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Section 0.0 Preliminaries

Subsection 0.0.1 Integer Addition and Subtraction

Here is a picture of the integers on a number line:

The number line actually goes on forever in both directions. We designate one point to represent “zero”, \(0,\) and use it as a reference point. Next, we choose a “unit” of measure. The number \(1\) is one “unit” to the right of zero. The number \(2\) is a distance of two units to the right of zero and so on. The positive integers, \(1, 2, 3, \dots\text{,}\) represent whole units of distances to the right of \(0.\) The negative integers, \(-1, -2, -3, \dots\text{,}\) represent whole units of distances to the left of \(0.\) Some facts to note are:

  • The integer \(0\) is neither positive nor negative. It is not to the right or left of itself.

  • As we move to the right on the number line, the integers increase in value.

    • \(3\) is greater than \(2\text{.}\) Note that it is to the right of \(2\) on the number line.
    • But \(-2\) is greater than \(-3\) since it is to the right of \(-3\) on the number line.

Recall that adding a positive integer results in moving to the right \(\rightarrow\) on the number line.

Example 0.0.1. Add \(3\) to \(1\).

Plot \(1+3\text{:}\) Start at \(1\) on the number line

Move three places to the right

Our Solution: \(4\checkmark\) (We knew that!)

Similarly, adding a negative integer results in moving to the left \(\leftarrow\) on the number line.

Example 0.0.2. Add \(-2\) to \(-1\).

Plot \((-1)+(-2)\text{:}\) Start at \(-1\) on the number line

Move two places to left

Our Solution: \(-3\,\checkmark\)

Notice in Example 0.0.2 that adding two negative numbers is equivalent to adding the two corresponding positive numbers, but the final result is negative: \(1+2=3\) and \(-1+(-2)=-3\text{.}\)

Example 0.0.3. Add Two Negative Integers.
\begin{align*} \text{Compute }-4+(-11) \amp\amp\amp \text{Adding two negatives: add positives}\\ 4+11=15\amp\amp\amp\text{Our result should be negative}\\ -4+(-11)=-15 \amp\amp\amp \text{Our Solution}\,\checkmark \end{align*}

If you were to sketch out a number line to solve Example 0.0.3, you would start at \(-4\) then move left \(11\) positions. The result would end up at \(-15\text{.}\)

Now let's consider adding two numbers with opposite signs.

Example 0.0.4. Add \(-3\) to \(1\).

Plot \(1+(-3)\text{:}\) Start at \(1\) on the number line

Move three places to left

Our Solution: \(-2\checkmark\)

When adding integers with opposite signs use the corresponding positive integers, we subtract the smaller from the larger, then use the sign from the larger number in our result. This means if the larger number is positive, the answer is positive. If the larger number is negative, the answer is negative. Consider the following examples:

Example 0.0.5. Add Integers of Different Signs.
\begin{align*} \text{Compute }-5+2 \amp\amp\amp \text{Different signs: subtract first}\\ 5-2=3\amp\amp\amp\text{use sign from bigger number (\(5\)), negative}\\ -3 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.6. Add Integers of Different Signs.
\begin{align*} \text{Compute }-14+20 \amp\amp\amp \text{Different signs: subtract first}\\ 20-14=6\amp\amp\amp\text{use sign from bigger number (\(20\)), positive}\\ 6 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

To subtract a signed number, we add the opposite of the number after the subtraction sign using the rules for addition described above.

Example 0.0.7. Subtract Integers: add the opposite.
\begin{align*} \text{Compute }7-(-6) \amp\amp\amp \text{ Add the opposite of \(-6\)}\\ 7+(6) \amp\amp\amp \text{Addition of positive integers}\\ 13 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.8. Subtract Integers: add the opposite.
\begin{align*} \text{Compute }6-8 \amp\amp\amp \text{Add the opposite of \(8\)}\\ 6+(-8) \amp\amp\amp \text{Different signs: subtract}\\ 8-6=2\amp\amp\amp\text{use sign from bigger number (\(8\)), negative}\\ -2 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Checkpoint 0.0.9.

