What Is Another Way to Write the Absolute Value Inequality P 12

Solving Linear Equations

Solve Absolute Value Inequalities

Learning Objectives

Past the end of this section, you volition be able to:

Solve accented value equations
Solve absolute value inequalities with "less than"
Solve absolute value inequalities with "greater than"
Solve applications with absolute value

Earlier you get started, take this readiness quiz.

Evaluate: $\text{−}|7|.$
If you missed this problem, review (Figure).
Fill in $\text{<},\text{>},$ or $=$ for each of the following pairs of numbers.
ⓐ $|-8|\text{___}-|-8|$ ⓑ $12\text{___}-|-12|$ ⓒ $|-6|\text{___}-6$ ⓓ $\text{−}\left(-15\right)\text{___}-|-15|$

If you missed this problem, review (Figure).
Simplify: $14-2|8-3\left(4-1\right)|.$
If you missed this problem, review (Figure).

Solve Absolute Value Equations

As nosotros ready to solve absolute value equations, we review our definition of accented value.

Absolute Value

The accented value of a number is its distance from zero on the number line.

The accented value of a number n is written every bit $|n|$ and $|n|\ge 0$ for all numbers.

Absolute values are e'er greater than or equal to zero.

Nosotros learned that both a number and its opposite are the same altitude from zero on the number line. Since they accept the aforementioned altitude from zero, they have the same absolute value. For example:

$\phantom{\rule{3em}{0ex}}-5$ is v units abroad from 0, so $|-5|=5.$

$\phantom{\rule{3.65em}{0ex}}5$ is 5 units abroad from 0, and so $|5|=5.$

(Effigy) illustrates this idea.

The numbers 5 and $-5$ are both five units abroad from zip.

The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.

For the equation $|x|=5,$ nosotros are looking for all numbers that brand this a truthful statement. We are looking for the numbers whose altitude from zero is five. We but saw that both 5 and $-5$ are five units from zero on the number line. They are the solutions to the equation.

$\begin{array}{cccccc}\text{If}\hfill & & & & & |x|=5\hfill \\ \text{then}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}x=-5\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}x=5\hfill \end{array}$

The solution tin can be simplified to a single statement past writing $x=\text{±}5.$ This is read, "x is equal to positive or negative 5".

Nosotros tin can generalize this to the following property for absolute value equations.

Absolute Value Equations

For whatever algebraic expression, u, and any positive real number, a,

$\begin{array}{cccccc}\text{if}\hfill & & & & & |u|=a\hfill \\ \text{then}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}u=\text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u=a\hfill \end{array}$

Retrieve that an absolute value cannot be a negative number.

Solve: ⓐ $|x|=8$ ⓑ $|y|=-6$ ⓒ $|z|=0$

ⓐ

$\begin{array}{cccccc}& & & & & |x|=8\hfill \\ \text{Write the equivalent equations.}\hfill & & & & & \phantom{\rule{0.45em}{0ex}}x=-8\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.4em}{0ex}}x=8\hfill \\ & & & & & \phantom{\rule{0.45em}{0ex}}x=\text{±}8\hfill \end{array}$

ⓑ

$\begin{array}{cccccc}& & & & & \phantom{\rule{12.6em}{0ex}}|y|=-6\hfill \\ & & & & & \phantom{\rule{12.6em}{0ex}}\text{No solution}\hfill \end{array}$

Since an accented value is ever positive, there are no solutions to this equation.

ⓒ

$\begin{array}{cccccc}& & & & & |z|=0\hfill \\ \text{Write the equivalent equations.}\hfill & & & & & \phantom{\rule{0.45em}{0ex}}z=-0\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}z=0\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}-0=0,\hfill & & & & & \phantom{\rule{0.45em}{0ex}}z=0\hfill \end{array}$

Both equations tell u.s.a. that $z=0$ and and then there is only one solution.

Solve: ⓐ $|x|=2$ ⓑ $|y|=-4$ ⓒ $|z|=0$

ⓐ $\text{±}2$ ⓑ no solution ⓒ 0

Solve: ⓐ $|x|=11$ ⓑ $|y|=-5$ ⓒ $|z|=0$

ⓐ $\text{±}11$ ⓑ no solution ⓒ 0

To solve an absolute value equation, we offset isolate the absolute value expression using the same procedures nosotros used to solve linear equations. Once we isolate the absolute value expression we rewrite information technology as the two equivalent equations.

How to Solve Absolute Value Equations

Solve: $|3x-5|-1=6.$

$x=4,x=-\frac{2}{3}$

Solve: $|4x-3|-5=2.$

$x=-1,x=\frac{5}{2}$

The steps for solving an absolute value equation are summarized hither.

Solve absolute value equations.

Isolate the absolute value expression.
Write the equivalent equations.
Solve each equation.
Cheque each solution.

Solve $2|x-7|+5=9.$

Solve: $3|x-4|-4=8.$

$x=8,x=0$

Solve: $2|x-5|+3=9.$

$x=8,x=2$

Recollect, an absolute value is e'er positive!

