% \renewcommand{\familydefault}{\sfdefault} % \renewcommand{\sfdefault}{phv} \documentclass[11pt]{report} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amsmath} \usepackage{multicol} \usepackage{hyperref} \usepackage{appendix} \usepackage{boxedminipage} \usepackage{color} \usepackage{fancyhdr} \usepackage{geometry} \usepackage{graphicx} \usepackage{times} \usepackage{fancybox} \usepackage{mathptmx} \usepackage{sectsty} \usepackage{setspace} \usepackage{newpnts} \usepackage[none]{hyphenat} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Thursday, December 29, 2016 04:53:04} %TCIDATA{LastRevised=Saturday, September 25, 2021 15:07:52} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=40 LaTeX Report.cst} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.3in} \pagestyle{fancy} \lhead{\color{blue} Lecture Notes} \chead{\color{black} \Large Fractions -- Part 4: Adding and Subtracting Fractions} \rhead{\small page \ \thepage} \cfoot{} \lfoot{\footnotesize \copyright $\;$ Hidegkuti, 2018} \rfoot{\footnotesize Last revised: September 16, 2018} \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \cfoot{} \renewcommand{\headrulewidth}{0.8pt} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} } \setstretch{1.04} \begin{document} \begin{center} {\Large Part 1 - Fractions as Part of Our Number System} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion Until now, we have only looked at fractions as expressing a part of a quantity. \ Before we define the four basic operations on fractions, it is best to start looking at them as part of our number system. \ We have defined what it means $\dfrac{3}{5}$ of somthing. \ What about $% \dfrac{3}{5}$ alone, as a number? \ We understand $\dfrac{3}{5}$ alone as $% \dfrac{3}{5}$ of the number $1$. \ So, we can actually place fractions on the number line. \ In case of $\dfrac{3}{5}$, we divide the unit between zero and one into five equal parts, and place $\dfrac{3}{5}$ after three such line segments.% %TCIMACRO{% %\FRAME{dtbpF}{4.5766in}{0.5016in}{0pt}{}{}{fractions4pic1.bmp}{% %\special{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 4.5766in;height 0.5016in;depth 0pt;original-width 7.4166in;original-height 0.7706in;cropleft "0";croptop "1.0967";cropright "1.0083";cropbottom "0";filename 'fractions4pic1.bmp';file-properties "XNPEU";}}}% %BeginExpansion \begin{center} \includegraphics[ natheight=0.7706in, natwidth=7.4166in, height=0.5016in, width=4.5766in] {//Mac/Dropbox/lnotes/1Prealgebra/fractions/fractions4/graphics/fractions4pic1__1.pdf}% \end{center} %EndExpansion We can also interpret all integers as fractions. \ For example, $4$ is the same as $\dfrac{8}{2}$ or $\dfrac{4}{1}$. \ The expression $\dfrac{4}{1}$ could mean division between the integers $4$ and $1$, and also, the fraction with numerator $4$ and denominator $1$. \ Thus duality is fundamental, and only allowed because all answers are always the same when worked with the two different meaning. \ How about negative numbers? \ Negative fractions? \ We can clearly think if $% -\dfrac{3}{5}$ as the opposite of $\dfrac{3}{5}$.% %TCIMACRO{% %\FRAME{dtbpF}{4.5766in}{0.5293in}{0pt}{}{}{fractions4pic2.bmp}{% %\special{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 4.5766in;height 0.5293in;depth 0pt;original-width 7.1356in;original-height 0.8337in;cropleft "0";croptop "1";cropright "1.1213";cropbottom "0";filename 'fractions4pic2.bmp';file-properties "XNPEU";}}}% %BeginExpansion \begin{center} \includegraphics[ natheight=0.8337in, natwidth=7.1356in, height=0.5293in, width=4.5766in] {//Mac/Dropbox/lnotes/1Prealgebra/fractions/fractions4/graphics/fractions4pic2__2.pdf}% \end{center} %EndExpansion %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Definition:} \ A \textit{simplified }or\textit{\ reduced} fraction cannot have more than one negative sign.