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

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For example, consider the inequality $3<7$ .

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

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

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

$\left(-8\right)3>\left(-8\right)7$

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

$\begin{array}{cc}\frac{3}{-8}>\frac{7}{-8}& \text{True}\end{array}\text{\hspace{0.17em}}\left(\text{since}\text{\hspace{0.17em}}-.375-.875\right)$

## Sample set b

Solve the following linear inequalities. Draw a number line and place a point at each solution.

$\begin{array}{ll}3x>15\hfill & \text{Divide}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{3}\text{.}\text{\hspace{0.17em}}\text{The}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{number},\text{\hspace{0.17em}}\text{so}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{need}\text{\hspace{0.17em}}\text{not}\text{\hspace{0.17em}}\text{reverse}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sense}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\text{.}\hfill \\ x>5\hfill & \hfill \end{array}$
Thus, all numbers strictly greater than 5 are solutions to the inequality $3x>15$ .

$\begin{array}{ll}2y-1\le 16\hfill & \text{Add}\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}.\hfill \\ 2y\le 17\hfill & \text{Divide}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}2.\hfill \\ y\le \frac{17}{2}\hfill & \hfill \end{array}$

$\begin{array}{ll}-8x+5<14\hfill & \text{Subtract}\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{.}\hfill \\ -8x<9\hfill & \text{Divide}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-8.\text{\hspace{0.17em}}\text{We}\text{\hspace{0.17em}}\text{must}\text{\hspace{0.17em}}\text{reverse}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sense}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\hfill \\ \hfill & \text{since}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{dividing}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{number}\text{.}\hfill \\ x>-\frac{9}{8}\hfill & \hfill \end{array}$

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

$\begin{array}{ll}\frac{2z+7}{-4}\ge -6\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-4\hfill \\ 2z+7\le 24\hfill & \text{Notice}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{change}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sense}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\text{.}\hfill \\ 2z\le 17\hfill & \hfill \\ z\le \frac{17}{2}\hfill & \hfill \end{array}$

## Practice set b

Solve the following linear inequalities.

$y-6\le 5$

$y\le 11$

$x+4>9$

$x>5$

$4x-1\ge 15$

$x\ge 4$

$-5y+16\le 7$

$y\ge \frac{9}{5}$

$7\left(4s-3\right)<2s+8$

$s<\frac{29}{2}$

$5\left(1-4h\right)+4<\left(1-h\right)2+6$

$h>\frac{1}{18}$

$18\ge 4\left(2x-3\right)-9x$

$x\ge -30$

$-\frac{3b}{16}\le 4$

$b\ge \frac{-64}{3}$

$\frac{-7z+10}{-12}<-1$

$z<-\frac{2}{7}$

$-x-\frac{2}{3}\le \frac{5}{6}$

$x\ge \frac{-3}{2}$

## Compound inequality

Another type of inequality is the compound inequality . A compound inequality is of the form:

$a

There are actually two statements here. The first statement is $a . The next statement is $x . When we read this statement we say " $a$ is less than $x$ ," then continue saying "and $x$ is less than $b$ ."

Just by looking at the inequality we can see that the number $x$ is between the numbers $a$ and $b$ . The compound inequality $a indicates "betweenness." Without changing the meaning, the statement $a can be read $x>a$ . (Surely, if the number $a$ is less than the number $x$ , the number $x$ must be greater than the number $a$ .) Thus, we can read $a as " $x$ is greater than $a$ and at the same time is less than $b$ ." For example:

1. $4 .
The letter $x$ is some number strictly between 4 and 9. Hence, $x$ is greater than 4 and, at the same time, less than 9. The numbers 4 and 9 are not included so we use open circles at these points.
2. $-2 .
The $z$ stands for some number between $-2$ and 0. Hence, $z$ is greater than $-2$ but also less than 0.
3. $1 .
The expression $x+6$ represents some number strictly between 1 and 8. Hence, $x+6$ represents some number strictly greater than 1, but less than 8.
4. $\frac{1}{4}\le \frac{5x-2}{6}\le \frac{7}{9}$ .
The term $\frac{5x-2}{6}$ represents some number between and including $\frac{1}{4}$ and $\frac{7}{9}$ . Hence, $\frac{5x-2}{6}$ represents some number greater than or equal to $\frac{1}{4}$ to but less than or equal to $\frac{7}{9}$ .

Consider problem 3 above, $1 . The statement says that the quantity $x+6$ is between 1 and 8. This statement will be true for only certain values of $x$ . For example, if $x=1$ , the statement is true since $1<1+6<8$ . However, if $x=4.9$ , the statement is false since $1<4.9+6<8$ is clearly not true. The first of the inequalities is satisfied since 1 is less than $10.9$ , but the second inequality is not satisfied since $10.9$ is not less than 8.

We would like to know for exactly which values of $x$ the statement $1 is true. We proceed by using the properties discussed earlier in this section, but now we must apply the rules to all three parts rather than just the two parts in a regular inequality.

## Sample set c

Solve $1 .