Solve: $|\frac{2}{3}x-4|+11=3.$

$\begin{array}{cccccc}& & & & & |\frac{2}{3}x-4|+11=3\hfill \\ \text{Isolate the absolute value term.}\hfill & & & & & \phantom{\rule{2.2em}{0ex}}|\frac{2}{3}x-4|=-8\hfill \\ \text{An absolute value cannot be negative.}\hfill & & & & & \phantom{\rule{1.2em}{0ex}}\text{No solution}\hfill \end{array}$

Solve: $|\frac{3}{4}x-5|+9=4.$

No solution

Solve: $|\frac{5}{6}x+3|+8=6.$

No solution

Some of our accented value equations could exist of the form $|u|=|v|$ where u and 5 are algebraic expressions. For case, $|x-3|=|2x+1|.$

How would we solve them? If ii algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The belongings for absolute value equations says that for whatsoever algebraic expression, u, and a positive real number, a, if $|u|=a,$ then $u=\text{−}a$ or $u=a.$

This tell us that

$\begin{array}{cccccccccccc}\text{if}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}|u|=|v|\hfill & & & & & & \\ \text{then}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}u=\text{−}v\hfill & \phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u=v& & \text{or}\hfill & & & u=v\hfill \end{array}$

This leads us to the post-obit property for equations with two absolute values.

Equations with Two Absolute Values

For any algebraic expressions, u and 5,

$\begin{array}{cccccc}\text{if}\hfill & & & & & |u|=|v|\hfill \\ \text{then}\hfill & & & & & \phantom{\rule{0.3em}{0ex}}u=\text{−}v\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u=v\hfill \end{array}$

When we take the contrary of a quantity, nosotros must be careful with the signs and to add parentheses where needed.

Solve Absolute Value Inequalities with "Less Than"

Let's look at present at what happens when we have an absolute value inequality. Everything we've learned about solving inequalities still holds, only we must consider how the accented value impacts our piece of work.

Again we will await at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation $|x|=5,$ we saw that both 5 and $-5$ are five units from zero on the number line. They are the solutions to the equation.

$\begin{array}{c}\hfill |x|=5\hfill \\ \hfill x=-5\phantom{\rule{4em}{0ex}}\text{or}\phantom{\rule{4em}{0ex}}x=5\hfill \end{array}$

What well-nigh the inequality $|x|\le 5?$ Where are the numbers whose distance is less than or equal to 5? We know $-5$ and 5 are both five units from nothing. All the numbers between $-5$ and 5 are less than 5 units from zilch. Meet (Figure).

The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.

In a more than full general way, we can see that if $|u|\le a,$ then $\text{−}a\le u\le a.$ Come across (Effigy).

The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.

This result is summarized here.

Absolute Value Inequalities with $<$ or $\le$

For any algebraic expression, u, and any positive existent number, a,

$\begin{array}{cccccccccc}\text{if}\hfill & & & & |u|<a,\hfill & & & & & \text{so}\phantom{\dominion{0.5em}{0ex}}\text{−}a<u<a\hfill \\ \text{if}\hfill & & & & |u|\le a,\hfill & & & & & \text{then}\phantom{\rule{0.5em}{0ex}}\text{−}a\le u\le a\hfill \end{array}$

Later on solving an inequality, information technology is often helpful to check some points to run across if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete bank check, information technology often helps verify the solution.

Solve Graph the solution and write the solution in interval annotation.


Write the equivalent inequality.
Graph the solution.
Write the solution using interval notation.

Cheque:

To verify, bank check a value in each section of the number line showing the solution. Choose numbers such equally $-8,$ i, and 9.

Graph the solution and write the solution in interval notation:

The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.

Graph the solution and write the solution in interval notation:

The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.

Solve $|5x-6|\le 4.$ Graph the solution and write the solution in interval notation.

Step 1. Isolate the accented value expression. Information technology is isolated.	$\|5x-6\|\le 4$
Footstep ii. Write the equivalent compound inequality.	$-4\le 5x-6\le 4$
Step iii. Solve the chemical compound inequality.	$2\le 5x\le 10$ $\frac{2}{5}\le x\le 2$
Step 4. Graph the solution.
Stride 5. Write the solution using interval notation.	$\left[\frac{2}{5},2\right]$
Check: The check is left to you.

Solve $|2x-1|\le 5.$ Graph the solution and write the solution in interval notation:

The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.

Solve $|4x-5|\le 3.$ Graph the solution and write the solution in interval note:

The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.

Solve absolute value inequalities with < or ≤.

Isolate the absolute value expression.
Write the equivalent compound inequality.
$\begin{array}{ccccccccccc}|u|<a\hfill & & & & & \text{is equivalent to}\hfill & & & & & \text{−}a<u<a\hfill \\ |u|\le a\hfill & & & & & \text{is equivalent to}\hfill & & & & & \text{−}a\le u\le a\hfill \end{array}$
Solve the compound inequality.
Graph the solution
Write the solution using interval notation.

Solve Absolute Value Inequalities with "Greater Than"

What happens for absolute value inequalities that have "greater than"? Over again nosotros will wait at our definition of absolute value. The absolute value of a number is its altitude from zero on the number line.

We started with the inequality $|x|\le 5.$ We saw that the numbers whose distance is less than or equal to five from cipher on the number line were $-5$ and 5 and all the numbers betwixt $-5$ and v. See (Effigy).

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.

At present nosotros want to look at the inequality $|x|\ge 5.$ Where are the numbers whose altitude from cypher is greater than or equal to five?

Again both $-5$ and five are v units from zero and and then are included in the solution. Numbers whose altitude from zero is greater than five units would exist less than $-5$ and greater than 5 on the number line. See (Effigy).

In a more general manner, we tin can run into that if $|u|\ge a,$ then $u\le \text{−}a$ or $u\le a.$ See (Figure).

This result is summarized here.

