \(t\). We substituted in values of \(d\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r\) and then used algebra to solve for \(t\).

If you had to do this often, you might wonder why there is not a formula that gives the value of \(t\) when you substitute in the values of \(d\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}r\). We can make a formula like this by solving the formula \(d=rt\) for \(t\).

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Example

Solve the formula \(A=\frac{1}{2}bh\) for \(h\):

when \(A=90\) and \(b=15\) in general

Solution

when \(A=90\) and \(b=15\)				in general
Write the formula.				Write the formula.
Substitute.
Clear the fractions.				Clear the fractions.
Simplify.				Simplify.
Solve for \(h\).				Solve for \(h\).

We can now find the height of a triangle, if we know the area and the base, by using the formula \(h=\frac{2A}{b}\).

The formula \(I=Prt\) is used to calculate simple interest, I, for a principal, P, invested at rate, r, for t years.

Example

Solve the formula \(I=Prt\) to find the principal, \(P\):

when \(I=\text{\$}5,600,r=4%,t=7\phantom{\rule{0.2em}{0ex}}\text{years}\phantom{\rule{0.2em}{0ex}}\) in general

Solution

\(I=\$5,600\), \(r=4%\), $t=7$ years				in general
Write the formula.				Write the formula.
Substitute.
Simplify.				Simplify.
Divide, to isolate P.				Divide, to isolate P.
Simplify.				Simplify.
The principal is

Later in this class, and in future algebra classes, you’ll encounter equations that relate two variables, usually x and y. You might be given an equation that is solved for y and need to solve it for x, or vice versa. In the following example, we’re given an equation with both x and y on the same side and we’ll solve it for y.

Example

Solve the formula \(3x+2y=18\) for y:

when \(x=4\) in general

Solution

when \(x=4\)				in general
Substitute.
Subtract to isolate the \(y\)-term.				Subtract to isolate the \(y\)-term.
Divide.				Divide.
Simplify.				Simplify.

In Examples 1.60 through 1.64 we used the numbers in part as a guide to solving in general in part . Now we will solve a formula in general without using numbers as a guide.

Example

Solve the formula \(P=a+b+c\) for \(a\).

Solution

We will isolate \(a\) on one side of the equation.
Both \(b\) and \(c\) are added to \(a\), so we subtract them from both sides of the equation.
Simplify.

Example

Solve the formula \(6x+5y=13\) for y.

Solution

Subtract \(6x\) from both sides to isolate the term with \(y\).
Simplify.
Divide by 5 to make the coefficient 1.
Simplify.

The fraction is simplified. We cannot divide \(13-6x\) by 5.