This is “Operations with Real Numbers”, section 1.2 from the book Advanced Algebra (v. 1.0).
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In this section, we continue to review the properties of real numbers and their operations. The result of adding real numbers is called the sumThe result of adding. and the result of subtracting is called the differenceThe result of subtracting.. Given any real numbers a, b, and c, we have the following properties of addition:
Additive Identity Property: |
$a+0=0+a=a\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$ |
Additive Inverse Property: |
$a+\left(-a\right)=\left(-a\right)+a=0$ |
Associative Property: |
$\left(a+b\right)+c=a+\left(b+c\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$ |
Commutative Property: |
$a+b=b+a\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$ |
Given any real number a, $a+0=0+a=a\hspace{0.17em}.$
Given any real number a, $a+\left(-a\right)=\left(-a\right)+a=0.$
Given real numbers a, b and c, $\left(a+b\right)+c=a+\left(b+c\right).$
Given real numbers a and b, $a+b=b+a.$
It is important to note that addition is commutative and subtraction is not. In other words, the order in which we add does not matter and will yield the same result. However, this is not true of subtraction.
$$\begin{array}{cc}\hfill \begin{array}{lll}\hfill 5+10& =& 10+5\hspace{1em}\hspace{1em}\hfill \\ \hfill 15& =& 15\hfill \end{array}& \begin{array}{lll}\hfill \hspace{1em}\hspace{1em}5-10& \ne & 10-5\hfill \\ \hfill -5& \ne & 5\hfill \end{array}\end{array}$$
We use these properties, along with the double-negative property for real numbers, to perform more involved sequential operations. To simplify things, make it a general rule to first replace all sequential operations with either addition or subtraction and then perform each operation in order from left to right.
Simplify: $-10-\left(-10\right)+\left(-5\right).$
Solution:
Replace the sequential operations and then perform them from left to right.
$$\begin{array}{cccc}\hfill -10-\left(-10\right)+\left(-5\right)& =& -10+10-5\text{\hspace{0.17em}\hspace{0.17em}}\hfill & {Replace\text{\hspace{0.17em}}-(-)\text{\hspace{0.17em}}with\text{\hspace{0.17em}}addition\text{\hspace{0.17em}}\left(\text{+}\right)\text{.}}\hfill \\ \hfill \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \hfill & \hfill & {Replace\text{\hspace{0.17em}}+(-)\text{\hspace{0.17em}}with\text{\hspace{0.17em}}subtraction\text{\hspace{0.17em}}(-)\text{.}}\hfill \\ \hfill & =& 0-5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\hfill \\ \hfill & =& -5\hfill \end{array}$$
Answer: −5
Adding or subtracting fractions requires a common denominatorA denominator that is shared by more than one fraction.. Assume the common denominator c is a nonzero integer and we have
$$\begin{array}{ccc}\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}& \text{\hspace{0.17em}}\frac{a}{c}-\frac{b}{c}=\frac{a-b}{c}\end{array}$$
Simplify: $\frac{2}{9}-\frac{1}{15}+\frac{8}{45}.$
Solution:
First determine the least common multiple (LCM) of 9, 15, and 45. The least common multiple of all the denominators is called the least common denominatorThe least common multiple of a set of denominators. (LCD). We begin by listing the multiples of each given denominator:
$$\begin{array}{ll}\left\{9,18,27,36,45,54,63,72,81,90,\mathrm{\dots}\right\}& {Multiples}\text{\hspace{0.17em}}{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{9}\hfill \\ \left\{15,30,45,60,75,90,\mathrm{\dots}\right\}& {Multiples\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}15}\hfill \\ \left\{45,90,135\mathrm{\dots}\right\}& {Multiples\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}45}\hfill \end{array}$$
Here we see that the LCM(9, 15, 45) = 45. Multiply the numerator and the denominator of each fraction by values that result in equivalent fractions with the determined common denominator.