\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.09in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion Let us consider the fraction $\dfrac{-2}{-7}$. \ Recall the fundamental property of fractions: that we can multiply both numerator and denominator by the same non-zero number without changing the value of the fraction. \ Let us proceed with $-1$.\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $% \dfrac{-2}{-7}=\dfrac{-2\left( -1\right) }{-7\left( -1\right) }=\dfrac{2}{7}$% \vspace{0.04in} So, a fraction with a negative number in the numerator and denominator can be simplified, and the result is simply a positive fraction. \ Notice that the rule is the same as with division between two integers. \ The negative sign of a fraction can be located in three places. \ All three of the fractions shown below are equaivalent. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ %TCIMACRO{\TeXButton{\LARGE}{\LARGE}}% %BeginExpansion \LARGE% %EndExpansion $-\dfrac{2}{3}$% %TCIMACRO{\TeXButton{\normalsize}{\normalsize} }% %BeginExpansion \normalsize %EndExpansion \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ %TCIMACRO{\TeXButton{\LARGE}{\LARGE}}% %BeginExpansion \LARGE% %EndExpansion $\dfrac{-2}{~3}$% %TCIMACRO{\TeXButton{\normalsize}{\normalsize} }% %BeginExpansion \normalsize %EndExpansion \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ %TCIMACRO{\TeXButton{\LARGE}{\LARGE}}% %BeginExpansion \LARGE% %EndExpansion $\dfrac{~2}{-3}$% %TCIMACRO{\TeXButton{\normalsize}{\normalsize}}% %BeginExpansion \normalsize% %EndExpansion The first form, $-\dfrac{2}{3}$ $\left( \text{ the opposite of }\dfrac{2}{3}% \right) $ is considered simplified, and it is an elegant way to present a fraction as final answer. \ However, it is not the recommended form for computing with signed fractions. \ The second form, $\dfrac{-2}{~3}$ \ (the numerator is $-2$, the denominator is $3$) \ is both simplified and highly recommended for computations. \ Indeed, when performing operations with signed fractions, we recommend that the negative sign is interpreted as belonging to the numerator. The third form, $\dfrac{~2}{-3}$ (the numerator is $2$, the denominator is $% -3$) \ is not considered simplified, and not recommended for computations either. \ Thus we recommend that when faced with such a fraction, we immediately multiply both numerator and denominator by $-1$, thereby switching to the second form, $\dfrac{-2}{~3}$. \ \textbf{Fractions with a negative sign in their denominators are not acceptable as final answers.% \vspace{0.05in}} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1. }\ Simplify each of the given fractions.\textbf{\vspace{% 0.05in}} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $\dfrac{10}{-28}$ \ \ \ \ \ \ \ \ \ \ \ b) \ $\dfrac{-12}{1}$ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{-3}{-18}$ \ \ \ \ \ \ \ \ \ \ \ d) \ $\dfrac{~4}{-7}$% \textbf{\vspace{0.05in} \ \ \ } %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ The fraction itself can be simplified by dividing both numerator and denominator by $2.