﻿ ﻿Unmasking Mr. X: Algebraic Equations - The X-Files: Introduction to Algebra - Basic Math & Pre-Algebra For Dummies

## Basic Math & Pre-Algebra For Dummies, 2nd Edition (2014)

### Chapter 22. Unmasking Mr. X: Algebraic Equations

In This Chapter Using variables (such as x) in equations Knowing some quick ways to solve for x in simple equations Understanding the balance scale method for solving equations Rearranging terms in an algebraic equation Isolating algebraic terms on one side of an equation Removing parentheses from an equation Cross-multiplying to remove fractions

When it comes to algebra, solving equations is the main event.

Solving an algebraic equation means finding out what number the variable (usually x) stands for. Not surprisingly, this process is called solving for x, and when you know how to do it, your confidence — not to mention your grades in your algebra class — will soar through the roof.

This chapter is all about solving for x. First, I show you a few informal methods to solve for x when an equation isn't too difficult. Then I show you how to solve more difficult equations by thinking of them as a balance scale.

The balance scale method is really the heart of algebra (yes, algebra has a heart, after all!). When you understand this simple idea, you're ready to solve more complicated equations, using all the tools I show you in Chapter 21, such as simplifying expressions and removing parentheses. You find out how to extend these skills to algebraic equations. Finally, I show you how cross-multiplying (see Chapter 9) can make solving algebraic equations with fractions a piece of cake.

By the end of this chapter, you'll have a solid grasp of a bunch of ways to solve equations for the elusive and mysterious x.

Understanding Algebraic Equations

An algebraic equation is an equation that includes at least one variable — that is, a letter (such as x) that stands for a number. Solving an algebraic equation means finding out what number x stands for.

In this section, I show you the basics of how a variable like x works its way into an equation in the first place. Then I show you a few quick ways to solve for x when an equation isn't too difficult.

Using x in equations

As you discover in Chapter 5, an equation is a mathematical statement that contains an equals sign. For example, here's a perfectly good equation: At its heart, a variable (such as x) is nothing more than a placeholder for a number. You're probably used to equations that use other placeholders: One number is purposely left as a blank or replaced by an underline or a question mark, and you're supposed to fill it in. Usually, this number comes after the equals sign. For example: As soon as you're comfortable with addition, subtraction, or whatever, you can switch the equation around a bit: When you stop using underlines and question marks and start using variables such as x to stand for the part of the equation you want to figure out, bingo! You have an algebra problem: Choosing among four ways to solve algebraic equations

You don't need to call an exterminator just to kill a bug. Similarly, algebra is strong stuff, and you don't always need it to solve an algebraic equation.

Generally, you have four ways to solve algebraic equations such as the ones I introduce earlier in this chapter. In this section, I introduce them in order of difficulty.

Eyeballing easy equations

You can solve easy problems just by looking at them. For example:

· 5 + x = 6

When you look at this problem, you can see that x = 1. When a problem is this easy and you can see the answer, you don't need to go to any particular trouble to solve it.

Rearranging slightly harder equations

When you can't see an answer just by looking at a problem, sometimes rearranging the problem helps to turn it into one that you can solve using a Big Four operation. For example:

· 6x = 96

You can rearrange this problem using inverse operations, as I show you in Chapter 4, changing multiplication to division: Now solve the problem by division (long or otherwise) to find that x = 16.

Guessing and checking equations

You can solve some equations by guessing an answer and then checking to see whether you're right. For example, suppose you want to solve the following equation:

· 3x + 7 = 19

To find out what x equals, start by guessing that x = 2. Now check to see whether you're right by substituting 2 for x in the equation:

· 3(2) + 7 = 13 WRONG! (13 is less than 19.)

· 3(5) + 7 = 22 19 WRONG! (22 is greater than 19.)

· 3(4) + 7 = 19 RIGHT!

With only three guesses, you found that x = 4.

Applying algebra to more difficult equations

When an algebraic equation gets hard enough, you find that looking at it and rearranging it just isn't enough to solve it. For example:

· 11x – 13 = 9x + 3

You probably can't tell what x equals just by looking at this problem. You also can't solve it just by rearranging it, using an inverse operation. And guessing and checking would be very tedious. Here's where algebra comes into play.