Subsection 0.0.2 Integer Multiplication and Division

Think of multiplying positive integers as repeated addition. Note that we could add \(4\) three times:

\begin{align*} 3 \cdot 4 \amp= 4 + 4 + 4\\ \amp= 12. \end{align*}

Or add \(3\) four times:

\begin{align*} 3 \cdot 4 \amp= 3 + 3 + 3 + 3 \\ \amp= 12. \end{align*}

How does this work with negative integers? Think of repeated subtraction. We could subtract \(2\) five times

\begin{align*} -5 \cdot 2 \amp= -(2)-(2)-(2)-(2)-(2) \\ \amp= -10 \end{align*}

or add \(-5\) two times

\begin{align*} -5 \cdot 2 \amp= (-5)+(-5)\\ \amp= -10. \end{align*}

With two negative integers, we could either subtract \(-4\) twice

\begin{align*} -4 \cdot -2 \amp= -(-4)-(-4) \\ \amp= 4 + 4 \\ \amp= 8 \end{align*}

or

\begin{align*} -4 \cdot -2 \amp= -(-2)-(-2)-(-2)-(-2) \\ \amp= 2 + 2 + 2 + 2\\ \amp= 8. \end{align*}

Notice the pattern when multiplying: The product of two numbers with the same sign is positive. The product of two numbers with opposite signs is negative. We do not need to do the multiplcation by writing out the additions or subtractions. Just multiply the numbers and make the result negative if one number is positive and the other negative. Otherwise, the result is positive.

Example 0.0.10. Multiply.
\begin{align*} \text{Compute }(4)(-6) \amp\amp\amp \text{Signs do not match, answer is negative}\\ -24 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.11. Multiply.
\begin{align*} \text{Compute }(-2)(-6) \amp\amp\amp \text{Signs match, answer is positive}\\ 12 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

The rules for division are the same as for multiplication. To see why, consider \(6 \div -2.\) The result of the division is the integer which when multiplied by \(-2\) produces \(6.\) To get a positive integer when multiplying a negative integer requires multiplication by another negative integer. Since \(-2 \cdot -3 = 6,\) the answer to the division is \(6 \div -2 = -3.\) Similarily, \(-6 \div -2 = 3\) since \(-2 \cdot 3 = -6.\)

Hence, the quotient of two numbers with the same sign is positive. The quotient of two numbers with opposite signs is negative.

Example 0.0.12. Divide.
\begin{align*} \text{Compute }-36\div -9 \amp\amp\amp \text{Signs match, answer is positive}\\ \amp\amp\amp\text{(Note that we could also write \(\frac{-36}{-9}\) for division)}\\ 4 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.13. Divide.
\begin{align*} \text{Compute }15\div -3 \amp\amp\amp \text{Signs do not match, answer is negative}\\ -5 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Checkpoint 0.0.14.

There is a special case we must consider. The integer zero is neither positive nor negative. It is neither to the right nor left of itself. When we consider multiplication by zero, it means add the other integer to itself zero times. If there is nothing to add, the result is zero. For example: \(0 \cdot 3 = 0.\) Or we could add zero to itself \(3\) times, \(3 \cdot 0 = 0 + 0 + 0 = 0.\) Either way, multiplication by zero results in zero.

What about dividing with zero? For \(0 \div 5,\) since \(5 \cdot 0 = 0\) the answer is \(0 \div 5 = 0.\) However, if we divide by zero we have a problem. Consider \(8 \div 0.\) The answer must be the integer that multiplies zero to give \(8.\) No number can multiply \(0\) to give \(8.\) Since there is no answer for \(8 \div 0,\) it is customary to say expression is not defined.

Checkpoint 0.0.15.

Exponents are used to denote repeated multiplication of a number times itself. We call the number the base of the exponent.

Example 0.0.16. Exponents: Evaluate \(6^2\).
\begin{align*} 6^2\amp=(6)(6) \amp\amp \text{The exponent \(2\) means the base \(6\) is multiplied by itself \(2\) times}\\ \amp\amp\amp\text{(Note that we could also write \(6\cdot6\) for multiplication)}\\ \amp=36 \amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.17. Exponents: Evaluate \((-5)^4\).
\begin{align*} (-5)^4\amp=(-5)(-5)(-5)(-5) \amp\amp \text{The exponent \(4\) means the base \(-5\) is multiplied by itself four times.}\\ \amp\amp\amp\text{(there are \(4\) negatives)}\\ \amp=125 \amp\amp \text{Our Solution}\checkmark \end{align*}

Compare the example above with the next example and note the difference the parentheses make.