$ \ But then $\dfrac{~5}{-14}$ is not simplified because the negative sign is in the wrong place. \ Multiplying both numerator and denominator by $-1$ will give us the answer in the correct form, $\dfrac{-5}{14}$. \ We can also present this as $-\dfrac{5}{14} $. \ \textbf{\vspace{0.05in}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $% \dfrac{10}{-28}=\dfrac{5}{-14}=\,$\fbox{$\dfrac{-5}{14}$ or \ $-\dfrac{5}{14} $}\textbf{\vspace{0.05in}} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion Eventually we will be able to speed up the process by dividing numerator and denominator of $\dfrac{10}{-28}$ by $-2$. %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ The fraction $\dfrac{-12}{1}$ is not simplified, because it can be simplified to $-12$. \ Integers should never be presented as fractions. \ Clearly $-12$ is simpler than $\dfrac{-12}{1}$. \ So, the correct answer is \fbox{$-12$}. %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ We can simplify $\dfrac{-3}{-18}$ by dividing numerator and denominator by $-3$. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $% \dfrac{-3}{-18}=\,$\fbox{$\dfrac{1}{6}$} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ The fraction $\dfrac{~4}{-7}$ is noot simplified only because the negative sign is in the denominator. \ We can fix that by multplying numerator and denominator by $-1$. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $% \dfrac{~4}{-7}=\dfrac{~4\cdot \left( -1\right) }{-7\cdot \left( -1\right) }% =\,$\fbox{$\dfrac{-4}{7}$ or $-\dfrac{4}{7}$} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2. }\ Re-write the mixed number $-4\dfrac{5}{8}$ as an improper fraction. %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ We simply ignore the negative sign, convert the mixed number to an improper fraction, and then put the negative sign back.\vspace{% 0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $% -4\dfrac{5}{8}=-\left( \dfrac{4\cdot 8+5}{8}\right) =-\dfrac{37}{8}$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion So the correct answer is \fbox{$-\dfrac{37}{8}$ or \ $\dfrac{-37}{8}$}.% \vspace{0.1in} \pagebreak %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \begin{center} {\Large Part 2 - Adding and Subtracting Fractions with the Same Denominator} \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion Recall that the numerator and denominator play very different roles. \ The numerator counts how many slices we have. \ The denominator only describes how large each slice is. \ For example, in the fraction $\dfrac{3}{8}$, we are dealing with three slices. \ Each slice is so large so that eight of them makes up a whole. \ So, $\dfrac{3}{8}$ \ and $\dfrac{3}{5}$ \ both express three slices, only the slices in $\dfrac{3}{5}$ are bigger that those in $\dfrac{3}{8}$. \ %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion %TCIMACRO{\TeXButton{hang}{\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\hangindent}{0pt}% %EndExpansion \textbf{To add or subtract fractions with the same denominator, we write a single denominator and express the addition or subtraction of the signed numbers in the numerator.}\vspace{0.04in}\vspace{0.04in} %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.09in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion For example, the addition \ $\dfrac{2}{7}+\dfrac{3}{7}$ \ is simple: we are talking about the same size of slices, and we add three slices to two slices. \ Thus \ $\dfrac{2}{7}+\dfrac{3}{7}=\dfrac{2+3}{7}=\dfrac{5}{7}$.% \vspace{0.04in} Consider now the addition $\dfrac{4}{5}+\dfrac{2}{5}$. \ Again, the same denominator indicates that the slices are of the same size, so we simpliy add two slices to the four. \ In short,\ \ $\dfrac{4}{5}+\dfrac{2}{5}=\dfrac{% 4+2}{5}=\dfrac{6}{5}$.