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One-Variable First-Order Equations for Physics Help

By — McGraw-Hill Professional
Updated on Sep 17, 2011

Introduction

In algebra, it is customary to classify equations according to the highest exponent, that is, the highest power to which the variables are raised. A one-variable first-order equation , also called a first-order equation in one variable , can be written in the following standard form:

ax + b = 0

where a and b are constants, and x is the variable. Equations of this type always have one real-number solution.

What’s A “real” Number?

A real number can be defined informally as any number that appears on a number line (Fig. 1-4). Pure mathematicians would call that an oversimplification, but it will do here. Examples of real numbers include 0, 5, −7, 22.55, the square root of 2, and π.

If you wonder what a “nonreal” number is like, consider the square root of −1. What real number can you multiply by itself and get −1? There is no such number. All the negative numbers, when squared, yield positive numbers; all the positive numbers also yield positive numbers; zero squared equals zero. The square root of −1 exists, but it lies somewhere other than on the number line shown in Fig. 1-4.

Equations, Formulas, and Vectors One-Variable First-Order Equations What’s A “real” Number?

Fig. 1-4 . The real numbers can be depicted graphically as points on a straight line.

Later in this chapter you will be introduced to imaginary numbers and complex numbers , which are, in a certain theoretical sense, “nonreal.” For now, however, let’s get back to the task at hand: first-order equations in one variable.

Some Examples

Any equation that can be converted into the preceding standard form is a one-variable first-order equation. Alternative forms are

cx = d

x = m/n

where c, d, m , and n are constants and n ≠ 0. Here are some examples of single-variable first-order equations:

4 x − 8 = 0

−π x = 22

3 ex = c

x = π/ c

In these equations, π, e , and c are known as physical constants , representing the circumference-to-diameter ratio of a circle, the natural exponential base, and the speed of light in free space, respectively. The constants π and e are not specified in units of any sort. They are plain numbers, and as such, they are called dimensionless constants:

π ≈ 3.14159

e ≈ 2.7 1828

The squiggly equals sign means “is approximately equal to.” The constant c does not make sense unless units are specified. It must be expressed in speed units of some kind, such as miles per second (mi/s) or kilometers per second (km/s):

c ≈ 186,282 mi/s

c ≈ 299,792 km/s

How To Solve

To solve a single-variable equation, it must in effect be converted into a formula. The variable should appear all by itself on the left-hand side of the equals sign, and the expression on the right-hand side should be reducible to a specific number. There are several techniques for getting such an equation into the form of a statement that expressly tells you the value of the variable:

Equations, Formulas, and Vectors One-Variable First-Order Equations How To Solve Add the same quantity to each side of the equation.

Equations, Formulas, and Vectors One-Variable First-Order Equations How To Solve Subtract the same quantity from each side of the equation. 

Equations, Formulas, and Vectors One-Variable First-Order Equations How To Solve Multiply each side of the equation by the same quantity.

Equations, Formulas, and Vectors One-Variable First-Order Equations How To Solve Divide each side of the equation by the same quantity.

The quantity involved in any of these processes can contain numbers, constants, variables—anything. There’s one restriction: You can’t divide by zero or by anything that can equal zero under any circumstances. The reason for this is simple: Division by zero is not defined.

Consider the four equations mentioned a few paragraphs ago. Let’s solve them. Listed them again, they are

4 x − 8 = 0

−π x = 22

3 ex = c

x = π/ c

The first equation is solved by adding 8 to each side and then dividing each side by 4:

4 x − 8 = 0

4 x = 8

x = 8/4 = 2

The second equation is solved by dividing each side by π and then multiplying each side by −1:

−π x = 22

x = 22/π

x = −22/π

x ≈ −22/3.14159

x ≈ −7.00282

The third equation is solved by first expressing c (the speed of light in free space) in the desired units, then dividing each side by e (where e ≈ 2.71828), and finally dividing each side by 3. Let’s consider c in kilometers per second; c ≈ 299,792 km/s. Then

3 ex = c

(3 × 2.71828) x ≈ 299,792 km/s

3 x ≈ (299,792/2.71828) km/s ≈ 110,287 km/s

x ≈ (110,287/3) km/s ≈ 36,762.3 km/s

Note that we must constantly keep the units in mind here. Unlike the first two equations, this one involves a variable having a dimension (speed).

The fourth equation doesn’t need solving for the variable, except to divide out the right-hand side. However, the units are tricky! Consider the speed of light in miles per second for this example; c ≈ 186,282 mi/s. Then

x = π/ c

x ≈ 3.14159/(186,282 mi/s)

When units appear in the denominator of a fractional expression, as they do here, they must be inverted. That is, we must take the reciprocal of the unit involved. In this case, this means changing miles per second into seconds per mile (s/mi). This gives us

x ≈ (3.14159/186,282) s/mi

x ≈ 0.0000168647 s/mi

This is not the usual way to express speed, but if you think about it, it makes sense. Whatever “object x ” might be, it takes about 0.0000168647 s to travel 1 mile.

 

Practice problems for these concepts can be found at:  Equations, Formulas, And Vectors for Physics Practice Test

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