Algebra is especially useful because you can follow mathematical rules to find your answer. Throughout the rest of this chapter, I show you how to use the rules of algebra to turn tough problems like this one into problems that you can solve.

The Balancing Act: Solving for x

As I show you in the preceding section, some problems are too complicated to find out what the variable (usually x) equals just by eyeballing it or rearranging it. For these problems, you need a reliable method for getting the right answer. I call this method the balance scale.

The balance scale allows you to solve for x — that is, find the number that x stands for — in a step-by-step process that always works. In this section, I show you how to use the balance scale method to solve algebraic equations.

Striking a balance The equals sign in any equation means that both sides balance. To keep that equals sign, you have to maintain that balance. In other words, whatever you do to one side of an equation, you have to do to the other.

For example, here's a balanced equation:

· Illustration by Wiley, Composition Services Graphics

If you add 1 to one side of the equation, the scale goes out of balance.

· Illustration by Wiley, Composition Services Graphics

But if you add 1 to both sides of the equation, the scale stays balanced:

· Illustration by Wiley, Composition Services Graphics

You can add any number to the equation, as long as you do it to both sides. And in math, any number means x:

· 1 + 2 + x = 3 + x

Remember that x is the same wherever it appears in a single equation or problem.

This idea of changing both sides of an equation equally isn't limited to addition. You can just as easily subtract an x, or even multiply or divide by x, as long as you do the same to both sides of the equation: Using the balance scale to isolate x

The simple idea of balance is at the heart of algebra, and it enables you to find out what x is in many equations. When you solve an algebraic equation, the goal is to isolate x — that is, to get x alone on one side of the equation and some number on the other side. In algebraic equations of middling difficulty, this is a three-step process:

1. Get all constants (non-x terms) on one side of the equation.

2. Get all x-terms on the other side of the equation.

3. Divide to isolate x.

For example, take a look at the following problem:

· 11x – 13 = 9x + 3

As you follow the steps, notice how I keep the equation balanced at each step:

1. Get all the constants on one side of the equation by adding 13 to both sides of the equation: Because you've obeyed the rules of the balance scale, you know that this new equation is also correct. Now the only non-x term (16) is on the right side of the equation.

2. Get all the x-terms on the other side by subtracting 9x from both sides of the equation: Again, the balance is preserved, so the new equation is correct.

3. Divide by 2 to isolate x: To check this answer, you can simply substitute 8 for x in the original equation: This checks out, so 8 is the correct value of x.

Rearranging Equations and Isolating x

When you understand how algebra works like a balance scale, as I show you in the preceding section, you can begin to solve more-difficult algebraic equations. The basic tactic is always the same: Changing both sides of the equation equally at every step, try to isolate x on one side of the equation.

In this section, I show you how to put your skills from Chapter 21 to work solving equations. First, I show you how rearranging the terms in an expression is similar to rearranging them in an algebraic equation. Next, I show you how removing parentheses from an equation can help you solve it. Finally, you discover how cross-multiplication is useful for solving algebraic equations with fractions.

Rearranging terms on one side of an equation

Rearranging terms becomes all-important when working with equations. For example, suppose you're working with this equation:

· 5x – 4 = 2x + 2

When you think about it, this equation is really two expressions connected with an equals sign. And of course, that's true of every equation. That's why everything you find out about expressions in Chapter 21 is useful for solving equations. For example, you can rearrange the terms on one side of an equation. So here's another way to write the same equation:

· – 4 + 5x = 2x + 2

And here's a third way:

· – 4 + 5x = 2 + 2x

This flexibility to rearrange terms comes in handy when you're solving equations.

Moving terms to the other side of the equals sign

Earlier in this chapter, I show you how an equation is similar to a balance scale. For example, take a look at Figure 22-1. Illustration by Wiley, Composition Services Graphics

Figure 22-1: Showing how an equation is similar to a balance scale.

To keep the scale balanced, if you add or remove anything on one side, you must do the same on the other side. For example: Now take a look at these two versions of this equation side by side:

· 2x – 3 = 11   –3 = 11 – 2x

In the first version, the term 2x is on the left side of the equals sign. In the second, the term –2x is on the right side. This example illustrates an important rule. When you move any term in an expression to the other side of the equals sign, change its sign (from plus to minus or from minus to plus).