Example 0.0.18. Exponents: Evaluate \(-5^4\).
\begin{align*} -5^4\amp=-(5)(5)(5)(5) \amp\amp \text{The exponent \(4\) means the base \(5\) is multiplied by itself four times.}\\ \amp\amp\amp\text{The minus sign is applied after exponentiation.}\\ \amp=-125 \amp\amp \text{Our Solution}\checkmark \end{align*}

Be careful when working with integers. For example, compare the expression \(-3-8\) with \(-3(-8)\text{.}\) The second expression is a product. If there is no operation symbol between parentheses, we assume we are multiplying. On the other hand, the expression \(-3-8\) is subtraction.

Also, be careful not to mix up the rules for adding and subtracting integers with the rules for multiplying and dividing integers. For example, \(-3+(-7)=-10\text{,}\) but \((-3)(-7)=21\text{.}\)

Subsection 0.0.3 Operations of Numbers

When simplifying an expression it is important that we perform the operations in the correct order. Consider the following problem done two different ways:

Example 0.0.19. WARNING! 0nly one below solution is correct:.
\begin{align*} \underbrace{2+5} \cdot 3 \amp\amp\amp \text{Add First} \amp\amp\amp \color{red}{\text{versus}} \amp\amp\amp 2+\underbrace{5 \cdot 3} \amp\amp\amp \text{Multiply First}\\ 7 \cdot 3 \amp\amp\amp \text{Then Multiply} \amp\amp\amp\amp\amp\amp 2+15\amp\amp\amp \text{Then Add}\\ 21 \amp\amp\amp \text{Solution??} \amp\amp\amp\amp\amp\amp 17 \amp\amp\amp \text{Solution??} \end{align*}

The previous Example 0.0.19 illustrates that if the same problem is done two different ways we will arrive at two different solutions. However, only one method can be correct. It turns out the second method, with solution \(17\text{,}\) is the correct method. This list then is the order of operations we will use to simplify expressions.

Order of Operations:
parentheses (Grouping)
Exponents
Multiply and Divide (Left to Right)
Add and Subtract (Left to Right)

Multiply and Divide are on the same level because they are the same operation (division is just multiplying by the reciprocal). This means they must be done left to right, so some problems we will divide first, others we will multiply first. The same is true for adding and subtracting (subtracting is just adding the opposite). Often students use the word PEMDAS to remember the order of operations, as the first letter of each operation creates the word PEMDAS. However, it is the author's suggestion to think about PEMDAS as a vertical word written as:

\(P\)
\(E\)
\(M\)
\(D\)
\(A\)
\(S\)

so we don't forget that multiplication and division are done left to right (same with addition and subtraction). Another way students remember the order of operations is to think of a phrase such as “Please Excuse My Dear Aunt Sally” where each word starts with the same letters as the order of operations start with.

Example 0.0.20. Order of Operations.
\begin{align*} \text{Compute }2+3(\underbrace{9 - 4})^2 \amp\amp\amp \text{Parentheses first (subtract)}\\ 2+3\underbrace{(5)^2} \amp\amp\amp \text{Exponents}\\ 2+ \underbrace{3(25)} \amp\amp\amp \text{Multiply}\\ \underbrace{2 + 75} \amp\amp\amp \text{Add}\\ 77 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

The most common mistake to make on a problem like this is to add the \(2\) and \(3\) first. Keep in mind that multiplication must be done before addition.

Checkpoint 0.0.21.

As the next example shows, it is very important to remember to multiply and divide from from left to right!