$$\begin{array}{ccc}\hfill \frac{2}{9}-\frac{1}{15}+\frac{8}{45}& =& \frac{2}{9}\cdot {\frac{5}{5}}-\frac{1}{15}\cdot {\frac{3}{3}}+\frac{8}{45}\hfill \\ \hfill & =& \frac{10}{45}-\frac{3}{45}+\frac{8}{45}\hfill \end{array}$$
Once we have equivalent fractions, with a common denominator, we can perform the operations on the numerators and write the result over the common denominator.
$$\begin{array}{ccc}\hfill & =& \frac{10-3+8}{45}\hfill \\ \hfill & =& \frac{15}{45}\hfill \end{array}$$
And then reduce if necessary,
$$\begin{array}{ccc}\hfill & =& \frac{15{\xf715}}{45{\xf715}}\hfill \\ \hfill & =& \frac{1}{3}\hfill \end{array}$$
Answer: $\frac{1}{3}$
Finding the LCM using lists of multiples, as described in the previous example, is often very cumbersome. For example, try making a list of multiples for 12 and 81. We can streamline the process of finding the LCM by using prime factors.
$$\begin{array}{ccc}\hfill 12& =& {2}^{2}\cdot 3\hfill \\ \hfill 81& =& {3}^{4}\hfill \end{array}$$
The least common multiple is the product of each prime factor raised to the highest power. In this case,
$$\text{LCM}(12,81)={2}^{2}\cdot {3}^{4}=324$$
Often we will find the need to translate English sentences involving addition and subtraction to mathematical statements. Below are some common translations.
$$\begin{array}{ll}n+2& {The\text{\hspace{0.17em}}sum\text{\hspace{0.17em}}of\text{\hspace{0.17em}}a\text{\hspace{0.17em}}number\text{\hspace{0.17em}}and\text{\hspace{0.17em}}2.}\hfill \\ 2-n& {The\text{\hspace{0.17em}}difference\text{\hspace{0.17em}}of\text{\hspace{0.17em}}2\text{\hspace{0.17em}}and\text{\hspace{0.17em}}a\text{\hspace{0.17em}}number\text{.}}\hfill \\ n-2& {Here\text{\hspace{0.17em}}2\text{\hspace{0.17em}}is\text{\hspace{0.17em}}subtracted\text{\hspace{0.17em}}from\text{\hspace{0.17em}}a\text{\hspace{0.17em}}number\text{.}}\hfill \end{array}$$
What is 8 subtracted from the sum of 3 and $\frac{1}{2}$?
Solution:
We know that subtraction is not commutative; therefore, we must take care to subtract in the correct order. First, add 3 and $\frac{1}{2}$ and then subtract 8 as follows:
Perform the indicated operations.
$$\begin{array}{ccc}\hfill \left(3+\frac{1}{2}\right)-8& =& \left(\frac{3}{1}\cdot {\frac{2}{2}}+\frac{1}{2}\right)-8\hfill \\ \hfill & =& \left(\frac{6+1}{2}\right)-8\hfill \\ \hfill & =& \frac{7}{2}-\frac{8}{1}\cdot {\frac{2}{2}}\hfill \\ \hfill & =& \frac{7-16}{2}\hfill \\ \hfill & =& -\frac{9}{2}\hfill \end{array}$$
Answer: $-\frac{9}{2}$
The result of multiplying real numbers is called the productThe result of multiplying. and the result of dividing is called the quotientThe result of dividing.. Given any real numbers a, b, and c, we have the following properties of multiplication:
Zero Factor Property: |
$a\cdot 0=0\cdot a=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}$ |
Multiplicative Identity Property: |
$a\cdot 1=1\cdot a=a\hspace{0.17em}$ |
Associative Property: |
$\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right)$ |
Commutative Property: |
$a\cdot b=b\cdot a$ |
Given any real number a, $a\cdot 0=0\cdot a=0\hspace{0.17em}.$
Given any real number a, $a\cdot 1=1\cdot a=a\hspace{0.17em}.$
Given any real numbers a, b and c, $\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right).$
Given any real numbers a and b, $a\cdot b=b\cdot a.$
It is important to note that multiplication is commutative and division is not. In other words, the order in which we multiply does not matter and will yield the same result. However, this is not true of division.