\vspace{0.04in} The answer is an improper fraction. \ However, we do not see anything imporper about it, as long as it is in lowest terms. \ We could present the final answer as $1\dfrac{1}{5}$ but $\dfrac{6}{5}$ is correct, simplified, and is actually preferred.\vspace{0.04in} In case of signed numbers, the negative sign should always be kept in the numerator. \ Then the addition or subtraction of signed fractions will be reduced to addition or subtraction of (signed) integers in the numerator.% \vspace{0.04in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3. }\ Perform the addition or subtraction of the fractions as indicated.\vspace{0.04in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $\dfrac{3}{8}-\dfrac{7}{8}$ \ \ \ \ \ \ \ \ b) \ $\dfrac{2}{5}-\dfrac{4% }{5}-\left( -\dfrac{3}{5}\right) $ \ \ \ \ \ \ \ \ c) \ $\dfrac{3}{7}-\dfrac{% 9}{7}$ \ \ \ \ \ \ \ d) \ $\dfrac{3}{10}-\dfrac{7}{10}+\left( -\dfrac{1}{10}% \right) $\textbf{\ \ \ \ \ \ \ }e) \ $2\dfrac{1}{3}-5\dfrac{2}{3}$\vspace{% 0.04in} %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ We re-write the subtraction as one in the numerator. \ Then, if we can, we simplify the answer. \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{8}-% \dfrac{7}{8}=\dfrac{3-7}{8}=\dfrac{-4}{8}=\,$\fbox{$\dfrac{-1}{2}$ or $-% \dfrac{1}{2}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ We re-write the subtraction with one big fraction bar and a subtraction of integers the same numerator. \ Then, if we can, we simplify the answer.% \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{2}{5}-\dfrac{4}{5}-\left( -\dfrac{3}{5}% \right) =\dfrac{2-4-\left( -3\right) }{5}=\dfrac{-2-\left( -3\right) }{5}=% \dfrac{-2+3}{5}=\,$\fbox{$\dfrac{1}{5}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ We re-write the subtraction with one big fraction bar and a subtraction of integers the same numerator. \ Then, if we can, we simplify the answer.% \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{7}-% \dfrac{9}{7}=\dfrac{3-9}{7}=\,$\fbox{$\dfrac{-6}{7}$ or $-\dfrac{6}{7}$}% \vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ We re-write the subtraction with one big fraction bar and a subtraction of integers the same numerator. \ Then, if we can, we simplify the answer.% \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{10}-\dfrac{7}{10}+\left( -\dfrac{1}{10}% \right) =\dfrac{3-7+\left( -1\right) }{10}=\dfrac{-4+\left( -1\right) }{10}=% \dfrac{-5}{10}=\,$\fbox{$\dfrac{-1}{2}$ or $-\dfrac{1}{2}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ In case of adding or subtracting mixed numbers, we recommend the use of improper fractions. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $2\dfrac{1}{3}-5\dfrac{2}{3}=\dfrac{7}{3}-\dfrac{17}{3}=\dfrac{% 7-17}{3}=\,$\fbox{$\dfrac{-10}{3}$ or $-\dfrac{10}{3}$}\ \vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion There are other ways of handling this, but we recommend the method shown above. \ However, there is a way to do this subtraction by separating integer and fraction parts. \ We need to be careful though to correctly interpret the negative signs and subtraction signs to both integer and fraction parts.\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $2\dfrac{1}{3}-5\dfrac{2}{3}=\underset{\text{integer part}}{2-5}+% \underset{\text{fraction part}}{\dfrac{1}{3}-\dfrac{2}{3}}=-3+\left( -\dfrac{% 1}{3}\right) =-3\dfrac{1}{3}$\vspace{0.04in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion We got lucky with the last problem because both integer and fraction part ended up with the same sign in the result. \ What about a subtraction such as $4\dfrac{2}{5}-1\dfrac{4}{5}$?