Absolute Value Inequalities with > or ≥

For whatsoever algebraic expression, u, and any positive real number, a,

$\begin{array}{ccccccccccc}\text{if}\hfill & & & & & |u|>a,\hfill & & & & & \text{then}\phantom{\rule{0.2em}{0ex}}u<-a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u>a\hfill \\ \text{if}\hfill & & & & & |u|\ge a,\hfill & & & & & \text{then}\phantom{\rule{0.2em}{0ex}}u\le \text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u\ge a\hfill \end{array}$

Solve $|x|>4.$ Graph the solution and write the solution in interval notation.

Solve $|x|>2.$ Graph the solution and write the solution in interval notation.

The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.

Solve $|x|>1.$ Graph the solution and write the solution in interval notation.

The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.

Solve $|2x-3|\ge 5.$ Graph the solution and write the solution in interval notation.

	$\|2x-3\|\ge 5$
Step i. Isolate the absolute value expression. It is isolated.
Step ii. Write the equivalent compound inequality.	$2x-3\le -5\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}2x-3\ge 5$
Step 3. Solve the compound inequality.	$2x\le \text{−}2\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}2x\ge 8$ $x\le \text{−}1\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}x\ge 4$
Step 4. Graph the solution.
Footstep 5. Write the solution using interval notation.	$\left(\text{−}\infty ,-1\right]\cup \left[4,\infty \right)$
Bank check: The cheque is left to you.

Solve $|4x-3|\ge 5.$ Graph the solution and write the solution in interval annotation.

The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis

Solve $|3x-4|\ge 2.$ Graph the solution and write the solution in interval notation.

The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

Solve absolute value inequalities with > or ≥.

Isolate the absolute value expression.
Write the equivalent compound inequality.
$\begin{array}{c}|u|>a\phantom{\rule{2em}{0ex}}\text{is equivalent to}\phantom{\rule{2em}{0ex}}u<-a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u>a\hfill \\ |u|\ge a\phantom{\rule{2em}{0ex}}\text{is equivalent to}\phantom{\rule{2em}{0ex}}u\le \text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u\ge a\hfill \end{array}$
Solve the compound inequality.
Graph the solution
Write the solution using interval notation.

Solve Applications with Absolute Value

Absolute value inequalities are often used in the manufacturing process. An particular must exist made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

$|\text{actual-ideal}|\le \text{tolerance}$

The ideal diameter of a rod needed for a car is lx mm. The actual diameter tin can vary from the ideal diameter past $0.075$ mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

$\begin{array}{cccccc}& & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}x=\text{the actual measurement.}\hfill \\ \text{Use an absolute value inequality to express this situation.}\hfill & & & & & |\text{actual-ideal}|\le \text{tolerance}\hfill \\ & & & & & \hfill |x-60|\le 0.075\hfill \\ \text{Rewrite as a compound inequality.}\hfill & & & & & \hfill \text{−}0.075\le x-60\le 0.075\hfill \\ \text{Solve the inequality.}\hfill & & & & & \hfill 59.925\le x\le 60.075\hfill \\ \text{Answer the question.}\hfill & & & & & \text{The diameter of the rod can be between}\hfill \\ & & & & & \text{59.925 mm and 60.075 mm.}\hfill \end{array}$

The ideal bore of a rod needed for a machine is eighty mm. The bodily diameter tin vary from the platonic diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

The diameter of the rod tin can be between 79.991 and eighty.009 mm.

The ideal diameter of a rod needed for a auto is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the client without causing the rod to be rejected?

The diameter of the rod can be between 74.95 and 75.05 mm.

Key Concepts

Absolute Value
The accented value of a number is its distance from 0 on the number line.

The absolute value of a number due north is written equally $|n|$ and $|n|\ge 0$ for all numbers.

Absolute values are always greater than or equal to goose egg.
Absolute Value Equations
For any algebraic expression, u, and any positive real number, a,

$\begin{array}{ccccc}\text{if}\hfill & & & & |u|=a\hfill \\ \text{then}\hfill & & & & \phantom{\rule{0.3em}{0ex}}u=\text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u=a\hfill \end{array}$

Remember that an accented value cannot be a negative number.
How to Solve Absolute Value Equations
1. Isolate the accented value expression.
2. Write the equivalent equations.
3. Solve each equation.
4. Check each solution.
Equations with 2 Absolute Values
For whatever algebraic expressions, u and five,

$\begin{array}{ccccc}\text{if}\hfill & & & & |u|=|v|\hfill \\ \text{then}\hfill & & & & \phantom{\rule{0.3em}{0ex}}u=\text{−}v\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u=v\hfill \end{array}$
Absolute Value Inequalities with $<$ or $\le$
For whatever algebraic expression, u, and whatsoever positive real number, a,

$\begin{assortment}{ccccccccccc}\text{if}\hfill & & & & & |u|<a,\hfill & & & & & \text{then}\phantom{\dominion{0.5em}{0ex}}\text{−}a<u<a\hfill \\ \text{if}\hfill & & & & & |u|\le a,\hfill & & & & & \text{then}\phantom{\rule{0.5em}{0ex}}\text{−}a\le u\le a\hfill \end{array}$
How To Solve Absolute Value Inequalities with or
1. Isolate the absolute value expression.
2. Write the equivalent compound inequality.
  $\begin{array}{ccccccccccc}\hfill |u|<a\hfill & & & & & \hfill \text{is equivalent to}\hfill & & & & & \hfill \text{−}a<u<a\hfill \\ \hfill |u|\le a\hfill & & & & & \hfill \text{is equivalent to}\hfill & & & & & \hfill \text{−}a\le u\le a\hfill \end{array}$
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval notation
Absolute Value Inequalities with $>$ or $\ge$
For any algebraic expression, u, and any positive existent number, a,