$$\begin{array}{cc}\hfill \begin{array}{lll}\hfill 5\cdot 10& =& 10\cdot 5\hspace{1em}\hspace{1em}\hfill \\ \hfill 50& =& 50\hfill \end{array}& \begin{array}{lll}\hfill \hspace{1em}\hspace{1em}5\xf710& \ne & 10\xf75\hfill \\ \hfill 0.5& \ne & 2\hfill \end{array}\end{array}$$
We will use these properties to perform sequential operations involving multiplication and division. Recall that the product of a positive number and a negative number is negative. Also, the product of two negative numbers is positive.
Multiply: $5\left(-3\right)\left(-2\right)\left(-4\right).$
Solution:
Multiply two numbers at a time as follows:
Answer: −120
Because multiplication is commutative, the order in which we multiply does not affect the final answer. However, when sequential operations involve multiplication and division, order does matter; hence we must work the operations from left to right to obtain a correct result.
Simplify: $10\xf7\left(-2\right)\left(-5\right).$
Solution:
Perform the division first; otherwise the result will be incorrect.
Notice that the order in which we multiply and divide does affect the result. Therefore, it is important to perform the operations of multiplication and division as they appear from left to right.
Answer: 25
The product of two fractions is the fraction formed by the product of the numerators and the product of the denominators. In other words, to multiply fractions, multiply the numerators and multiply the denominators:
$$\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$$
Multiply: $-\frac{4}{5}\cdot \frac{25}{12}.$
Solution:
Multiply the numerators and multiply the denominators. Reduce by dividing out any common factors.
$$\begin{array}{ccc}\hfill -\frac{4}{5}\cdot \frac{25}{12}& =& -\frac{4\cdot 25}{5\cdot 12}\hfill \\ \hfill & =& -\frac{\stackrel{1}{\overline{)4}}\cdot \stackrel{5}{\overline{)25}}}{\underset{1}{\overline{)5}}\cdot \underset{3}{\overline{)12}}}\hfill \\ \hfill & =& -\frac{5}{3}\hfill \end{array}$$
Answer: $-\frac{5}{3}$
Two real numbers whose product is 1 are called reciprocalsTwo real numbers whose product is 1.. Therefore, $\frac{a}{b}$ and $\frac{b}{a}$ are reciprocals because $\hspace{0.17em}\frac{a}{b}\cdot \frac{b}{a}=\frac{ab}{ab}=1.$ For example,
$$\frac{2}{3}\cdot \frac{3}{2}=\frac{6}{6}=1$$
Because their product is 1, $\frac{2}{3}$ and $\frac{3}{2}$ are reciprocals. Some other reciprocals are listed below:
$$\frac{5}{8}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{and}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{8}{5}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{and}\hspace{0.17em}\hspace{0.17em}\frac{1}{7}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-\frac{4}{5}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{and}\hspace{0.17em}\hspace{0.17em}-\frac{5}{4}$$
This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor.
$$\hspace{0.17em}\frac{a}{b}\xf7{\frac{c}{d}}=\frac{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{a}{b}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\frac{c}{d}}\cdot \frac{\text{\hspace{0.17em}}{\frac{d}{c}}\text{\hspace{0.17em}}}{{\frac{d}{c}}}=\frac{\frac{a}{b}\cdot \frac{d}{c}}{1}=\frac{a}{b}\cdot {\frac{d}{c}}$$
In general,
$$\frac{a}{b}\xf7{\frac{c}{d}}=\frac{a}{b}\cdot {\frac{d}{c}}=\frac{ad}{bc}$$
Simplify: $\frac{5}{4}\xf7\frac{3}{5}\cdot \frac{1}{2}.$
Solution:
Perform the multiplication and division from left to right.