$\begin{array}{ll}1-6

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

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

$\begin{array}{ll}-3\left(5\right)<\frac{-2x-7}{5}\left(5\right)<8\left(5\right)\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{each}\text{\hspace{0.17em}}\text{part}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}5.\hfill \\ -15<-2x-7<40\hfill & \text{Add}\text{\hspace{0.17em}}7\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{parts}.\hfill \\ -8<-2x<47\hfill & \text{Divide}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{parts}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-2.\hfill \\ 4>x>-\frac{47}{2}\hfill & \text{Remember}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{reverse}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{direction}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\hfill \\ \hfill & \text{signs}\text{.}\text{\hspace{0.17em}}\hfill \\ -\frac{47}{2}

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

## Practice set c

Find the values of $x$ that satisfy the given continued inequality.

$4

$9

$-3<7y+1<18$

$-\frac{4}{7}

$0\le 1-6x\le 7$

$-1\le x\le \frac{1}{6}$

$-5\le \frac{2x+1}{3}\le 10$

$-8\le x\le \frac{29}{2}$

$9<\frac{-4x+5}{-2}<14$

$\frac{23}{4}

Does $4 have a solution?

no

## Exercises

For the following problems, solve the inequalities.

$x+7<12$

$x<5$

$y-5\le 8$

$y+19\ge 2$

$y\ge -17$

$x-5>16$

$3x-7\le 8$

$x\le 3$

$9y-12\le 6$

$2z+8<7$

$z<-\frac{1}{2}$

$4x-14>21$

$-5x\le 20$

$x\ge -4$

$-8x<40$

$-7z<77$

$z>-11$

$-3y>39$

$\frac{x}{4}\ge 12$

$x\ge 48$

$\frac{y}{7}>3$

$\frac{2x}{9}\ge 4$

$x\ge 18$

$\frac{5y}{2}\ge 15$

$\frac{10x}{3}\le 4$

$x\le \frac{6}{5}$

$\frac{-5y}{4}<8$

$\frac{-12b}{5}<24$

$b>-10$

$\frac{-6a}{7}\le -24$

$\frac{8x}{-5}>6$

$x<-\frac{15}{4}$

$\frac{14y}{-3}\ge -18$

$\frac{21y}{-8}<-2$

$y>\frac{16}{21}$

$-3x+7\le -5$

$-7y+10\le -4$

$y\ge 2$

$6x-11<31$

$3x-15\le 30$

$x\le 15$

$-2y+\frac{4}{3}\le -\frac{2}{3}$

$5\left(2x-5\right)\ge 15$

$x\ge 4$

$4\left(x+1\right)>-12$

$6\left(3x-7\right)\ge 48$

$x\ge 5$

$3\left(-x+3\right)>-27$

$-4\left(y+3\right)>0$

$y<-3$

$-7\left(x-77\right)\le 0$

$2x-1

$x<6$

$6y+12\le 5y-1$

$3x+2\le 2x-5$

$x\le -7$

$4x+5>5x-11$

$3x-12\ge 7x+4$

$x\le -4$

$-2x-7>5x$

$-x-4>-3x+12$

$x>8$

$3-x\ge 4$

$5-y\le 14$

$y\ge -9$

$2-4x\le -3+x$

$3\left[4+5\left(x+1\right)\right]<-3$

$x<-2$

$2\left[6+2\left(3x-7\right)\right]\ge 4$

$7\left[-3-4\left(x-1\right)\right]\le 91$

$x\ge -3$

$-2\left(4x-1\right)<3\left(5x+8\right)$

$-5\left(3x-2\right)>-3\left(-x-15\right)+1$

$x<-2$

$-.0091x\ge 2.885x-12.014$

What numbers satisfy the condition: twice a number plus one is greater than negative three?

$x>-2$

What numbers satisfy the condition: eight more than three times a number is less than or equal to fourteen?

One number is five times larger than another number. The difference between these two numbers is less than twenty-four. What are the largest possible values for the two numbers? Is there a smallest possible value for either number?

First number: any number strictly smaller that 6.
Second number: any number strictly smaller than 30.
No smallest possible value for either number.
No largest possible value for either number.

The area of a rectangle is found by multiplying the length of the rectangle by the width of the rectangle. If the length of a rectangle is 8 feet, what is the largest possible measure for the width if it must be an integer (positive whole number) and the area must be less than 48 square feet?

## Exercises for review

( [link] ) Simplify ${\left({x}^{2}{y}^{3}{z}^{2}\right)}^{5}$ .

${x}^{10}{y}^{15}{z}^{10}$

( [link] ) Simplify $-\left[-\left(-|-8|\right)\right]$ .

( [link] ) Find the product. $\left(2x-7\right)\text{\hspace{0.17em}}\left(x+4\right)$ .

$2{x}^{2}+x-28$

( [link] ) Twenty-five percent of a number is $12.32$ . What is the number?

( [link] ) The perimeter of a triangle is 40 inches. If the length of each of the two legs is exactly twice the length of the base, how long is each leg?

16 inches

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