$\begin{array}{ccccccccccc}\text{if}\hfill & & & & & |u|>a,\hfill & & & & & \text{then}\phantom{\rule{0.2em}{0ex}}u<\text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u>a\hfill \\ \text{if}\hfill & & & & & |u|\ge a,\hfill & & & & & \text{then}\phantom{\rule{0.2em}{0ex}}u\le \text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u\ge a\hfill \end{array}$
How To Solve Absolute Value Inequalities with or
1. Isolate the absolute value expression.
2. Write the equivalent compound inequality.
  $\begin{array}{ccccccccc}\hfill |u|>a\hfill & & & & \hfill \text{is equivalent to}\hfill & & & & \hfill u<\text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u>a\hfill \\ \hfill |u|\ge a\hfill & & & & \hfill \text{is equivalent to}\hfill & & & & \hfill u\le \text{−}a\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}u\ge a\hfill \end{array}$
3. Solve the compound inequality.
4. Graph the solution
5. Write the solution using interval annotation

Section Exercises

Practice Makes Perfect

Solve Accented Value Equations

In the following exercises, solve.

ⓐ $x=4,x=-4$ ⓑ no solution ⓒ $z=0$

ⓐ $x=3,x=-3$ ⓑ no solution ⓒ $z=0$

$|2x-3|-4=1$

$|4x-1|-3=0$

$x=1,x=-\frac{1}{2}$

$|3x-4|+5=7$

$|4x+7|+2=5$

$x=-1,x=-\frac{5}{2}$

$4|x-1|+2=10$

$3|x-4|+2=11$

$x=7,x=1$

$3|4x-5|-4=11$

$3|x+2|-5=4$

$x=1,x=-5$

$-2|x-3|+8=-4$

$-3|x-4|+4=-5$

$x=7,x=1$

$|\frac{3}{4}x-3|+7=2$

$|\frac{3}{5}x-2|+5=2$

no solution

$|\frac{1}{2}x+5|+4=1$

$|\frac{1}{4}x+3|+3=1$

no solution

$|3x-2|=|2x-3|$

$|4x+3|=|2x+1|$

$x=-1,x=-\frac{2}{3}$

$|6x-5|=|2x+3|$

$|6-x|=|3-2x|$

$x=-3,x=3$

Solve Absolute Value Inequalities with "less than"

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

$|x|<5$

The solution is negative 1 is less than x which is less than 1. The number line shows an open circle at negative 1, an open circle at 1, and shading between the circles. The interval notation is negative 1 to 1 within parentheses.

$|x|\le 8$

$|x|\le 3$

The solution is negative 3 is less than or equal to x which is less than or equal to 3. The number line shows a closed circle at negative 3, a closed circle at 3, and shading between the circles. The interval notation is negative 3 to 3 within brackets.

$|3x-3|\le 6$

$|2x-5|\le 3$

The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading between the circles. The interval notation is 1 to 4 within brackets.

$|3x-7|+iii<1$

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

$|6x-5|<7$

The solution is negative one-third is less than x which is less than 2. The number line shows an open circle at negative one-half, an open circle at 2, and shading between the circles. The interval notation is negative one-third to 2 within parentheses.

$|x-4|\le -1$

$|5x+1|\le -2$

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

Solve Absolute Value Inequalities with "greater than"

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

$|x|>3$

$|x|>6$

The solution is x is less than negative 6 or x is greater than 6. The number line shows an open circle at negative 6 with shading to its left and an open circle at 6 with shading to its right. The interval notation is the union of negative infinity to negative 6 within parentheses and 6 to infinity within parentheses

$|x|\ge 2$

$|x|\ge 5$

The solution is x is less than negative 5 or x is greater than 5. The number line shows an open circle at negative 5 with shading to its left and an open circle at 5 with shading to its right. The interval notation is the union of negative infinity to negative 5 within parentheses and 5 to infinity within parentheses.

$|3x-8|>\text{−}1$

$|x-5|>\text{−}2$

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

$|3x-2|>4$

$|2x-1|>5$

The solution is x is less than negative 2 or x is greater than 3. The number line shows an open circle at negative 2 with shading to its left and an open circle at 3 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 3 to infinity within parentheses.

$|x+3|\ge 5$

$|x-7|\ge 1$

The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to 6 within parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.

$3|x|+4\ge 1$

$5|x|+6\ge 1$

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

In the post-obit exercises, solve. For each inequality, too graph the solution and write the solution in interval notation.

$2|x+6|+4=8$

$|3x-4|\ge 2$

$x=4,x=\frac{2}{7}$

$|6x-5|=|2x+3|$

$x=3,x=2$

$|2x-5|+2=3$

$|3x+1|-3=7$

$x=3,x=-\frac{11}{3}$

$|7x+two|+viii<4$

$5|2x-1|-3=7$

$x=\frac{3}{2},x=-\frac{1}{2}$

$|x-7|>\text{−}3$

$|8-x|=|4-3x|$

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

Solve Applications with Absolute Value

In the post-obit exercises, solve.

A chicken subcontract ideally produces 200,000 eggs per twenty-four hours. Simply this total tin vary by as much every bit 25,000 eggs. What is the maximum and minimum expected production at the farm?

An organic juice bottler ideally produces 215,000 bottle per 24-hour interval. Only this total can vary by as much as 7,500 bottles. What is the maximum and minimum expected production at the bottling company?

The minimum to maximum expected product is 207,500 to 2,225,000 bottles

In society to insure compliance with the police, Miguel routinely overshoots the weight of his tortillas by 0.5 gram. He just received a report that told him that he could be losing every bit much equally ?100,000 per year using this practise. He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches. If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will exist guaranteed?