\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $4\dfrac{2}{5}-1\dfrac{4}{5}=\underset{\text{integer part}}{4-1}+% \underset{\text{fraction part}}{\dfrac{2}{5}-\dfrac{4}{5}}=3+\left( -\dfrac{2% }{5}\right) $\vspace{0.04in} We can not present a mixed number with mixed signs. \ The sign of the entire mixed number is determined by the integer part, because it has more value than the fraction part. \ This answer should be positive. \ How do we express $3$ and $-\dfrac{2}{5}$ as a mixed number? \ We borrow $1$ from the integer part, express it as $\dfrac{5}{5}$ and throw it in with the integer part. \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $3+\left( -\dfrac{2}{5}\right) =2+1+\left( -\dfrac{2}{5}\right) =2+\dfrac{5}{5}+\left( -\dfrac{2}{5}\right) =2+\dfrac{5+\left( -2\right) }{5}% =2+\dfrac{3}{5}=2\dfrac{3}{5}$\vspace{0.04in} This can get quite complicated, which is why we recommend to use improper fractions instead of mixed numbers.\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $4\dfrac{2}{5}-1\dfrac{4}{5}=% \dfrac{22}{5}-\dfrac{9}{5}=\dfrac{22-9}{5}=\dfrac{13}{5}$ \ \ \ $\ \ \ \left( \text{same as }2\dfrac{3}{5}\right) \vspace{0.15in}$ \begin{center} {\Large Part 3 - Adding and Subtracting Fractions with Different Denominators% } \end{center} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\setstretch{1.3}}{\setstretch{1.3}}}% %BeginExpansion \setstretch{1.3}% %EndExpansion How do we add the fractions $\dfrac{5}{6}$ and $\dfrac{3}{4}$? \ It is true that we are adding five slices and three slices, but \ they do not have the same size. \ \ So, how do we add fractions with different denominators? \ Well, we just don't. \ Instead, we use the fundamental property of fractions to bring the fractions to a common denominator. The fundamental property of fractions is that we can multiply both numerator and denominator by the same non-zero number without changing the value of the fraction. \ Therefore, we can find fractions equivalent to $\dfrac{5}{6}$ with all denominators that are multiples of $6.$ \ For example, \ $\dfrac{5}{% 6}=\dfrac{10}{12}=\dfrac{15}{18}=\dfrac{20}{24}=\dfrac{30}{36}=\dfrac{50}{60} $\vspace{0.04in} \ and so on. \ Similarly, we can find fractions\vspace{% 0.04in} equivalent to $\dfrac{3}{4}$ with all denominators that are multiples of $4.$ \ For example, \ $\dfrac{3}{4}=\dfrac{6}{8}=\dfrac{9}{12}=% \dfrac{12}{20}=\dfrac{15}{20}=\dfrac{18}{24}$, and so on.\vspace{0.04in} \ If we multiply $6$ and $4$, we get a common multiple, $24$. \ However, we should always attempt to work with \textbf{the least common multiple}, and that is $12$ in this case. \ To bring $\dfrac{5}{6}$ to a denominator $12$, we multiply numerator and denominator by $2$. \ To bring $\dfrac{3}{4}$ to a denominator $12$, we multiply numerator and denominator by $3$.\vspace{0.09in% } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{5}{6}=\dfrac{5\cdot 2}{6\cdot 2}=% \dfrac{10}{12}$ \ \ and \ \ $\dfrac{3}{4}=\dfrac{3\cdot 3}{4\cdot 3}=\dfrac{9% }{12}$\vspace{0.04in} \ Thus we have that \vspace{0.04in}\ $\dfrac{5}{6}+% \dfrac{3}{4}=\dfrac{10}{12}+\dfrac{9}{12}=\dfrac{10+9}{12}=\dfrac{19}{12}$% \vspace{0.04in} This method also works with adding fractions with integers, because we can always re-write an integer as a fraction with denominator $1$. %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4. }\ Perform the addition or subtraction of the fractions as indicated.