Basic Math & Pre-Algebra For Dummies, 2nd Edition (2014)
Part V. The X-Files: Introduction to Algebra
For more info on simplifying algebraic expressions and to discover a trick that makes factoring quadratic polynomials easy, go to www.dummies.com/extras/basicmathandprealgebra.
In this part…
· Evaluate, simplify, and factor algebraic expressions
· Keep algebraic equations balanced and solve them by isolating the variable
· Use algebra to solve word problems too difficult to solve with just arithmetic
Chapter 21. Enter Mr. X: Algebra and Algebraic Expressions
In This Chapter
Meeting Mr. X head-on
Understanding how a variable such as x stands for a number
Using substitution to evaluate an algebraic expression
Identifying and rearranging the terms in any algebraic expression
Simplifying algebraic expressions
You never forget your first love, your first car, or your first x. Unfortunately for some folks, remembering their first x in algebra is similar to remembering their first love who stood them up at the prom or their first car that broke down someplace in Mexico.
The most well-known fact about algebra is that it uses letters — like x — to represent numbers. So if you have a traumatic x-related tale, all I can say is that the future will be brighter than the past.
What good is algebra? That question is a common one, and it deserves a decent answer. Algebra is used for solving problems that are just too difficult for ordinary arithmetic. And because number crunching is so much a part of the modern world, algebra is everywhere (even if you don't see it): architecture, engineering, medicine, statistics, computers, business, chemistry, physics, biology, and, of course, higher math. Anywhere that numbers are useful, algebra is there. That fact is why virtually every college and university insists that you leave (or enter) with at least a passing familiarity with algebra.
In this chapter, I introduce (or reintroduce) you to that elusive little fellow, Mr. X, in a way that's bound to make him seem a little friendlier. Then I show you how algebraic expressions are similar to and different from the arithmetic expressions that you're used to working with. (For a refresher on arithmetic expressions, see Chapter 5.)
Seeing How X Marks the Spot
In math, x stands for a number — any number. Any letter that you use to stand for a number is a variable, which means that its value can vary — that is, its value is uncertain. In contrast, a number in algebra is often called a constant because its value is fixed.
Sometimes you have enough information to find out the identity of x. For example, consider the following:
· 2 + 2 = x
Obviously, in this equation, x stands for the number 4. But other times, what the number x stands for stays shrouded in mystery. For example:
· x > 5
In this inequality, x stands for some number greater than 5 — maybe 6, maybe , maybe 542.002.
Expressing Yourself with Algebraic Expressions
In Chapter 5, I introduce you to arithmetic expressions: strings of numbers and operators that can be evaluated or placed on one side of an equation. For example:
In this chapter, I introduce you to another type of mathematical expression: the algebraic expression. An algebraic expression is any string of mathematical symbols that can be placed on one side of an equation and that includes at least one variable.
Here are a few examples of algebraic expressions:
As you can see, the difference between arithmetic and algebraic expressions is simply that an algebraic expression includes at least one variable.
In this section, I show you how to work with algebraic expressions. First, I demonstrate how to evaluate an algebraic expression by substituting the values of its variables. Then I show you how to separate an algebraic expression into one or more terms, and I walk through how to identify the coefficient and the variable part of each term.
Evaluating algebraic expressions
To evaluate an algebraic expression, you need to know the numerical value of every variable. For each variable in the expression, substitute, or plug in, the number that it stands for and then evaluate the expression.
In Chapter 5, I show you how to evaluate an arithmetic expression. Briefly, this means finding the value of that expression as a single number (flip to Chapter 5 for more on evaluating).
Knowing how to evaluate arithmetic expressions comes in handy for evaluating algebraic expressions. For example, suppose you want to evaluate the following expression:
· 4x – 7
Note that this expression contains the variable x, which is unknown, so the value of the whole expression is also unknown.
An algebraic expression can have any number of variables, but you usually don't work with expressions that have more than two or maybe three, at the most. You can use any letter as a variable, but x, y, and z tend to get a lot of mileage.