At Lilly's Baker, the ideal weight of a loaf of bread is 24 ounces. By constabulary, the actual weight can vary from the ideal by one.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

The acceptable weight is 22.5 to 25.5 ounces.

Writing Exercises

Write a graphical clarification of the absolute value of a number.

In your own words, explain how to solve the accented value inequality, $|3x-2|\ge 4.$

Answers will vary.

Self Bank check

ⓐ After completing the exercises, utilise this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and five rows. The first row is a header and it labels each column,

ⓑ What does this checklist tell yous well-nigh your mastery of this section? What steps will you have to improve?

Affiliate Review Exercises

Employ a General Strategy to Solve Linear Equations

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, determine whether each number is a solution to the equation.

$10x-1=5x,x=\frac{1}{5}$

$-12n+5=8n,n=-\frac{5}{4}$

In the post-obit exercises, solve each linear equation.

$6\left(x+6\right)=24$

$\text{−}\left(s+4\right)=18$

$s=-22$

$23-3\left(y-7\right)=8$

$\frac{1}{3}\left(6m+21\right)=m-7$

$m=-14$

$4\left(3.5y+0.25\right)=365$

$0.25\left(q-8\right)=0.1\left(q+7\right)$

$q=18$

$8\left(r-2\right)=6\left(r+10\right)$

$5+7\left(2-5x\right)=2\left(9x+1\right)-\left(13x-57\right)$

$x=-1$

$\left(9n+5\right)-\left(3n-7\right)=20-\left(4n-2\right)$

$2\left[-16+5\left(8k-6\right)\right]=8\left(3-4k\right)-32$

$k=\frac{3}{4}$

Allocate Equations

In the following exercises, classify each equation equally a conditional equation, an identity, or a contradiction and so land the solution.

$17y-3\left(4-2y\right)=11\left(y-1\right)+12y-1$

$9u+32=15\left(u-4\right)-3\left(2u+21\right)$

contradiction; no solution

$-8\left(7m+4\right)=-6\left(8m+9\right)$

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation.

$\frac{2}{5}n-\frac{1}{10}=\frac{7}{10}$

$n=2$

$\frac{3}{4}a-\frac{1}{3}=\frac{1}{2}a+\frac{5}{6}$

$\frac{1}{2}\left(k+3\right)=\frac{1}{3}\left(k+16\right)$

$k=23$

$\frac{5y-1}{3}+4=\frac{-8y+4}{6}$

$0.8x-0.3=0.7x+0.2$

$x=5$

$0.10d+0.05\left(d-4\right)=2.05$

Use a Problem-Solving Strategy

Use a Problem Solving Strategy for Word Problems

In the following exercises, solve using the problem solving strategy for word issues.

3-fourths of the people at a concert are children. If there are 87 children, what is the total number of people at the concert?

There are 116 people.

There are ix saxophone players in the band. The number of saxophone players is i less than twice the number of tuba players. Find the number of tuba players.

Solve Number Give-and-take Problems

In the following exercises, solve each number word problem.

The sum of a number and 3 is forty-ane. Observe the number.

One number is ix less than another. Their sum is negative xx-seven. Notice the numbers.

One number is two more than than four times another. Their sum is negative thirteen. Notice the numbers.

$-3,-10$

The sum of two consecutive integers is $-135.$ Notice the numbers.

Find 3 consecutive even integers whose sum is 234.

76, 78, 80

Notice three consecutive odd integers whose sum is 51.

Koji has ?5,502 in his savings business relationship. This is ?30 less than six times the amount in his checking account. How much money does Koji accept in his checking account?

?922

Solve Percent Applications

In the following exercises, translate and solve.

What number is 67% of 250?

12.5% of what number is twenty?

160

What per centum of 125 is 150?

In the following exercises, solve.

The bill for Dino's lunch was ?nineteen.45. He wanted to leave 20% of the full bill equally a tip. How much should the tip be?

$\text{?}3.89$

Dolores bought a crib on auction for ?350. The auction price was twoscore% of the original price. What was the original price of the crib?

Jaden earns ?2,680 per month. He pays ?938 a calendar month for rent. What percent of his monthly pay goes to rent?

35%

Angel received a raise in his almanac salary from ?55,400 to ?56,785. Find the percent change.

Rowena's monthly gasoline bill dropped from ?83.75 last month to ?56.95 this month. Observe the percent alter.

32%

Emmett bought a pair of shoes on sale at 40% off from an original price of ?138. Find ⓐ the amount of discount and ⓑ the sale price.

Lacey bought a pair of boots on auction for ?95. The original cost of the boots was ?200. Find ⓐ the amount of discount and ⓑ the discount charge per unit. (Circular to the nearest tenth of a per centum, if needed.)

ⓐ ?105 ⓑ $52.5\text{%}$

Nga and Lauren bought a chest at a flea market place for ?50. They re-finished it and then added a 350% mark-up. Find ⓐ the amount of the mark-up and ⓑ the list price.

Solve Simple Interest Applications

In the post-obit exercises, solve.

Winston deposited ?iii,294 in a depository financial institution account with interest rate 2.6% How much interest was earned in v years?

?428.22

Moira borrowed ?4,500 from her gramps to pay for her first year of higher. 3 years subsequently, she repaid the ?4,500 plus ?243 interest. What was the rate of interest?

Jaime's refrigerator loan statement said he would pay ?1,026 in interest for a four-twelvemonth loan at xiii.5%. How much did Jaime infringe to purchase the refrigerator?