$$\begin{array}{ccccc}\hfill \frac{5}{4}\xf7{\frac{3}{5}}\cdot \frac{1}{2}& & =& \hfill & \frac{5}{4}\cdot {\frac{5}{3}}\cdot \frac{1}{2}\hfill \\ \hfill & & =& \hfill & \frac{5\cdot 5\cdot 1}{4\cdot 3\cdot 2}\hfill \\ \hfill & & =& \hfill & \frac{25}{24}\hfill \end{array}$$
In algebra, it is often preferable to work with improper fractions. In this case, we leave the answer expressed as an improper fraction.
Answer: $\frac{25}{24}$
In a computation where more than one operation is involved, grouping symbols help tell us which operations to perform first. The grouping symbolsParentheses, brackets, braces, and the fraction bar are the common symbols used to group expressions and mathematical operations within a computation. commonly used in algebra are:
$$\begin{array}{ll}\left(\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}& {Parentheses}\hfill \\ \left[\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}& {Brackets}\hfill \\ \left\{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\right\}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}& {Braces}\hfill \\ \text{\hspace{0.17em}}\frac{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}& {Fraction\text{\hspace{0.17em}}bar}\hfill \end{array}$$
All of the above grouping symbols, as well as absolute value, have the same order of precedence. Perform operations inside the innermost grouping symbol or absolute value first.
Simplify: $2-\left(\frac{4}{5}-\frac{2}{15}\right).$
Solution:
Perform the operations within the parentheses first.
$$\begin{array}{ccc}\hfill 2-\left(\frac{4}{5}-\frac{2}{15}\right)& =& 2-\left(\frac{4}{5}\cdot {\frac{3}{3}}-\frac{2}{15}\right)\hfill \\ \hfill & =& 2-\left(\frac{12}{15}-\frac{2}{15}\right)\hfill \\ \hfill & =& 2-\left(\frac{10}{15}\right)\hfill \\ \hfill & =& \frac{2}{1}\cdot {\frac{3}{3}}-\frac{2}{3}\hfill \\ \hfill & =& \frac{6-2}{3}\hfill \\ \hfill & =& \frac{4}{3}\hfill \end{array}$$
Answer: $\frac{4}{3}$
Simplify: $\frac{5-\left|4-(-3)\right|}{\left|-3\right|-\left(5-7\right)}.$
Solution:
The fraction bar groups the numerator and denominator. Hence, they should be simplified separately.
$$\begin{array}{ccc}\hfill \frac{5-\left|4-\left(-3\right)\right|}{\left|-3\right|-\left(5-7\right)}& =& \frac{5-\left|4+3\right|}{\left|-3\right|-\left(-2\right)}\hfill \\ \hfill & =& \frac{5-\left|7\right|}{\left|-3\right|+2}\hfill \\ \hfill & =& \frac{5-7}{3+2}\hfill \\ \hfill & =& \frac{-2}{5}\hfill \\ \hfill & =& -\frac{2}{5}\hfill \end{array}$$
Answer: $-\frac{2}{5}$
If a number is repeated as a factor numerous times, then we can write the product in a more compact form using exponential notationThe compact notation ${a}^{n}$ used when a factor a is repeated n times.. For example,
$$5\cdot 5\cdot 5\cdot 5={5}^{4}$$
The baseThe factor a in the exponential notation ${a}^{n}.$ is the factor and the positive integer exponentThe positive integer n in the exponential notation ${a}^{n}$ that indicates the number of times the base is used as a factor. indicates the number of times the base is repeated as a factor. In the above example, the base is 5 and the exponent is 4. Exponents are sometimes indicated with the caret (^) symbol found on the keyboard, 5^4 = 5*5*5*5. In general, if a is the base that is repeated as a factor n times, then