Suppose in this case that x = 2. To evaluate the expression, substitute 2 for x everywhere it appears in the expression:
· 4(2) – 7
After you make the substitution, you're left with an arithmetic expression, so you can finish your calculations to evaluate the expression:
· = 8 – 7 = 1
So given x = 2, the algebraic expression 4x – 7 = 1.
Now suppose you want to evaluate the following expression, where x = 4:
Again, the first step is to substitute 4 for x everywhere this variable appears in the expression:
Now evaluate according to the order of operations explained in Chapter 5. You do powers first, so begin by evaluating the exponent 4^{2}, which equals 4 × 4:
Now proceed to the multiplication, moving from left to right:
Then evaluate the subtraction, again from left to right:
So given x = 4, the algebraic expression 2x^{2} – 5x – 15 = –3.
You aren't limited to expressions of only one variable when using substitution. As long as you know the value of every variable in the expression, you can evaluate algebraic expressions with any number of variables. For example, suppose you want to evaluate this expression:
To evaluate it, you need the values of all three variables:
The first step is to substitute the equivalent value for each of the three variables wherever you find them:
Now use the rules for order of operations from Chapter 5. Begin by evaluating the exponent 3^{2}:
Next, evaluate the multiplication from left to right (if you need to know more about the rules for multiplying negative numbers, check out Chapter 4):
· = 27 + (–12) – (–30)
Now all that's left is addition and subtraction. Evaluate from left to right, remembering the rules for adding and subtracting negative numbers in Chapter 4:
· = 15 – (–30) = 15 + 30 = 45
So given the three values for x, y, and z, the algebraic expression 3x^{2} + 2xy – xyz = 45.
For practice, copy this expression and the three values on a separate piece of paper, close the book, and see whether you can substitute and evaluate on your own to get the same answer.
Coming to algebraic terms
A term in an algebraic expression is any chunk of symbols set off from the rest of the expression by either addition or subtraction. As algebraic expressions get more complex, they begin to string themselves out in more terms. Here are some examples:
Expression |
Number of Terms |
Terms |
5x |
One |
5x |
–5x + 2 |
Two |
–5x and 2 |
Four |
, , –xyz, and 8 |
No matter how complicated an algebraic expression gets, you can always separate it out into one or more terms.
When separating an algebraic expression into terms, group the plus or minus sign with the term that it immediately precedes.
When a term has a variable, it's called an algebraic term. When it doesn't have a variable, it's called a constant. For example, look at the following expression:
The first three terms are algebraic terms, and the last term is a constant. As you can see, in algebra, constant is just a fancy word for number.
Terms are really useful to know about because you can follow rules to move them, combine them, and perform the Big Four operations on them. All these skills are important for solving equations, which I explain in the next chapter. But for now, this section explains a bit about terms and some of their traits.
Making the commute: Rearranging your terms
When you understand how to separate an algebraic expression into terms, you can go one step further by rearranging the terms in any order you like. Each term moves as a unit, kind of like a group of people carpooling to work together — everyone in the car stays together for the whole ride.
For example, suppose you begin with the expression –5x + 2. You can rearrange the two terms of this expression without changing its value. Notice that each term's sign stays with that term, although dropping the plus sign at the beginning of an expression is customary:
· = 2 – 5x
Rearranging terms in this way doesn't affect the value of the expression because addition is commutative — that is, you can rearrange things that you're adding without changing the answer. (See Chapter 4 for more on the commutative property of addition.)
For example, suppose x = 3. Then the original expression and its rearrangement evaluate as follows (using the rules that I outline earlier in “Evaluating algebraic expressions”):
Rearranging expressions in this way becomes handy later in this chapter, when you simplify algebraic expressions. As another example, suppose you have this expression:
· 4x – y + 6
You can rearrange it in a variety of ways:
Because the term 4x has no sign, it's positive, so you can write in a plus sign as needed when rearranging terms.
As long as each term's sign stays with that term, rearranging the terms in an expression has no effect on its value.