?1,900

Solve a formula for a Specific Variable

Solve a Formula for a Specific Variable

In the following exercises, solve the formula for the specified variable.

Solve the formula

$V=LWH$ for L.

Solve the formula

$h=48t+\frac{1}{2}a{t}^{2}$ for t.

Solve the formula

$4x-3y=12$ for y.

$y=\frac{4x}{3}-4$

Use Formulas to Solve Geometry Applications

In the following exercises, solve using a geometry formula.

What is the peak of a triangle with surface area $67.5$ square meters and base 9 meters?

The measure of the smallest angle in a right triangle is $45\text{°}$ less than the measure of the adjacent larger bending. Find the measures of all 3 angles.

$22.5\text{°},\phantom{\rule{0.2em}{0ex}}67.5\text{°},90\text{°}$

The perimeter of a triangle is 97 feet. One side of the triangle is eleven feet more than the smallest side. The tertiary side is six feet more than than twice the smallest side. Find the lengths of all sides.

Observe the length of the hypotenuse.

The figure is a right triangle with a base of 10 units and a height of 24 units.

Discover the length of the missing side. Round to the nearest tenth, if necessary.

The figure is a right triangle with a height of 15 units and a hypotenuse of 17 units.

Sergio needs to attach a wire to hold the antenna to the roof of his house, as shown in the figure. The antenna is eight feet alpine and Sergio has ten feet of wire. How far from the base of the antenna tin can he attach the wire? Estimate to the nearest tenth, if necessary.

The figure is a right triangle with a height of 8 feet and a hypotenuse of 10 feet.

6 feet

Seong is building shelving in his garage. The shelves are 36 inches broad and 15 inches tall. He wants to put a diagonal brace beyond the back to stabilize the shelves, as shown. How long should the brace be?

The figure illustrates rectangular shelving whose width of 36 inch and height of 15 inches forms a right triangle with a diagonal brace.

The length of a rectangle is 12 cm more than the width. The perimeter is 74 cm. Notice the length and the width.

$24.5$ cm, $12.5$ cm

The width of a rectangle is three more than twice the length. The perimeter is 96 inches. Find the length and the width.

The perimeter of a triangle is 35 feet. 1 side of the triangle is five feet longer than the 2d side. The tertiary side is 3 feet longer than the 2d side. Find the length of each side.

9 ft, 14 ft, 12 ft

Solve Mixture and Uniform Move Applications

Solve Coin Give-and-take Problems

In the post-obit exercises, solve.

Paulette has ?140 in ?5 and ?10 bills. The number of ?ten bills is one less than twice the number of ?five bills. How many of each does she have?

Lenny has ?3.69 in pennies, dimes, and quarters. The number of pennies is three more than the number of dimes. The number of quarters is twice the number of dimes. How many of each money does he have?

ix pennies, vi dimes, 12 quarters

Solve Ticket and Stamp Word Problems

In the following exercises, solve each ticket or stamp give-and-take trouble.

Tickets for a basketball game cost ?2 for students and ?v for adults. The number of students was three less than 10 times the number of adults. The total amount of money from ticket sales was ?619. How many of each ticket were sold?

125 tickets were sold for the jazz ring concert for a full of ?1,022. Pupil tickets cost ?6 each and general admission tickets cost ?10 each. How many of each kind of ticket were sold?

57 students, 68 adults

Yumi spent ?34.fifteen buying stamps. The number of ?0.56 stamps she bought was 10 less than 4 times the number of ?0.41 stamps. How many of each did she purchase?

Solve Mixture Word Problems

In the following exercises, solve.

Marquese is making 10 pounds of trail mix from raisins and basics. Raisins cost ?three.45 per pound and basics toll ?7.95 per pound. How many pounds of raisins and how many pounds of nuts should Marquese use for the trail mix to cost him ?six.96 per pound?

$2.2$ lbs of raisins, $7.8$ lbs of nuts

Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square anxiety of tile. She will use basic tiles that cost ?eight per square pes and decorator tiles that cost ?20 per square pes. How many foursquare anxiety of each tile should she utilise then that the overall cost of the backsplash will exist ?10 per square foot?

Enrique borrowed ?23,500 to buy a car. He pays his uncle ii% interest on the ?four,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the full ?23,500? (Round your respond to the nearest tenth of a percent.)

*** QuickLaTeX cannot compile formula: 9.seven\text{%}  *** Error bulletin: File ended while scanning use of \text@. Emergency stop.

Solve Uniform Move Applications

In the following exercises, solve.

When Gabe drives from Sacramento to Redding it takes him 2.2 hours. Information technology takes Elsa two hours to drive the aforementioned distance. Elsa's speed is 7 miles per hour faster than Gabe'due south speed. Find Gabe'due south speed and Elsa's speed.

Louellen and Tracy met at a eatery on the road between Chicago and Nashville. Louellen had left Chicago and drove three.ii hours towards Nashville. Tracy had left Nashville and drove 4 hours towards Chicago, at a speed 1 mile per hour faster than Louellen's speed. The distance between Chicago and Nashville is 472 miles. Find Louellen's speed and Tracy'due south speed.

Louellen 65 mph, Tracy 66 mph

Two busses leave Amarillo at the aforementioned time. The Albuquerque bus heads due west on the I-40 at a speed of 72 miles per hour, and the Oklahoma City omnibus heads east on the I-40 at a speed of 78 miles per 60 minutes. How many hours will it take them to be 375 miles apart?