As another example, suppose you're working with this equation:

· 4x – 2 = 3x + 1

You have x's on both sides of the equation, so say you want to move the 3x. When you move the term 3x from the right side to the left side, you have to change its sign from plus to minus (technically, you're subtracting 3x from both sides of the equation).

· 4x – 2 – 3x = 1

After that, you can simplify the expression on the left side of the equation by combining like terms:

· x – 2 = 1

At this point, you can probably see that x = 3 because 3 – 2 = 1. But just to be sure, move the –2 term to the right side and change its sign: To check this result, substitute a 3 wherever x appears in the original equation: As you can see, moving terms from one side of an equation to the other can be a big help when you're solving equations.

Removing parentheses from equations

Chapter 21 gives you a treasure trove of tricks for simplifying expressions, and they come in handy when you're solving equations. One key skill from that chapter is removing parentheses from expressions. This tactic is also indispensable when you're solving equations.

For example, suppose you have the following equation:

· 5x + (6x – 15) = 30 – (x – 7) + 8

Your mission is to get all the x terms on one side of the equation and all the constants on the other. As the equation stands, however, x terms and constants are “locked together” inside parentheses. In other words, you can't isolate the x terms from the constants. So before you can isolate terms, you need to remove the parentheses from the equation.

Recall that an equation is really just two expressions connected by an equals sign. So you can start working with the expression on the left side. In this expression, the parentheses begin with a plus sign (+), so you can just remove them:

· 5x + 6x – 15 = 30 – (x – 7) + 8

Now move on to the expression on the right side. This time, the parentheses come right after a minus sign (–). To remove them, change the sign of both terms inside the parentheses: x becomes –x, and –7 becomes 7:

· 5x + 6x – 15 = 30 x + 7 + 8

Bravo! Now you can isolate x terms to your heart's content. Move the –x from the right side to the left, changing it to x:

· 5x + 6x – 15 + x = 30 + 7 + 8

Next, move –15 from the left side to the right, changing it to 15:

· 5x + 6x + x = 30 + 7 + 8 + 15

Now combine like terms on both sides of the equation: Finally, get rid of the coefficient 12 by dividing: As usual, you can check your answer by substituting 5 into the original equation wherever x appears: Here's one more example:

· 11 + 3(–3x + 1) = 25 – (7x – 3) – 12

As in the preceding example, start out by removing both sets of parentheses. This time, however, on the left side of the equation, you have no sign between 3 and (–3x + 1). But again, you can put your skills from Chapter 21 to use. To remove the parentheses, multiply 3 by both terms inside the parentheses:

· 11 – 9x + 3 = 25 – (7x – 3) – 12

On the right side, the parentheses begin with a minus sign, so remove the parentheses by changing all the signs inside the parentheses:

· 11 – 9x + 3 = 25 – 7x + 3 – 12

Now you're ready to isolate the x terms. I do this in one step, but take as many steps as you want:

· –9x + 7x = 25 + 3 – 12 – 11 – 3

At this point, you can combine like terms:

· –2x = 2

To finish, divide both sides by –2:

· x = –1

Copy this example, and work through it a few times with the book closed.

Cross-multiplying

In algebra, cross-multiplication helps to simplify equations by removing unwanted fractions (and, honestly, when are fractions ever wanted?). As I discuss in Chapter 9, you can use cross-multiplication to find out whether two fractions are equal. You can use this same idea to solve algebra equations with fractions, like this one: This equation looks hairy. You can't do the division or cancel anything out because the fraction on the left has two terms in the denominator, and the fraction on the right has two terms in the numerator (see Chapter 21 for info on dividing algebraic terms). However, an important piece of information that you have is that the fraction equals the fraction. So if you cross-multiply these two fractions, you get two results that are also equal:

· x(4x) = (2x + 3)(2x – 2)

At this point, you have something you know how to work with. The left side is easy: The right side requires a bit of FOILing (flip to Chapter 21 for details): Now all the parentheses are gone, so you can isolate the x terms. Because most of these terms are already on the right side of the equation, isolate them on that side: Combining like terms gives you a pleasant surprise:

· 6 = 2x

The two x2 terms cancel each other out. You may be able to eyeball the correct answer, but here's how to finish:  