For example, suppose that x = 2 and y = 3. Here's how to evaluate the original expression and the two rearrangements:
Identifying the coefficient and variable
Every term in an algebraic expression has a coefficient. The coefficient is the signed numerical part of a term in an algebraic expression — that is, the number and the sign (+ or –) that goes with that term. For example, suppose you're working with the following algebraic expression:
The following table shows the four terms of this expression, with each term's coefficient:
Term |
Coefficient |
Variable |
–4x^{3} |
–4 |
x^{3} |
x^{2} |
1 |
x^{2} |
–x |
–1 |
x |
–7 |
–7 |
none |
Notice that the sign associated with the term is part of the coefficient. So the coefficient of –4x^{3} is –4.
When a term appears to have no coefficient, the coefficient is actually 1. So the coefficient of x^{2} is 1, and the coefficient of –x is –1. And when a term is a constant (just a number), that number with its associated sign is the coefficient. So the coefficient of the term –7 is simply –7.
By the way, when the coefficient of any algebraic term is 0, the expression equals 0 no matter what the variable part looks like:
·
0x = 0 |
0xyz = 0 |
0x^{3}y^{4}z^{10} = 0 |
In contrast, the variable part of an expression is everything except the coefficient. The previous table shows the four terms of the same expression, with each term's variable part.
Identifying like terms
Like terms (or similar terms) are any two algebraic terms that have the same variable part — that is, both the letters and their exponents have to be exact matches. Here are some examples:
As you can see, in each example, the variable part in all three like terms is the same. Only the coefficient changes, and it can be any real number: positive or negative, whole number, fraction, or decimal — or even an irrational number such as π. (For more on real numbers, see Chapter 25.)
Considering algebraic terms and the Big Four
In this section, I get you up to speed on how to apply the Big Four to algebraic expressions. For now, just think of working with algebraic expressions as a set of tools that you're collecting, for use when you get on the job. You find how useful these tools are in Chapter 22, when you begin solving algebraic equations.
Adding terms
Add like terms by adding their coefficients and keeping the same variable part.
For example, suppose you have the expression 2x + 3x. Remember that 2x is just shorthand for x + x, and 3x means simply x + x + x. So when you add them up, you get the following:
· = x + x + x + x + x = 5x
As you can see, when the variable parts of two terms are the same, you add these terms by adding their coefficients: 2x + 3x = (2 + 3)x. The idea here is roughly similar to the idea that 2 apples + 3 apples = 5 apples.
You cannot add nonlike terms. Here are some cases in which the variables or their exponents are different:
In these cases, you can't simplify the expression. You're faced with a situation that's similar to 2 apples + 3 oranges. Because the units (apples and oranges) are different, you can't combine terms. (See Chapter 4 for more on how to work with units.)
Subtracting terms
Subtraction works much the same as addition. Subtract like terms by finding the difference between their coefficients and keeping the same variable part.
For example, suppose you have 3x – x. Recall that 3x is simply shorthand for x + x + x. So doing this subtraction gives you the following:
· x + x + x – x = 2x
No big surprises here. You simply find (3 – 1)x. This time, the idea roughly parallels the idea that $3 – $1 = $2.
Here's another example:
· 2x – 5x
Again, no problem, as long as you know how to work with negative numbers (see Chapter 4 if you need details). Just find the difference between the coefficients:
· = (2 – 5)x = –3x
In this case, recall that $2 – $5 = –$3 (that is, a debt of $3).
You cannot subtract nonlike terms. For example, you can't subtract either of the following:
As with addition, you can't do subtraction with different variables. Think of this as trying to figure out $7 – 4 pesos. Because the units in this case (dollars versus pesos) are different, you're stuck. (See Chapter 4 for more on working with units.)
Multiplying terms
Unlike adding and subtracting, you can multiply nonlike terms. Multiply any two terms by multiplying their coefficients and combining — that is, by collecting or gathering up — all the variables in each term into a single term, as I show you next.
