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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
$\begin{array}{cc}8(3)<8(7)& \\ 24<56& \text{True}\end{array}$
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
$\begin{array}{cc}\frac{3}{-8}>\frac{7}{-8}& \text{True}\end{array}\text{\hspace{0.17em}}(\text{since}\text{\hspace{0.17em}}-.375-.875)$
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
$3x>15$ .
$\begin{array}{ll}2y-1\le 16\hfill & \text{Add}\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}.\hfill \\ 2y\le 17\hfill & \text{Divide}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}2.\hfill \\ y\le \frac{17}{2}\hfill & \hfill \end{array}$
$\begin{array}{ll}-8x+5<14\hfill & \text{Subtract}\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{.}\hfill \\ -8x<9\hfill & \text{Divide}\text{\hspace{0.17em}}\text{both}\text{\hspace{0.17em}}\text{sides}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-8.\text{\hspace{0.17em}}\text{We}\text{\hspace{0.17em}}\text{must}\text{\hspace{0.17em}}\text{reverse}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sense}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\hfill \\ \hfill & \text{since}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{dividing}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{negative}\text{\hspace{0.17em}}\text{number}\text{.}\hfill \\ x>-\frac{9}{8}\hfill & \hfill \end{array}$
$\begin{array}{l}5-3(y+2)<6y-10\hfill \\ 5-3y-6<6y-10\hfill \\ -3y-1<6y-10\hfill \\ -9y<-9\hfill \\ y>1\hfill \end{array}$
$\begin{array}{ll}\frac{2z+7}{-4}\ge -6\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-4\hfill \\ 2z+7\le 24\hfill & \text{Notice}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{change}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sense}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\text{.}\hfill \\ 2z\le 17\hfill & \hfill \\ z\le \frac{17}{2}\hfill & \hfill \end{array}$
Solve the following linear inequalities.
$a<x<b$
There are actually two statements here. The first statement is $a<x$ . The next statement is $x<b$ . 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 $a<x<b$ indicates "betweenness." Without changing the meaning, the statement $a<x$ 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 $a<x<b$ as " $x$ is greater than $a$ and at the same time is less than $b$ ." For example:
Consider problem 3 above, $1<x+6<8$ . 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 $1<x+6<8$ 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.
Solve $1<x+6<8$ .
$\begin{array}{ll}1-6<x+6-6<8-6\hfill & \text{Subtract}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\text{from}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{parts}\text{.}\hfill \\ -5<x<2\hfill & \hfill \end{array}$
Thus, if $x$ is any number strictly between $-5$ and 2, the statement $1<x+6<8$ will be true.
Solve $-3<\frac{-2x-7}{5}<8$ .
$\begin{array}{ll}-3(5)<\frac{-2x-7}{5}(5)<8(5)\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{each}\text{\hspace{0.17em}}\text{part}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}5.\hfill \\ -15<-2x-7<40\hfill & \text{Add}\text{\hspace{0.17em}}7\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{parts}.\hfill \\ -8<-2x<47\hfill & \text{Divide}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}\text{three}\text{\hspace{0.17em}}\text{parts}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}-2.\hfill \\ 4>x>-\frac{47}{2}\hfill & \text{Remember}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{reverse}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{direction}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\hfill \\ \hfill & \text{signs}\text{.}\text{\hspace{0.17em}}\hfill \\ -\frac{47}{2}<x<4\hfill & \text{It}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{customary}\text{\hspace{0.17em}}\text{(but}\text{\hspace{0.17em}}\text{not}\text{\hspace{0.17em}}\text{necessary)}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{write}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{inequality}\hfill \\ \hfill & \text{so}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{inequality}\text{\hspace{0.17em}}\text{arrows}\text{\hspace{0.17em}}\text{point}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{left}\text{.}\hfill \end{array}$
Thus, if $x$ is any number between $-\frac{47}{2}$ and 4, the original inequality will be satisfied.
Find the values of $x$ that satisfy the given continued inequality.
For the following problems, solve the inequalities.
$y-5\le 8$
$9y-12\le 6$
$4x-14>21$
$\frac{y}{7}>3$
$\frac{5y}{2}\ge 15$
$\frac{-5y}{4}<8$
$\frac{-6a}{7}\le -24$
$\frac{14y}{-3}\ge -18$
$-3x+7\le -5$
$6x-11<31$
$-2y+\frac{4}{3}\le -\frac{2}{3}$
$4(x+1)>-12$
$3(-x+3)>-27$
$-7(x-77)\le 0$
$6y+12\le 5y-1$
$4x+5>5x-11$
$-2x-7>5x$
$3-x\ge 4$
$2-4x\le -3+x$
$2[6+2(3x-7)]\ge 4$
$-2(4x-1)<3(5x+8)$
$-.0091x\ge 2.885x-12.014$
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?
( [link] ) Simplify ${({x}^{2}{y}^{3}{z}^{2})}^{5}$ .
${x}^{10}{y}^{15}{z}^{10}$
( [link] ) Simplify $-[-(-\left|-8\right|)]$ .
( [link] ) Find the product. $(2x-7)\text{\hspace{0.17em}}(x+4)$ .
$2{x}^{2}+x-28$
( [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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