Kyle rowed his gunkhole upstream for 50 minutes. Information technology took him xxx minutes to row back downstream. His speed going upstream is two miles per 60 minutes slower than his speed going downstream. Find Kyle'south upstream and downstream speeds.

upstream three mph, downstream 5 mph

At six:30, Devon left her house and rode her bike on the flat route until 7:30. Then she started riding uphill and rode until 8:00. She rode a total of 15 miles. Her speed on the apartment route was three miles per hour faster than her speed going uphill. Find Devon'due south speed on the flat road and riding uphill.

Anthony collection from New York Metropolis to Baltimore, which is a distance of 192 miles. He left at 3:45 and had heavy traffic until 5:30. Traffic was lite for the rest of the drive, and he arrived at 7:30. His speed in lite traffic was four miles per hour more than than twice his speed in heavy traffic. Find Anthony's driving speed in heavy traffic and lite traffic.

heavy traffic 32 mph, low-cal traffic 66 mph

Solve Linear Inequalities

Graph Inequalities on the Number Line

In the following exercises, graph the inequality on the number line and write in interval annotation.

$x<\text{−}1$

$x\ge -2.5$

The solution is x is greater than or equal to negative 2.5. The number line shows a left bracket at negative 2.5 with shading to its right. The interval notation is negative 2.5 to infinity within a bracket and a parenthesis.

$x\le \frac{5}{4}$

$x>2$

The solution is x is greater than 2. The number line shows a left parenthesis at 2 with shading to its right. The interval notation is 2 to infinity within parentheses.

$-ii<ten<0$

$-five\le 10<\text{−}3$

The solution is negative 5 is less than or equal to x which is less than negative 3. The number line shows a closed circle at negative 5, an open circle at negative 3, and shading between the circles. The interval notation is negative 5 to negative 3 within a bracket and a parenthesis.

$0\le x\le 3.5$

Solve Linear Inequalities

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval annotation.

$n-12\le 23$

The solution is n is less than or equal to 35. The number line shows a a right bracket at 35 with shading to its left. The interval notation is negative infinity to 35 within a parenthesis and a bracket.

$a+\frac{2}{3}\ge \frac{7}{12}$

$9x>54$

The solution is x is greater than 6. The number line shows a left parenthesis at 6 with shading to its right. The interval notation is 6 to infinity within parentheses.

$\frac{q}{-2}\ge -24$

$6p>15p-30$

The solution is p is less than ten-thirds. The number line shows a right parenthesis at ten-thirds with shading to its left. The interval notation is negative infinity to ten-thirds within parentheses.

$9h-7\left(h-1\right)\le 4h-23$

$5n-15\left(4-n\right)<10\left(n-6\right)+10n$

The solution is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.

$\frac{3}{8}a-\frac{1}{12}a>\frac{5}{12}a+\frac{3}{4}$

Translate Words to an Inequality and Solve

In the following exercises, translate and solve. So write the solution in interval notation and graph on the number line.

Five more than z is at most xix.

The inequality is z plus 5 is less than or equal to 19. Its solution is z is less than or equal to 14. The number line shows a right bracket at 14 with shading to its left. The interval notation is negative infinity to 14 within a parenthesis and a bracket.

Three less than c is at least 360.

Nine times n exceeds 42.

The inequality is 9 n is greater than 42. Its solution is n is greater than fourteen-thirds. The number line shows a left parentheses at fourteen-thirds with shading to its right. The interval notation is fourteen-thirds to infinity within parentheses.

Negative 2 times a is no more than than eight.

Solve Applications with Linear Inequalities

In the following exercises, solve.

Julianne has a weekly nutrient budget of ?231 for her family. If she plans to budget the aforementioned corporeality for each of the vii days of the week, what is the maximum amount she can spend on food each day?

?33 per day

Rogelio paints watercolors. He got a ?100 gift carte du jour to the art supply store and wants to utilize it to purchase 12″ × 16″ canvases. Each canvas costs ?10.99. What is the maximum number of canvases he can buy with his souvenir bill of fare?

Briana has been offered a sales job in another metropolis. The offering was for ?42,500 plus 8% of her total sales. In order to make it worth the move, Briana needs to take an annual bacon of at to the lowest degree ?66,500. What would her total sales need to be for her to move?

at least ?300,000

Renee's car costs her ?195 per month plus ?0.09 per mile. How many miles can Renee drive so that her monthly car expenses are no more than than ?250?

Costa is an accountant. During revenue enhancement season, he charges ?125 to practise a simple taxation return. His expenses for buying software, renting an office, and advertising are ?6,000. How many taxation returns must he do if he wants to brand a profit of at to the lowest degree ?8,000?

at least 112 jobs

Jenna is planning a five-24-hour interval resort holiday with three of her friends. It will toll her ?279 for airfare, ?300 for food and entertainment, and ?65 per day for her share of the hotel. She has ?550 saved towards her vacation and can earn ?25 per 60 minutes as an assistant in her uncle'south photography studio. How many hours must she work in order to have plenty money for her holiday?

Solve Chemical compound Inequalities

Solve Compound Inequalities with "and"

In each of the following exercises, solve each inequality, graph the solution, and write the solution in interval notation.

$4x-2\le 4$ and

$7x-1>\text{−}8$

$\frac{3}{4}\left(x-8\right)\le 3$ and

$\frac{1}{5}\left(x-5\right)\le 3$

$-v\le 4x-1<7$

Solve Compound Inequalities with "or"

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$3\left(2x-3\right)<\text{−}5$ or

$4x-1>3$

$2\left(x+3\right)\ge 0$ or

$3\left(x+4\right)\le 6$

Solve Applications with Compound Inequalities

In the following exercises, solve.

