5.6 Linear inequalities in one variable  (Page 2/2)

For $3<7$ , if 8 is added to both sides, we get

$\begin{array}{cc}3+8<7+8.& \\ 11<15& \text{True}\end{array}$

For $3<7$ , if 8 is subtracted from both sides, we get

$\begin{array}{cc}3-8<7-8.& \\ -5<-1& \text{True}\end{array}$

For $3<7$ , if both sides are multiplied by 8 (a positive number), we get

$\begin{array}{cc}8(3)<8(7)& \\ 24<56& \text{True}\end{array}$

For $3<7$ , if both sides are multiplied by $-8$ (a negative number), we get

$(-8)3>(-8)7$

Notice the change in direction of the inequality sign.

$\begin{array}{cc}-24>-56& \text{True}\end{array}$

If we had forgotten to reverse the direction of the inequality sign we would have obtained the incorrect statement $-24<-56$ .

For $3<7$ , if both sides are divided by 8 (a positive number), we get

$\begin{array}{cc}\frac{3}{8}<\frac{7}{8}& \text{True}\end{array}$

For $3<7$ , if both sides are divided by $-8$ (a negative number), we get

$\begin{array}{cc}\frac{3}{-8}>\frac{7}{-8}& \text{True}\end{array}(\text{since}-.375-.875)$

$\begin{array}{ll}3x>15\hfill & \text{Divide}\text{both}\text{sides}\text{by}\text{3}\text{.}\text{The}3\text{is}\text{a}\text{positive}\text{number},\text{so}\text{we}\text{need}\text{not}\text{reverse}\text{the}\text{sense}\text{of}\text{the}\text{inequality}\text{.}\hfill \\ x>5\hfill & \hfill \end{array}$
Thus, all numbers strictly greater than 5 are solutions to the inequality $3x>15$ .

$\begin{array}{ll}2y-1\le 16\hfill & \text{Add}\text{1}\text{to}\text{both}\text{sides}.\hfill \\ 2y\le 17\hfill & \text{Divide}\text{both}\text{sides}\text{by}2.\hfill \\ y\le \frac{17}{2}\hfill & \hfill \end{array}$

$\begin{array}{ll}-8x+5<14\hfill & \text{Subtract}5\text{from}\text{both}\text{sides}\text{.}\hfill \\ -8x<9\hfill & \text{Divide}\text{both}\text{sides}\text{by}-8.\text{We}\text{must}\text{reverse}\text{the}\text{sense}\text{of}\text{the}\text{inequality}\hfill \\ \hfill & \text{since}\text{we}\text{are}\text{dividing}\text{by}\text{a}\text{negative}\text{number}\text{.}\hfill \\ x>-\frac{9}{8}\hfill & \hfill \end{array}$

$\begin{array}{l}5-3(y+2)<6y-10\hfill \\ 5-3y-6<6y-10\hfill \\ -3y-1<6y-10\hfill \\ -9y<-9\hfill \\ y>1\hfill \end{array}$

$\begin{array}{ll}\frac{2z+7}{-4}\ge -6\hfill & \text{Multiply}\text{by}-4\hfill \\ 2z+7\le 24\hfill & \text{Notice}\text{the}\text{change}\text{in}\text{the}\text{sense}\text{of}\text{the}\text{inequality}\text{.}\hfill \\ 2z\le 17\hfill & \hfill \\ z\le \frac{17}{2}\hfill & \hfill \end{array}$

Solve $1<x+6<8$ .

$\begin{array}{ll}1-6<x+6-6<8-6\hfill & \text{Subtract}6\text{from}\text{all}\text{three}\text{parts}\text{.}\hfill \\ -5<x<2\hfill & \hfill \end{array}$

Thus, if $x$ is any number strictly between $-5$ and 2, the statement $1<x+6<8$ will be true.

Solve $-3<\frac{-2x-7}{5}<8$ .

$\begin{array}{ll}-3(5)<\frac{-2x-7}{5}(5)<8(5)\hfill & \text{Multiply}\text{each}\text{part}\text{by}5.\hfill \\ -15<-2x-7<40\hfill & \text{Add}7\text{to}\text{all}\text{three}\text{parts}.\hfill \\ -8<-2x<47\hfill & \text{Divide}\text{all}\text{three}\text{parts}\text{by}-2.\hfill \\ 4>x>-\frac{47}{2}\hfill & \text{Remember}\text{to}\text{reverse}\text{the}\text{direction}\text{of}\text{the}\text{inequality}\hfill \\ \hfill & \text{signs}\text{.}\hfill \\ -\frac{47}{2}<x<4\hfill & \text{It}\text{is}\text{customary}\text{(but}\text{not}\text{necessary)}\text{to}\text{write}\text{the}\text{inequality}\hfill \\ \hfill & \text{so}\text{that}\text{inequality}\text{arrows}\text{point}\text{to}\text{the}\text{left}\text{.}\hfill \end{array}$

Thus, if $x$ is any number between $-\frac{47}{2}$ and 4, the original inequality will be satisfied.