\vspace{0.04in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $\dfrac{3}{10}-\dfrac{4}{5}$ \ \ \ \ \ \ \ \ b) \ $4-\dfrac{2}{3}% -\left( -\dfrac{1}{2}\right) $ \ \ \ \ \ \ \ \ c) \ $\dfrac{2}{5}-3$ \ \ \ \ \ \ \ d) \ $\dfrac{2}{3}-\dfrac{5}{6}-\left( -\dfrac{1}{2}\right) $\textbf{\ \ \ \ \ \ \ }e) \ $4\dfrac{1}{6}-3\dfrac{2}{3}$\vspace{0.04in} %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ The least common denominator between $10$ and $5$ is $10$. \ We re-write $\dfrac{4}{5}$\vspace{0.04in} as a fraction with denominator $10$. by multiplying both numerator and denominator by $2$. \ \ \ \ $\dfrac{4}{5}=\dfrac{4\cdot 2}{5\cdot 2}=\dfrac{8}{10}$\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{3}{10}-\dfrac{4}{5}=\dfrac{3}{% 10}-\dfrac{8}{10}=\dfrac{3-8}{10}=\dfrac{-5}{10}=\,$\fbox{$\dfrac{-1}{2}$ or $-\dfrac{1}{2}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ The integer $4$ is not a fraction, but that can be easily fixed, by re-writing it as $\dfrac{4}{1}$. \ We are now looking for the least common multiple of $1$, $3$, and $2$. \ Clearly, $6$ will be good as the least common multiple. \ Then $\dfrac{4}{1}$ can be re-written as $\dfrac{4}{1}=% \dfrac{4\cdot 6}{1\cdot 6}=\dfrac{24}{6}$. \ Similarly, $\dfrac{2}{3}=\dfrac{% 2\cdot 2}{3\cdot 2}=\dfrac{4}{6}$ \ and \ $\dfrac{-1}{2}=\dfrac{-1\cdot 3}{% 2\cdot 3}=\dfrac{-3}{6}$.\vspace{0.04in} \ Once the fractions have the same denominator, we perform the additions and subtractions in the numerator. \ If there is a negative sign, we keep it in the numerator. \ \vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $4-\dfrac{2}{3}-\left( -\dfrac{1}{2}% \right) =\dfrac{4}{1}-\dfrac{2}{3}-\left( \dfrac{-1}{2}\right) =\dfrac{24}{6}% -\dfrac{4}{6}-\left( \dfrac{-3}{6}\right) =\dfrac{24-4-\left( -3\right) }{6}=% \dfrac{20-\left( -3\right) }{6}=\,$\fbox{$\dfrac{23}{6}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion c) \ We re-write $3$ as $\dfrac{3}{1}.$ \ The least common multiple of $1$ and $5$ is $5.$ \ \ To re-write $\dfrac{3}{1}$ with a denominator $5$, we multiply both numerator and denominator by $5$. \ $\dfrac{3}{1}=\dfrac{% 3\cdot 5}{1\cdot 5}=\dfrac{15}{5}$\vspace{0.04in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{2}{5}-3=\dfrac{2}{5}-\dfrac{3}{1}=% \dfrac{2}{5}-\dfrac{15}{5}=\dfrac{2-15}{5}=\,$\fbox{$\dfrac{-13}{5}$ or $-% \dfrac{13}{5}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion We can express the answer as a mixed number, but there is no need, unless we are specifically asked for it.\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion d) \ The least common multiple of $3$, $6$, and $2$ is $6$. \ So we will re-write each fraction with a denominator of $6$. \ If there is a negative sign, we will keep it in the numerator.\vspace{0.04in} $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $So,$% \ \ \dfrac{2}{3}=\dfrac{2\cdot 2}{3\cdot 2}=\dfrac{4}{6}$ \ \ \ and \ $% \dfrac{5}{6}=\dfrac{5}{6}$ \ \ and \ $-\dfrac{1}{2}=\dfrac{-1}{2}=\dfrac{% -1\cdot 3}{2\cdot 3}=\dfrac{-3}{6}$ \vspace{0.04in} %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion We can now perform the subtractions, left to right. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \dfrac{2}{3}-\dfrac{5}{6}% -\left( -\dfrac{1}{2}\right) =\dfrac{4}{6}-\dfrac{5}{6}-\left( \dfrac{-3}{6}% \right) =\dfrac{4-5-\left( -3\right) }{6}=\dfrac{-1-\left( -3\right) }{6}=% \dfrac{-1+3}{6}=\dfrac{2}{6}=\,$\fbox{$\dfrac{1}{3}$}\vspace{0.04in} %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion e) \ In case of mixed numbers, \ we again recommend to switch to improper fractions. \ We re-write $4\dfrac{1}{6}$ as $\dfrac{25}{6}.