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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