Skip to main content
Mathematics LibreTexts

5.7E: Exercises

  • Page ID
    104866

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Practice Makes Perfect

    Determine if Whether an Ordered Pair is a Solution

    In the following exercises, determine if the following points are solutions to the given system of equations.

    1. \(\left\{ \begin{array} {l} 2x−6y=0 \\ 3x−4y=5 \end{array} \right.\)

    ⓐ \((3,1)\) ⓑ \((−3,4)\)

    1. \(\left\{ \begin{array} {l} x+y=2 \\ y=x \end{array} \right.\)

    ⓐ \((8,7)\) ⓑ \((1,1)\)

    Answer
    1. ⓐ yes ⓑ no
    2. ⓐ no ⓑ yes
    Solve by Substitution

    In the following exercises, solve the systems of equations by substitution.

    1. \(\left\{ \begin{array} {l} 2x+y=−2\\ 3x−y=7 \end{array} \right.\)
    2. \(\left\{ \begin{array} {l} x−3y=−9 \\ 2x+5y=4 \end{array} \right.\)
    3. \(\left\{ \begin{array} {l} −2x+2y=6 \\ y=−3x+1 \end{array} \right.\)
    4. \(\left\{ \begin{array} {l} 3x+4y=1 \\ y=−\frac{2}{5}x+2 \end{array} \right.\)
    5. \(\left\{ \begin{array} {l} 4x+y=10 \\ x−2y=−20 \end{array} \right.\)
    6. \(\left\{ \begin{array} {l} y=x−6 \\ y=−\frac{3}{2}x+4 \end{array} \right.\)
    7. \(\left\{ \begin{array} {l} 2x−16y=8 \\ −x−8y=−4 \end{array} \right.\)
    8. \(\left\{ \begin{array} {l} y=−\frac{2}{3}x+5 \\ 2x+3y=11 \end{array} \right.\)
    9. \(\left\{ \begin{array} {l} x=−3y+4 \\ 2x+6y=8 \end{array} \right.\)
    Answer
    1. \((1,−4)\)
    2. \((−3,2)\)
    3. \((−1/2,5/2)\)
    4. \((−5,4)\)
    5. \((0,10)\)
    6. \((4,−2)\)
    7. \((4,0)\)
    8. no solution
    9. infinite solutions with solution set: \(\big\{ (x,y) | y=-\frac{1}{3}x+4 \big\}\)
    Solve by Elimination

    In the following exercises, solve the systems of equations by elimination.

    1. \(\left\{ \begin{array} {l} 6x−5y=−1 \\ 2x+y=13 \end{array} \right.\)
    2. \(\left\{ \begin{array} {l} 5x−3y=−1 \\ 2x−y=2 \end{array} \right.\)
    3. \(\left\{ \begin{array} {l} 4x−3y=3 \\ 2x+5y=−31 \end{array} \right.\)
    4. \(\left\{ \begin{array} {l} 11x+9y=−5 \\ 7x+5y=−1 \end{array} \right.\)
    5. \(\left\{ \begin{array} {l} 2x+9y=−4 \\ 3x+13y=−7 \end{array} \right.\)
    6. \(\left\{ \begin{array} {l} x+\frac{1}{2}y=\frac{3}{2} \\ \frac{1}{5}x−\frac{1}{5}y=3 \end{array} \right.\)
    7. \(\left\{ \begin{array} {l} \frac{1}{3}x−y=−3 \\ \frac{2}{3}x+\frac{5}{2}y=3 \end{array} \right.\)
    8. \(\left\{ \begin{array} {l} x−4y=−1 \\ −3x+12y=3 \end{array} \right.\)
    Answer
    1. \((4,5)\)
    2. \((7,12)\)
    3. \((−3,−5)\)
    4. \((2,−3)\)
    5. \((−11,2)\)
    6. \((6,-9)\)
    7. \((−3,2)\)
    8. infinitely many solutions with solution set: \(\big\{ (x,y) | x−4y=−1 \big\}\)

    Writing Exercises

    Writing Exercises
    1. In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.
    2. Solve the system of equations by substitution and explain all your steps in words: \(\left\{ \begin{array} {l} 3x+y=1 \\ 2x=y−8 \end{array} \right. \)
    Answer

    Answers will vary


    This page titled 5.7E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Stanislav A. Trunov and Elizabeth J. Hale via source content that was edited to the style and standards of the LibreTexts platform.