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}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
Practice Makes Perfect
In the following exercises, determine if the following points are solutions to the given system of equations.
- \(\left\{ \begin{array} {l} 2x−6y=0 \\ 3x−4y=5 \end{array} \right.\)
ⓐ \((3,1)\) ⓑ \((−3,4)\)
- \(\left\{ \begin{array} {l} x+y=2 \\ y=x \end{array} \right.\)
ⓐ \((8,7)\) ⓑ \((1,1)\)
- Answer
-
- ⓐ yes ⓑ no
- ⓐ no ⓑ yes
In the following exercises, solve the systems of equations by substitution.
- \(\left\{ \begin{array} {l} 2x+y=−2\\ 3x−y=7 \end{array} \right.\)
- \(\left\{ \begin{array} {l} x−3y=−9 \\ 2x+5y=4 \end{array} \right.\)
- \(\left\{ \begin{array} {l} −2x+2y=6 \\ y=−3x+1 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 3x+4y=1 \\ y=−\frac{2}{5}x+2 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 4x+y=10 \\ x−2y=−20 \end{array} \right.\)
- \(\left\{ \begin{array} {l} y=x−6 \\ y=−\frac{3}{2}x+4 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 2x−16y=8 \\ −x−8y=−4 \end{array} \right.\)
- \(\left\{ \begin{array} {l} y=−\frac{2}{3}x+5 \\ 2x+3y=11 \end{array} \right.\)
- \(\left\{ \begin{array} {l} x=−3y+4 \\ 2x+6y=8 \end{array} \right.\)
- Answer
-
- \((1,−4)\)
- \((−3,2)\)
- \((−1/2,5/2)\)
- \((−5,4)\)
- \((0,10)\)
- \((4,−2)\)
- \((4,0)\)
- no solution
- infinite solutions with solution set: \(\big\{ (x,y) | y=-\frac{1}{3}x+4 \big\}\)
In the following exercises, solve the systems of equations by elimination.
- \(\left\{ \begin{array} {l} 6x−5y=−1 \\ 2x+y=13 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 5x−3y=−1 \\ 2x−y=2 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 4x−3y=3 \\ 2x+5y=−31 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 11x+9y=−5 \\ 7x+5y=−1 \end{array} \right.\)
- \(\left\{ \begin{array} {l} 2x+9y=−4 \\ 3x+13y=−7 \end{array} \right.\)
- \(\left\{ \begin{array} {l} x+\frac{1}{2}y=\frac{3}{2} \\ \frac{1}{5}x−\frac{1}{5}y=3 \end{array} \right.\)
- \(\left\{ \begin{array} {l} \frac{1}{3}x−y=−3 \\ \frac{2}{3}x+\frac{5}{2}y=3 \end{array} \right.\)
- \(\left\{ \begin{array} {l} x−4y=−1 \\ −3x+12y=3 \end{array} \right.\)
- Answer
-
- \((4,5)\)
- \((7,12)\)
- \((−3,−5)\)
- \((2,−3)\)
- \((−11,2)\)
- \((6,-9)\)
- \((−3,2)\)
- infinitely many solutions with solution set: \(\big\{ (x,y) | x−4y=−1 \big\}\)
Writing Exercises
- In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.
- 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