When the exponent is 2 we call the result a squareThe result when the exponent of any real number is 2., and when the exponent is 3 we call the result a cubeThe result when the exponent of any real number is 3.. For example,
$$\begin{array}{cccc}\hfill {5}^{2}& =& 5\cdot 5=25\hfill & {\text{\u201c}5\text{\hspace{0.17em}}squared\text{\u201d}}\hfill \\ \hfill {5}^{3}& =& 5\cdot 5\cdot 5=125\hfill & {\text{\u201c}5\text{\hspace{0.17em}}cubed\text{\u201d}}\hfill \end{array}$$
If the exponent is greater than 3, then the notation ${a}^{n}$ is read, “a raised to the nth power.” The base can be any real number,
$$\begin{array}{ccc}\hfill {\left(2.5\right)}^{2}& =& \left(2.5\right)\left(2.5\right)=6.25\hfill \\ \hfill {\left(-\frac{2}{3}\right)}^{3}& =& \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right)=-\frac{8}{27}\hfill \\ \hfill {\left(-2\right)}^{4}& =& \left(-2\right)\left(-2\right)\left(-2\right)\left(-2\right)=16\hfill \\ \hfill -{2}^{4}& =& -1\cdot 2\cdot 2\cdot 2\cdot 2=-16\hfill \end{array}$$
Notice that the result of a negative base with an even exponent is positive. The result of a negative base with an odd exponent is negative. These facts are often confused when negative numbers are involved. Study the following four examples carefully:
The base is (−3). |
The base is 3. |
---|---|
$\begin{array}{ccc}\hfill {\left(-3\right)}^{4}& =& \left(-3\right)\left(-3\right)\left(-3\right)\left(-3\right)=+81\hfill \\ \hfill {\left(-3\right)}^{3}& =& \left(-3\right)\left(-3\right)\left(-3\right)=-27\hfill \end{array}$ |
$\begin{array}{ccc}\hfill -{3}^{4}& =& -1\cdot 3\cdot 3\cdot 3\cdot 3=-81\hfill \\ \hfill -{3}^{3}& =& -1\cdot 3\cdot 3\cdot 3=-27\hfill \end{array}$ |
The parentheses indicate that the negative number is to be used as the base.
Calculate:
Solution:
Here $-\frac{1}{3}$ is the base for both problems.
Use the base as a factor three times.
$$\begin{array}{ccc}\hfill {\left(-\frac{1}{3}\right)}^{3}& =& \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\hfill \\ \hfill & =& -\frac{1}{27}\hfill \end{array}$$
Use the base as a factor four times.
$$\begin{array}{ccc}\hfill {\left(-\frac{1}{3}\right)}^{4}& =& \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\hfill \\ \hfill & =& +\frac{1}{81}\hfill \end{array}$$
Answers:
When several operations are to be applied within a calculation, we must follow a specific order to ensure a single correct result.
Note that multiplication and division should be worked from left to right. Because of this, it is often reasonable to perform division before multiplication.
Simplify: ${5}^{3}-24\xf76\cdot \frac{1}{2}+2.$
Solution:
First, evaluate ${5}^{3}$ and then perform multiplication and division as they appear from left to right.
$$\begin{array}{ccc}\hfill \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{5}^{3}-24\xf76\cdot \frac{1}{2}+2& =& {5}^{3}-24\xf76\cdot \frac{1}{2}+2\hfill \\ \hfill & =& 125-24\xf76\cdot \frac{1}{2}+2\hfill \\ \hfill & =& 125-4\cdot \frac{1}{2}+2\hfill \\ \hfill & =& 125-2+2\hfill \\ \hfill & =& 123+2\hfill \\ \hfill & =& 125\hfill \end{array}$$
Multiplying first would have led to an incorrect result.
