1.2E Exercises
- Page ID
- 152854
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)These exercises will combine everything you've learned about working with fractions, which means finding common denominators, adding and subtracting, multiplying and dividing, cancelling and simplifying, and in some cases remembering how Order of Operations works!
Write the fraction in simplest form (meaning the numerator and denominator have no common factors).
1. \( \frac{8}{4} \)
2. \( \frac{6}{2} \)
3. \( \frac{ 2}{16} \)
4. \( \frac{ 17}{34} \)
5. \( \frac{25}{30} \)
6. \( \frac{42}{35}\)
7. \( \frac{-8}{40} \)
8. \( -\frac{36}{12} \)
- Answer
-
- \(2\)
- \(3\)
- \( \frac{ 1}{8} \)
- \( \frac{ 1}{2} \)
- \( \frac{5}{6} \)
- \( \frac{6}{5} \)
- \( - \frac{1}{5} \)
- \( - 3\)
Simplify:
1. \( \frac{3}{4} + \frac{1}{2} \)
2. \( \frac{5}{6} - \frac{1}{3} \)
3. \( \frac{2}{5} \cdot \frac{3}{4} \)
4. \( \dfrac{\frac{3}{8}}{\frac{1}{2} }\)
5. \( \frac{2}{3} + \frac{4}{9} - \frac{1}{6} \)
6. \( \dfrac{\frac{5}{12}}{\frac{2}{3}} \cdot \frac{4}{5} \)
7. \( \frac{3}{4} - \frac{1}{4} \cdot \frac{2}{3} \)
8. \( \dfrac{\frac{5}{6}}{\frac{2}{5}} + \frac{1}{3} \)
9. \( \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{2}{3} \)
10. \( \frac{2}{3} - \frac{5}{9} \cdot \frac{3}{4} \)
11. \( \frac{7}{-5} - \frac{ -5}{4 }\)
12. \( - \frac{7}{2} + \frac{6}{-7} \)
- Answer
-
- \( \frac{5}{4} \)
- \( \frac{1}{2} \)
- \( \frac{3}{10} \)
- \( \frac{3}{4} \)
- \( \frac{17}{18}\)
- \( \frac{1}{2} \)
- \( \frac{7}{12} \)
- \( \frac{29}{12}\)
- \( \frac{8}{35}\)
- \(\frac{1}{4}\)
- \( - \frac{3}{20} \)
- \(- \frac{61}{14} \)
Simplify:
1. \( \dfrac{\frac{3}{4}}{\frac{1}{2}} \)
2. \( \dfrac{\frac{5}{6} + \frac{1}{3}}{\frac{2}{3}} \)
3. \( \dfrac{\frac{2}{5} \cdot \frac{3}{4}}{\frac{3}{8}} \)
4. \( \dfrac{\frac{1}{2}}{\frac{3}{4} - \frac{1}{3}} \)
5. \( \dfrac{\frac{3}{4} - \frac{1}{4}}{\frac{2}{3}} \cdot \frac{5}{6} \)
6. \( \dfrac{\frac{5}{6}}{\frac{2}{5}\cdot \frac{1}{3}} \)
7. \( \dfrac{\frac{1}{3} + \frac{4}{5}\cdot \frac{10}{3} }{ 2} \)
- Answer
-
1. \( \frac{3}{2} \)
2. \( \frac{7}{4} \)
3. \( \frac{4}{5} \)
4. \( \frac{6}{5} \)
5. \( \frac{5}{16}\)
6. \( \frac{25}{4} \)
7. \( \frac{3}{2} \)
1. Convert \( \frac{5}{2} \) to a mixed number.
2. Convert \( 3 \frac{1}{4} \) to an improper fraction.
3. Convert \( 6 \frac{3}{5} \) to an improper fraction.
4. Convert \( \frac{11}{3} \) to a mixed number.
5. Convert \( 4 \frac{2}{7} \) to an improper fraction.
6. Convert \( \frac{7}{2} \) to a mixed number.
7. Convert \( 2 \frac{5}{8} \) to an improper fraction.
8. Convert \( \frac{17}{4} \) to a mixed number.
- Answer
-
- \( 2 \frac{1}{2} \)
- \( \frac{13}{4}\)
- \( \frac{33}{5} \)
- \( 3\frac{2}{3} \)
- \( \frac{30}{7} \)
- \( 3\frac{1}{2} \)
- \( \frac{21}{8} \)
- \( 4 \frac{1}{4} \)
1. Remember I said you can only split up a fraction like \( \frac{a+b}{c} = \frac{a}{c}+\frac{b}{c} \) because the addition is in the numerator? Simplify the expression \( \frac{a}{b} + \frac{a}{c} \) and see what you get. Conclude that \( \frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c} \).
2. We have to be careful with negative signs when working with messier fractions. For example, this statement is NOT TRUE: \( \frac{a}{d} - \frac{b+c}{d} \overset{?}{=} \frac{a-b+c}{d} \). What is the correct result of \( \frac{a}{d} - \frac{b+c}{d} \)? Try choosing some numbers for \(a, b, c, \) and \(d\) to get the juices flowing or confirm...
- Answer
-
1. Hint: you need to get a common denominator, and since you don't know what \(b\) and \(c\) are, just use \(bc\) for that.
2. Let's choose \(a = 2, b= 1, c = 3,\) and \( d = 4 \) and see what happens. The left hand side is \( \frac{a}{d} - \frac{b+c}{d} = \frac{2}{4} - \frac{1+3}{4} = \frac{1}{2} - \frac{4}{4} = -\frac{1}{2} \), but the hypothesized right hand side would be \( \frac{2 - 1 + 3}{4} = \frac{4}{4} = 1.\) Not the same! Try rewriting
\[ \frac{a}{d} - \frac{b+c}{d} =\frac{a}{d} - \frac{(b+c)}{d} \notag \]
and proceeding from there.