1.1E: Exercises - Solving Linear Equations in One Variable
- Page ID
- 147251
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Determine whether or not the given value is a solution.
- \(−5x + 4 = −1 ; x = −1\)
- \(4x − 3 = −7 ; x = −1\)
- \(3y − 4 = 5; y = \frac{9}{3}\)
- \(−2y + 7 = 12 ; y = −\frac{5}{2}\)
- \(3a − 6 = 18 − a; a = −3\)
- \(5 (2t − 1) = 2 − t; t = 2\)
- Answer
-
1. No
3. Yes
5. No
Solve.
- \(5x − 3 = 27\)
- \(6x − 7 = 47\)
- \(9a + 10 = 10\)
- \(5 − 3a = 5\)
- \(−8t + 5 = 15\)
- \(−9t + 12 = 33\)
- \(7 − y = 22\)
- \(6 − y = 12\)
- \(\frac{2}{3} x + \frac{1}{2} = 1\)
- \(\frac{3}{8} x + \frac{5}{4} = \frac{3}{2}\)
- \(\frac{1 − 3y}{5} = 2\)
- \(\frac{2 − 5y}{6} = −8\)
- Answer
-
1. \(6\)
3. \(0\)
5. \(−\frac{5}{4}\)
7. \(−15\)
9. \(\frac{3}{4}\)
11. \(−3\)
Solve.
- \(6x − 5 + 2x = 19\)
- \(7 − 2x + 9 = 24\)
- \(5y − 6 − 9y = 3 − 2y + 8\)
- \(7 − 9y + 12 = 3y + 11 − 11y\)
- \(\frac{1}{3} x −\frac{3}{2} + \frac{5}{2} x = \frac{5}{6} x + \frac{1}{4}\)
- \(\frac{5}{8} + \frac{1}{5} x −\frac{3}{4} = \frac{3}{10} x − \frac{1}{4}\)
- \(5 (y + 2) = 3 (2y − 1) + 10\)
- \(7 (y − 3) = 4 (2y + 1) − 21\)
- \(7 − 5 (3t − 9) = 22\)
- \(10 − 5 (3t + 7) = 20\)
- \(4 (4a − 1) = 5 (a − 3) + 2 (a − 2)\)
- \(6 (2b − 1) + 24b = 8 (3b − 1)\)
- \(\frac{2}{3} (x + 18) + 2 = \frac{1}{3} x − 13\)
- \(\frac{2}{5} x − \frac{1}{2} (6x − 3) = \frac{4}{3}\)
- \(\frac{1}{5} (2a + 3) −\frac{1}{2} = \frac{1}{3} a + \frac{1}{10}\)
- \(\frac{3}{2} a = \frac{3}{4} (1 + 2a) −\frac{1}{5} (a + 5)\)
- Answer
-
1. \(3\)
3. \(−\frac{17}{2}\)
5. \(\frac{7}{8}\)
7. \(3\)
9. \(2\)
11. \(−\frac{5}{3}\)
13. \(−81\)
15. \(0\)
Set up an algebraic equation then solve.
Number Problems
- When \(3\) is subtracted from the sum of a number and \(10\) the result is \(2\). Find the number.
- The sum of \(3\) times a number and \(12\) is equal to \(3\). Find the number.
- Three times the sum of a number and \(6\) is equal to \(5\) times the number. Find the number.
- Twice the sum of a number and \(4\) is equal to \(3\) times the sum of the number and \(1\). Find the number.
- Answer
-
1. \(−5\)
3. \(9\)
Footnotes
138Linear expressions related with the symbols \(≤, <, ≥,\) and \(>\).
139A real number that produces a true statement when its value is substituted for the variable.
140Properties used to obtain equivalent inequalities and used as a means to solve them.
141Inequalities that share the same solution set.
142Two or more inequalities in one statement joined by the word “and” or by the word “or.”