For example, suppose you want to multiply 5x(3y). To get the coefficient, multiply 5 × 3. To get the algebraic part, combine the variables x and y:
· = 5(3)xy = 15xy
Now suppose you want to multiply 2x(7x). Again, multiply the coefficients and collect the variables into a single term:
· = 7(2)xx = 14xx
Remember that x^{2} is shorthand for xx, so you can write the answer more efficiently:
· =14x^{2}
Here's another example. Multiply all three coefficients together and gather up the variables:
As you can see, the exponent 3 that's associated with x is just the count of how many x's appear in the problem. The same is true of the exponent 2 associated with y.
A fast way to multiply variables with exponents is to add the exponents together. For example:
In this example, I added the exponents of the x's (4 + 2 + 6 = 12) to get the exponent of x in the expression. Similarly, I added the exponents of the y's (3 + 5 + 1 = 9 — don't forget that y = y^{1}!) to get the exponent of y in the expression.
Dividing terms
It's customary to represent division of algebraic expressions as a fraction instead of using the division sign (÷). So division of algebraic terms really looks like reducing a fraction to lowest terms (see Chapter 9 for more on reducing).
To divide one algebraic term by another, follow these steps:
1. Make a fraction of the two terms.
Suppose you want to divide 3xy by 12x^{2}. Begin by turning the problem into a fraction:
2. Cancel out factors in coefficients that are in both the numerator and the denominator.
In this case, you can cancel out a 3. Notice that when the coefficient in xy becomes 1, you can drop it:
3. Cancel out any variable that's in both the numerator and the denominator.
You can break x^{2} out as xx:
Now you can clearly cancel an x in both the numerator and the denominator:
As you can see, the resulting fraction is really a reduced form of the original.
As another example, suppose you want to divide –6x^{2}yz^{3} by –8x^{2}y^{2}z. Begin by writing the division as a fraction:
First, reduce the coefficients. Notice that, because both coefficients were originally negative, you can cancel out both minus signs as well:
Now you can begin canceling variables. I do this in two steps, as before:
At this point, just cross out any occurrence of a variable that appears in both the numerator and the denominator:
You can't cancel out variables or coefficients if either the numerator or the denominator has more than one term in it. This is a very common mistake in algebra, so don't let it happen to you!
Simplifying Algebraic Expressions
As algebraic expressions grow more complex, simplifying them can make them easier to work with. Simplifying an expression means (quite simply!) making it smaller and easier to manage. You see how important simplifying expressions becomes when you begin solving algebraic equations.
For now, think of this section as a kind of algebra toolkit. Here I show you how to use these tools. In Chapter 22, I show you when to use them.
Combining like terms
When two algebraic terms contain like terms (when their variables match), you can add or subtract them (see the earlier section “Considering algebraic terms and the Big Four”). This feature comes in handy when you're trying to simplify an expression. For example, suppose you're working with the following expression:
· 4x – 3y + 2x + y – x + 2y
As it stands, this expression has six terms. But three terms have the variable x and the other three have the variable y. Begin by rearranging the expression so that all like terms are grouped together:
· = 4x + 2x – x – 3y + y + 2y
Now you can add and subtract like terms. I do this in two steps, first for the x terms and then for the y terms:
Notice that the x terms simplify to 5x, and the y terms simplify to 0y, which is 0, so the y terms drop out of the expression altogether.
Here's a somewhat more complicated example that has variables with exponents:
This time, you have four different types of terms. As a first step, you can rearrange these terms so that groups of like terms are all together (I underline these four groups so you can see them clearly):
Now combine each set of like terms:
This time, the x^{2} terms add up to 0, so they drop out of the expression altogether:
· = 5x + 9xy + 8y
Removing parentheses from an algebraic expression
Parentheses keep parts of an expression together as a single unit. In Chapter 5, I show you how to handle parentheses in an arithmetic expression. This skill is also useful with algebraic expressions. As you find when you begin solving algebraic equations in Chapter 22, getting rid of parentheses is often the first step toward solving a problem. In this section, I show how to handle the Big Four operations with ease.