Answer: 125
Simplify: $-10-{5}^{2}+{\left(-3\right)}^{4}.$
Solution:
Take care to correctly identify the base when squaring.
$$\begin{array}{ccc}\hfill \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-10-{5}^{2}+{\left(-3\right)}^{4}& =& -10-25+81\hfill \\ \hfill & =& -35+81\hfill \\ \hfill & =& 46\hfill \end{array}$$
Answer: 46
We are less likely to make a mistake if we work one operation at a time. Some problems may involve an absolute value, in which case we assign it the same order of precedence as parentheses.
Simplify: $7-5\left|-{2}^{2}+{\left(-3\right)}^{2}\right|.$
Solution:
Begin by performing the operations within the absolute value first.
$$\begin{array}{ccc}\hfill 7-5\left|-{2}^{2}+{\left(-3\right)}^{2}\right|& =& 7-5\left|-4+9\right|\hfill \\ \hfill & =& 7-5\left|5\right|\hfill \\ \hfill & =& 7-5\cdot 5\hfill \\ \hfill & =& 7-25\hfill \\ \hfill & =& -18\hfill \end{array}$$
Subtracting $7-5$ first will lead to incorrect results.
Answer: −18
Try this! Simplify: $-{6}^{2}-\left[-15-{\left(-2\right)}^{3}\right]-{\left(-2\right)}^{4}.$
Answer: −45
Perform the operations. Reduce all fractions to lowest terms.
$33-\left(-15\right)+\left(-8\right)$
$-10-9+\left(-6\right)$
$-23+\left(-7\right)-\left(-10\right)$
$-1-\left(-1\right)-1$
$\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$-\frac{1}{5}+\frac{1}{2}-\frac{1}{10}$
$\frac{2}{3}-\left(-\frac{1}{4}\right)-\frac{1}{6}$
$-\frac{3}{2}-\left(-\frac{2}{9}\right)-\frac{5}{6}$
$\frac{3}{4}-\left(-\frac{1}{2}\right)-\frac{5}{8}$
$-\frac{1}{5}-\frac{3}{2}-\left(-\frac{7}{10}\right)$
Subtract 3 from 10.
Subtract −2 from 16.
Subtract $-\frac{5}{6}$ from 4.
Subtract $-\frac{1}{2}$ from $\frac{3}{2}.$
Calculate the sum of −10 and 25.
Calculate the sum of −30 and −20.
Find the difference of 10 and 5.
Find the difference of −17 and −3.
The formula $d=\left|b-a\right|$ gives the distance between any two points on a number line. Determine the distance between the given numbers on a number line.
10 and 15
6 and 22
0 and 12
−8 and 0
−5 and −25
−12 and −3
Determine the reciprocal of the following.
$\frac{1}{3}$
$\frac{2}{5}$
$-\frac{3}{4}$
−12
a where $a\ne 0$
$\frac{1}{a}$
$\frac{a}{b}$ where $a\ne 0$
$\frac{1}{ab}$
Perform the operations.
$-4\left(-5\right)\xf72$
$\left(-15\right)\left(-3\right)\xf7\left(-9\right)$
$-22\xf7\left(-11\right)\left(-2\right)$
$50\xf7\left(-25\right)\left(-4\right)$
$\frac{2}{3}\left(-\frac{9}{10}\right)$
$-\frac{5}{8}\left(-\frac{16}{25}\right)$
$\frac{7}{6}\left(-\frac{6}{7}\right)$
$-\frac{15}{9}\left(\frac{9}{5}\right)$
$\frac{4}{5}\left(-\frac{2}{5}\right)\xf7\frac{16}{25}$
$\left(-\frac{9}{2}\right)\left(-\frac{3}{2}\right)\xf7\frac{27}{16}$
$\frac{8}{5}\xf7\frac{5}{2}\cdot \frac{15}{40}$
$\frac{3}{16}\xf7\frac{5}{8}\cdot \frac{1}{2}$
Find the product of 12 and 7.