Example 0.0.22. Order of Operations.

\begin{align*} \text{Compute }\underbrace{30 \div 3} \cdot 2 \amp\amp\amp \text{Divide first (left to right!)}\\ \underbrace{10 \cdot 2} \amp\amp\amp \text{Multiply}\\ 20 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.23. Order of Operations: Exponents and Minus Signs.
\begin{align*} \text{Compute }3(\underbrace{1-6})^2-(\underbrace{3+1})^2 \amp\amp\amp \text{Evaluate inside parentheses first}\\ 3\underbrace{(-5)^2}-\underbrace{4^2} \amp\amp\amp \text{Evaluate exponents}\\ \underbrace{3(25)}-16 \amp\amp\amp \text{Multiply next}\\ 75-16 \amp\amp\amp \text{Subtract}\\ 59 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

This example illustrates an important point about exponents. Exponents apply only to the number they are attached to. This means when we see \(- 4^2\text{,}\) only the \(4\) is squared, giving us \(-(4^2)\) or \(-16\text{.}\) But when the negative is in parentheses, such as \((-5)^2\) the negative is part of the number and is also squared giving us a positive solution, \(25\text{.}\)

If there are several parentheses in a problem we will start with the inner most parentheses and work our way out.

Example 0.0.24. Order of Operations.
\begin{align*} \text{Compute }8 \div (-2) + 9 -7[\underbrace{(-4-5)}-3] \amp\amp\amp \text{Inner most parentheses}\\ 8 \div (-2) + 9 -7[\underbrace{-9-3}] \amp\amp\amp \text{Subtract inside the parentheses}\\ \underbrace{8 \div (-2)} + 9 -7(-12) \amp\amp\amp \text{Divide}\\ -4 + 9 - \underbrace{7(-12)} \amp\amp\amp \text{Multiply}\\ \underbrace{-4+9}-(-84) \amp\amp\amp \text{Add/subtract from left to right}\\ 5-(-84) \amp\amp\amp \text{Subtracting a negative is the same as adding the positive}\\ 5+84 \amp\amp\amp \text{Finish adding}\\ 89 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

As the above example illustrates, it can take several steps to simplify an expression. The key to follow the order of operations successfully is to take the time to show your work and do one step at a time. This will reduce the chance of making a mistake along the way.

In algebra we will often need to simplify an expression to make it easier to use. There are three basic forms of simplifying which we will review here.

Subsection 0.0.4 Working with Variables

We can evaluate an expresson when we know what number each variable in the expression represents. We replace each variable with the equivalent number and simplify what remains using order of operations.

Example 0.0.25. Evaluate Expression.
\begin{align*} \text{\(p(q +6)\) when \(p=3\) and \(q=5\)} \amp\amp\amp \text{Replace \(p\) with \(3\) and \(q\) with \(5\)}\\ 3(5+6) \amp\amp\amp \text{Evaluate parentheses}\\ 3(11)\amp\amp\amp \text{Multiply}\\ 33 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

When a negative number is substituted into an expression, it may be helpful to put it in parenthesis so that the steps for simplying are clear.

Example 0.0.26. Evaluate Expression.
\begin{align*} \text{\(x+zx(3-z)-(x-2)\) when \(x=-6\) and \(z =-2\)} \amp\amp\amp \text{Replace all \(x\)'s with \(-6\) and \(z\)'s with \(-2\)}\\ (-6)+(-2)(-6)(3-(-2))-((-6)-2) \amp\amp\amp \text{Simplify in parenthesis}\\ (-6)+(-2)(-6)(3+2)-(-6-2) \amp\amp\amp \text{Evaluate in parenthesis}\\ -6+(-2)(-6)(5)-(-8) \amp\amp\amp \text{Multiply left to right}\\ -6+(12)(5)+8 \amp\amp\amp \text{Multiply left to right}\\ -6+60+8 \amp\amp\amp \text{Add \(-6\) and \(60\)}\\ 54+8 \amp\amp\amp \text{Add}\\ 62 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Checkpoint 0.0.27.

It will be more common in our study of algebra that we do not know the value of the variables. In this case, we will have to simplify what we can and leave the variables in our final solution.