Drop everything: Parentheses with a plus sign
When an expression contains parentheses that come right after a plus sign (+), you can just remove the parentheses. Here's an example:
Now you can simplify the expression by combining like terms:
· = 5x + 4y
When the first term inside the parentheses is negative, when you drop the parentheses, the minus sign replaces the plus sign. For example:
Sign turnabout: Parentheses with a minus sign
Sometimes an expression contains parentheses that come right after a minus sign (–). In this case, change the sign of every term inside the parentheses to the opposite sign; then remove the parentheses.
Consider this example:
· 6x – (2xy – 3y) + 5xy
A minus sign is in front of the parentheses, so you need to change the signs of both terms in the parentheses and remove the parentheses. Notice that the term 2xy appears to have no sign because it's the first term inside the parentheses. This expression really means the following:
· = 6x – (+2xy – 3y) + 5xy
You can see how to change the signs:
· = 6x – 2xy + 3y + 5xy
At this point, you can combine the two xy terms:
· = 6x + 3xy + 3y
Distribution: Parentheses with no sign
When you see nothing between a number and a set of parentheses, it means multiplication. For example,
·
2(3) = 6 |
4(4) = 16 |
10(15) = 150 |
This notation becomes much more common with algebraic expressions, replacing the multiplication sign (×) to avoid confusion with the variable x:
·
3(4x) = 12x |
4x(2x) = 8x^{2} |
3x(7y) = 21xy |
To remove parentheses without a sign, multiply the term outside the parentheses by every term inside the parentheses; then remove the parentheses. When you follow those steps, you're using the distributive property.
Here's an example:
· 2(3x – 5y + 4)
In this case, multiply 2 by each of the three terms inside the parentheses:
· = 2(3x) + 2(–5y) + 2(4)
For the moment, this expression looks more complex than the original one, but now you can get rid of all three sets of parentheses by multiplying:
· = 6x – 10y + 8
Multiplying by every term inside the parentheses is simply distribution of multiplication over addition — also called the distributive property — which I discuss in Chapter 4.
As another example, suppose you have the following expression:
Begin by multiplying –2x by the three terms inside the parentheses:
The expression looks worse than when you started, but you can get rid of all the parentheses by multiplying:
Now you can combine like terms:
· = x^{2} – 12x
Parentheses by FOILing
Sometimes, expressions have two sets of parentheses next to each other without a sign between them. In that case, you need to multiply every term inside the first set by every term inside the second.
When you have two terms inside each set of parentheses, you can use a process called FOILing. This is really just the distributive property, as I show you below. The word FOIL is an acronym to help you make sure you multiply the correct terms. It stands for First, Outside, Inside, and Last.
Here's how the process works. In this example, you're simplifying the expression (2x – 2)(3x – 6):
1. Start out by multiplying the two First terms in the parentheses.
The first term in the first set of parentheses is 2x, and 3x is the first term in the second set of parentheses: (2x – 2)(3x – 6).
o F: Multiply the first terms:
2. Multiply the two Outside terms.
The two outside terms, 2x and –6, are on the ends: (2x – 2)(3x – 6 )
o O: Multiply the outside terms: 2x(– 6) = –12x
3. Multiply the two Inside terms.
The two terms in the middle are –2 and 3x: (2x – 2)(3x – 6)
o I: Multiply the middle terms: –2(3x) = –6x
4. Multiply the two Last terms.
The last term in the first set of parentheses is –2, and –6 is the last term in the second set: (2x – 2)(3x – 6)
o L: Multiply the last terms: –2(–6) = 12
Add these four results together to get the simplified expression:
In this case, you can simplify this expression still further by combining the like terms –12x and –6x:
Notice that, during this process, you multiply every term inside one set of parentheses by every term inside the other set. FOILing just helps you keep track and make sure you've multiplied everything.
FOILing is really just an application of the distributive property, which I discuss in the section preceding this one. In other words, (2x – 2)(3x – 6) is really the same as 2x(3x – 6) + –2(3x – 6) when distributed. Then distributing again gives you .