1.23: Linear Inequalities
- Page ID
- 47196
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we solve linear inequalities.
Consider the inequality \(2 x-1>2\). To solve this means to find all values of \(x\) that satisfy the inequality (so that when you plug in those values for \(x\) you get a true statement). For example, since \(2 \cdot 5-1=9>2, x=5\) is a solution and since \(2 \cdot 1-1=1 \not>2, x=1\) is not a solution.
We can solve an inequality in much the same way as we solve an equality with one important exception:
Multiplication or division by a negative number reverses the direction of the inequality.
For example \(-3 x>9 \Longleftrightarrow \frac{-3 x}{-3}<\frac{9}{-3} \Longleftrightarrow x<-3\). We see that
\(x=-10\) satisfies all of these inequalities and \(x=3\) satisfies none of them. We can graph the solution on the number line:
But \(3 x>9 \Longleftrightarrow \frac{3 x}{3}>\frac{9}{3} \Longleftrightarrow x>3\).
For instance, check that \(x=5\) (which is greater than 3 ) satisfies the inequality.
Solve the given inequality and represent the solution on the number line:
a) \(5 x>10 \Longleftrightarrow x>\frac{10}{5} \Longleftrightarrow x>2\)
b) \(-10 x \leq-5 \Longleftrightarrow x \geq \frac{-5}{-10} \Longleftrightarrow x \geq \frac{1}{2}\)
c) \(-x>-2 \Longleftrightarrow x<\frac{-2}{-1} \Longleftrightarrow x<2\)
d) \(2-x \geq 2 x-5 \Longleftrightarrow 7-x \geq 2 x \Longleftrightarrow 7 \geq 3 x \Longleftrightarrow \frac{7}{3} \geq x\) (or \(\left.x \leq \frac{7}{3}\right)\)
e) \(2-3 x \geq-2 x+7 \Longleftrightarrow-5-3 x \geq-2 x \Longleftrightarrow-5 \geq x\) (or \(x \leq-5)\)
f) \(3(x-2)+5 \leq 5-2(x+1) \Longleftrightarrow 3 x-6+5 \leq 5-2 x-2 \Longleftrightarrow 3 x-1 \leq 3-2 x \Longleftrightarrow 3 x+2 x \leq 3+1 \Longleftrightarrow 5 x \leq 4 \Longleftrightarrow x \leq \frac{4}{5}\)
Note: There is more than one way to do a problem. For example:
\(-3 x-2<1 \Longleftrightarrow-3 x<2+1 \Longleftrightarrow-3 x<3 \Longleftrightarrow \frac{-3 x}{-3}>\frac{3}{-3} \Longleftrightarrow x>-1 \)
or,
\(-3 x-2<1 \Longleftrightarrow-2<1+3 x \Longleftrightarrow-2-1<3 x \Longleftrightarrow-3<3 x \Longleftrightarrow-1<x(\) which is the same as \(x>-1)\)
Solve the inequality and show the graph of the solution:
\[7 x+4 \leq 2 x-6\nonumber\]