Find the product of $-\frac{2}{3}$ and 12.
Find the quotient of −36 and 12.
Find the quotient of $-\frac{3}{4}$ and 9.
Subtract 10 from the sum of 8 and −5.
Subtract −2 from the sum of −5 and −3.
Joe earns $18.00 per hour and “time and a half” for every hour he works over 40 hours. What is his pay for 45 hours of work this week?
Billy purchased 12 bottles of water at $0.75 per bottle, 5 pounds of assorted candy at $4.50 per pound, and 15 packages of microwave popcorn costing $0.50 each for his party. What was his total bill?
James and Mary carpooled home from college for the Thanksgiving holiday. They shared the driving, but Mary drove twice as far as James. If Mary drove for 210 miles, then how many miles was the entire trip?
A $6\frac{3}{4}$ foot plank is to be cut into 3 pieces of equal length. What will be the length of each piece?
A student earned 72, 78, 84, and 90 points on her first four algebra exams. What was her average test score? (Recall that the average is calculated by adding all the values in a set and dividing that result by the number of elements in the set.)
The coldest temperature on Earth, −129° F, was recorded in 1983 at Vostok Station, Antarctica. The hottest temperature on Earth, 136° F, was recorded in 1922 at Al’ Aziziyah, Libya. Calculate the temperature range on Earth.
Perform the operations.
$7-\left\{3-\left[-6-\left(10\right)\right]\right\}$
$-\left(9-12\right)-\left[6-\left(-8-3\right)\right]$
$\frac{1}{2}\left\{5-\left(10-3\right)\right\}$
$\frac{2}{3}\left\{-6+\left(6-9\right)\right\}$
Perform the operations.
${12}^{2}$
${\left(-12\right)}^{2}$
$-{12}^{2}$
$-{\left(-12\right)}^{2}$
$-{5}^{4}$
${\left(-5\right)}^{4}$
${\left(-\frac{1}{2}\right)}^{3}$
$-{\left(-\frac{1}{2}\right)}^{3}$
$-{\left(-\frac{3}{4}\right)}^{2}$
$-{\left(-\frac{5}{2}\right)}^{3}$
${\left(-1\right)}^{22}$
${\left(-1\right)}^{13}$
$-{\left(-1\right)}^{12}$
$-{\left(-1\right)}^{5}$
$-{10}^{2}$
$-{10}^{4}$
Simplify.
$5-3\left(4-{3}^{2}\right)$
$8-5\left(3-{3}^{2}\right)$
${\left(-5\right)}^{2}+3\left(2-{4}^{2}\right)$
Calculate the sum of the squares of the first three consecutive positive odd integers.
Calculate the sum of the squares of the first three consecutive positive even integers.
What is 6 subtracted from the sum of the squares of 5 and 8?
What is 5 subtracted from the sum of the cubes of 2 and 3?
What is PEMDAS and what is it missing?
Does 0 have a reciprocal? Explain.
Explain why we need a common denominator in order to add or subtract fractions.
Explain why ${\left(-10\right)}^{4}$ is positive and $-{10}^{4}$ is negative.
40
−20
$\frac{2}{3}$
$\frac{3}{4}$
$\frac{5}{8}$
7
$\frac{29}{6}$
15
5
5 units
12 units
20 units
3
$-\frac{4}{3}$
$\frac{1}{a}$
$\frac{b}{a}$
10
−4
$-\frac{3}{5}$
−1
$-\frac{1}{2}$
$\frac{6}{25}$
84
−3
−7
$855
315 miles
81 points
−12
−1
75
−3
0
144
−144
−625
$-\frac{1}{8}$
$-\frac{9}{16}$
1
−1
−100
20
−17
41
35
$\frac{9}{7}$
$\frac{3}{5}$
$-\frac{5}{2}$
−35
11
−36
150
$-\frac{11}{18}$
35
83
Answer may vary
Answer may vary