$ \ Notice that the addition $4+\dfrac{1}{6}$\vspace{0.04in} gives us exactly the same result. \ $3\dfrac{2}{3}$ \ can be re-written as $\dfrac{11}{3}$. \ The least common denominator between $6$ and $3$ is $6$. \ \vspace{0.04in} $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\dfrac{1}{6}-3\dfrac{2}{3}=\dfrac{25}{6}-\dfrac{11}{3}=\dfrac{25}{6}-% \dfrac{22}{6}=\dfrac{25-22}{6}=\dfrac{3}{6}=\,$\fbox{$\dfrac{1}{2}$}\vspace{% 0.04in}\vspace{0.2in} %TCIMACRO{% %\TeXButton{vissza, matek}{\vissza %\matek}}% %BeginExpansion \vissza \matek% %EndExpansion %TCIMACRO{% %\FRAME{itbpF}{0.7169in}{0.7437in}{0.1807in}{}{}{work.jpg}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7169in;height 0.7437in;depth 0.1807in;original-width 4.6977in;original-height 4.875in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'work.jpg';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.1807000000in}{ \includegraphics[ natheight=4.875in, natwidth=4.6977in, height=0.7437in, width=0.7169in] {//Mac/Dropbox/lnotes/1Prealgebra/fractions/fractions4/graphics/work__3.jpg}% }% %EndExpansion \ \ {\large Practice Problems} Perform the indicated operations. \ Present your answer as an integer or a reduced fraction. \ You do not need to convert improper numbers to mixed numbers.% %TCIMACRO{\TeXButton{4col begin}{\begin{multicols}{4}}}% %BeginExpansion \begin{multicols}{4}% %EndExpansion \begin{enumerate} \item $\dfrac{2}{3}+\dfrac{5}{6}\vspace{0.04in}$ \item $\dfrac{2}{3}-\dfrac{5}{6}\vspace{0.04in}$ \item $-2-\dfrac{2}{3}\vspace{0.04in}$ \item $-\dfrac{2}{5}-\left( -\dfrac{5}{2}\right) \vspace{0.04in}$ \item $\dfrac{3}{4}-\left( -\dfrac{5}{8}\right) \vspace{0.04in}$ \item $\dfrac{5}{4}-\dfrac{1}{6}\vspace{0.04in}$ \item $2-\dfrac{2}{3}\vspace{0.04in}$ \item $-2+\dfrac{5}{8}\vspace{0.04in}$ \item $2\dfrac{4}{5}-3\dfrac{1}{2}\vspace{0.04in}$ \item $-3\dfrac{1}{4}-2\dfrac{5}{6}\vspace{0.04in}$ \item $\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{5}{12}\vspace{0.04in}$ \item $-\dfrac{3}{4}-\dfrac{5}{6}\vspace{0.04in}$ \item $\dfrac{5}{12}-\dfrac{3}{4}\vspace{0.04in}$ \item $\dfrac{19}{3}-2\dfrac{5}{6}\vspace{0.04in}$ \item $\dfrac{5}{2}-\dfrac{7}{10}+\dfrac{6}{5}$ \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak %TCIMACRO{% %\FRAME{itbpF}{0.7766in}{0.7766in}{0.2006in}{}{}{answers.jpg}{% %\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7766in;height 0.7766in;depth 0.2006in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}}}% %BeginExpansion \raisebox{-0.2006000000in}{ \includegraphics[ natheight=8.6455in, natwidth=8.6455in, height=0.7766in, width=0.7766in] {//Mac/Dropbox/lnotes/1Prealgebra/fractions/fractions4/graphics/answers__4.jpg}% }% %EndExpansion \ \ \ {\large Answers} \begin{enumerate} \item[1.] $\dfrac{3}{2}$ \ \ \ \ \ 2. \ $-\dfrac{1}{6}$ \ \ \ \ \ 3. \ $-% \dfrac{8}{3}$ \ \ \ \ \ 4. \ $\dfrac{21}{10}$ \ \ \ \ \ 5. \ $\dfrac{11}{8}$ \ \ \ \ \ 6. \ $\dfrac{13}{12}$ \ \ \ \ \ 7. \ $\dfrac{4}{3}$ \ \ \ \ \ 8. \ $-\dfrac{11}{8}$ \ \ \ \ \ 9. \ $-\dfrac{7}{10}$ \ \ \ \ \ \ 10. \ $-\dfrac{% 73}{12}$ \item[11.] $0$ \ \ \ \ \ \ 12. \ $-\dfrac{19}{12}$ \ \ \ \ \ \ 13. $\ -% \dfrac{1}{3}$ \ \ \ \ 14. \ $\dfrac{7}{2}$ \ \ \ \ 15. \ $3$\vspace{0.26in}% \vspace{0.6in}\vspace{4in}\vspace{0.2in} \vspace{1.5in}\vspace{0.7in} \end{enumerate} {\Large %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}