One way we can simplify expressions is to combine like terms. “Like terms” are terms where the variables match exactly (exponents included). For example, \(3x^2y\) and \(-7x^2y\) are like terms since the variables and their powers are the same, but \(4xy\) does not have the same powers and is not a like term with the others. Note that \(3x^2y\) means there are three \(x^2y\)s and \(-7x^2y\) means negative seven \(x^2y\)s. If we combine the like terms, there are negative four \(x^2y\)s: \(3x^2y + (-7x^2y) = -4x^2y\text{.}\)

To combine like terms add (or subtract) the numbers in front of the variables, then keep the variables the same. This is shown in the following examples:

Example 0.0.28. Combine Like Terms.
\begin{align*} 5x-2y -8x+7y \amp\amp\amp \text{Combine like terms }5x-8x\text{ and }-2y +7y\\ -3x+5y \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.29. Combine Like Terms.
\begin{align*} 8x^2-3x+7-2x^2+4x-3 \amp\amp\amp \text{Combine like terms \(8x^2-2x^2\) and \(-3x+4x\) and \(7-3\)}\\ 6x^2+x+4 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
WeBWorK: Entering Exponents.

Type x^2 for \(x^2\)

When we combine like terms we interpret minus signs as part of the term. This means the following term is a negative term, the sign always stays with the term. We then add the terms together.

Another simplification technique uses the distributive property.

\begin{gather} \text{Distributive Property: } a(b+c)=ab+ac\label{distributive-property}\tag{0.0.1} \end{gather}

Several examples of using the distributive property are given below.

Example 0.0.30. Distribute a Positive.
\begin{align*} \text{Simplify }4(2t-7) \amp\amp\amp \text{Multiply each term by \(4\)}\\ 8t-28 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.31. Distribute a Negative.
\begin{align*} \text{Simplify }-7(5x-6) \amp\amp\amp \text{Multiply each term by \(-7\)}\\ -35x+42 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

In the previous Example 0.0.31 we again use the fact that the sign applies to the following number. This means we treat the \(-6\) as a negative number, this gives \((-7)(-6) = 42\text{,}\) a positive number. The most common error in distributing is a sign error. Be very careful with your signs!

It is possible to distribute just a negative through parentheses. If we have a negative in front of parentheses we can think of it like a \(-1\) in front and distribute the \(-1\) through.

Example 0.0.32. Distribute a Negative.
\begin{align*} \text{Simplify }-(4x-5y +6) \amp\amp\amp \text{Negative can be thought of as \(-1\)}\\ -1(4x-5y +6) \amp\amp\amp \text{Multiply each term by \(-1\)}\\ -4x+5y -6 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

We may need to use both the distributive property and combining like terms to simplify an expression. Order of operations tells us to multiply (distribute) first then add or subtract last (combine like terms). Thus we simplify in two steps, distribute then combine.

Example 0.0.33. Simplify.
\begin{align*} 5+3(2w-4) \amp\amp\amp \text{Distribute \(3\), multiplying each term}\\ 5+6w-12 \amp\amp\amp \text{Combine like terms \(5-12\)}\\ -7+6w \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.34. Simplify.
\begin{align*} 3x-2(4x-5) \amp\amp\amp \text{Distribute \(-2\), multiplying each term}\\ 3x-8x+10 \amp\amp\amp \text{Combine like terms \(3x-8x\)}\\ -5x+10 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

In the previous Example 0.0.34 we distributed \(-2\text{,}\) not just \(2\text{.}\) This is because a negative sign goes with the number after it. Note that \(-2\) times \(-5\) is positive \(10.\)

Checkpoint 0.0.35.

Following are more involved examples of distributing and combining like terms.

Example 0.0.36. Simplify.

\begin{align*} 2(5x-8)-6(4x+3) \amp\amp\amp \text{Distribute \(2\) into first parentheses and \(-6\) into second}\\ 10x-16-24x-18 \amp\amp\amp \text{Combine like terms \(10x-24x\) and \(-16-18\)}\\ -14x-34 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
Example 0.0.37. Simplify.

\begin{align*} 4(3x-8)-(2x-7) \amp\amp\amp \text{Negative (subtract) in middle can be thought of as \(-1\)}\\ 4(3x-8)-1(2x-7) \amp\amp\amp \text{Distribute \(4\) into first parentheses and \(-1\) into second}\\ 12x-32-2x+7 \amp\amp\amp \text{Combine like terms \(12x-2x\) and \(-32+7\)}\\ 10x-25 \amp\amp\amp \text{Our Solution}\checkmark \end{align*}
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