Liam is playing a number game with his sister Audry. Liam is thinking of a number and wants Audry to guess it. Five more than three times her number is between ii and 32. Write a chemical compound inequality that shows the range of numbers that Liam might be thinking of.

Elouise is creating a rectangular garden in her dorsum thou. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than than 48 feet. Apply a compound inequality to find the range of values for the width of the garden.

$6\le w\le 12$

Solve Absolute Value Inequalities

Solve Absolute Value Equations

In the following exercises, solve.

$|x|=8$

$|y|=-14$

no solution

$|z|=0$

$|3x-4|+5=7$

$x=2,x=\frac{2}{3}$

$4|x-1|+2=10$

$-2|x-3|+8=-4$

$x=9,x=-3$

$|\frac{1}{2}x+5|+4=1$

$|6x-5|=|2x+3|$

$x=2,x=\frac{1}{4}$

Solve Absolute Value Inequalities with "less than"

In the following exercises, solve each inequality. Graph the solution and write the solution in interval notation.

$|x|\le 8$

$|2x-5|\le 3$

The solution is 1 is less than or equal to x which is less than or equal to 4. The number line shows a closed circle at 1, a closed circle at 4, and shading in between the circles. The interval notation is 1 to 4 within brackets.

$|5x+1|\le -2$

The solution is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

Solve Accented Value Inequalities with "greater than"

In the following exercises, solve. Graph the solution and write the solution in interval notation.

$|x|>6$

$|x|\ge 2$

The solution is x is less than negative 2 or x is greater than 6. The number line shows a closed circle at negative 2 with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

$|x-5|>\text{−}2$

$|x-7|\ge 1$

The solution is x is less than or equal to 6 or x is greater than or equal to 8. The number line shows a closed circle at 6 with shading to its left and a closed circle at 8 with shading to its right. The interval notation is the union of negative infinity to negative 6 within a parenthesis and a bracket and 8 to infinity within a bracket and a parenthesis.

$3|x|+4\ge 1$

Solve Applications with Absolute Value

In the following exercises, solve.

A craft beer brewer needs 215,000 bottle per twenty-four hours. Only this full can vary past equally much as five,000 bottles. What is the maximum and minimum expected usage at the bottling company?

The minimum to maximum expected usage is 210,000 to 220,000 bottles

At Fancy Grocery, the ideal weight of a loaf of bread is sixteen ounces. By law, the actual weight can vary from the ideal by 1.5 ounces. What range of weight will be acceptable to the inspector without causing the bakery being fined?

Exercise Examination

In the post-obit exercises, solve each equation.

$-5\left(2x+1\right)=45$

$x=-5$

$\frac{1}{4}\left(12m+28\right)=6+2\left(3m+1\right)$

$8\left(3a+5\right)-7\left(4a-3\right)=20-3a$

$a=41$

$0.1d+0.25\left(d+8\right)=4.1$

$14n-3\left(4n+5\right)=-9+2\left(n-8\right)$

contradiction; no solution

$3\left(3u+2\right)+4\left[6-8\left(u-1\right)\right]=3\left(u-2\right)$

$\frac{3}{4}x-\frac{2}{3}=\frac{1}{2}x+\frac{5}{6}$

$x=6$

$|3x-4|=8$

$|2x-1|=|4x+3|$

$x=-2,x=-\frac{1}{3}$

Solve the formula

$x+2y=5$ for y.

In the following exercises, graph the inequality on the number line and write in interval note.

$x\ge -3.5$

The inequality is x is greater than or equal to negative 3.5. The number line shows a left bracket at negative 3.5 and shading to the right. The interval notation is negative 3.5 to infinity within a bracket and a parenthesis.

$x<\frac{11}{4}$

$-2\le x<5$

The inequality is negative two is less than or equal to x which is less than 5. The number line shows a closed circle at negative 2 and an open circle at 5 with shading between the circles. The interval notation is negative 2 to 5 within a bracket and a parenthesis.

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval note.

$8k\ge 5k-120$

$3c-10\left(c-ii\correct)<5c+16$

The solution is c is greater than one-third. The number line shows a left parenthesis at one-third with shading to its right. The interval notation is one-third to infinity within parentheses.

$\frac{3}{4}x-5\ge -2$ and

$-3\left(x+1\right)\ge 6$

$\frac{1}{2}x-3\le 4$ or

$\frac{1}{3}\left(x-6\right)\ge -2$

$|4x-3|\ge 5$

The solution is x is less than or equal to negative one-half or x is greater than or equal to 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and bracket and 2 to infinity within a bracket and a parenthesis.

In the post-obit exercises, interpret to an equation or inequality and solve.

Four less than twice x is 16.

Find the length of the missing side.

The figure is a right triangle with a base of 6 units and a height of 9 units.

$10.8$

One number is four more than twice some other. Their sum is $-47.$ Find the numbers.

The sum of ii consecutive odd integers is $-112.$ Find the numbers.

$-57,-55$

Marcus bought a television on auction for ?626.l The original cost of the television was ?895. Discover ⓐ the amount of discount and ⓑ the discount rate.

Bonita has ?two.95 in dimes and quarters in her pocket. If she has v more than dimes than quarters, how many of each coin does she have?

12 dimes, seven quarters

Kim is making eight gallons of punch from fruit juice and soda. The fruit juice costs ?six.04 per gallon and the soda costs ?4.28 per gallon. How much fruit juice and how much soda should she use and then that the punch costs ?v.71 per gallon?

The mensurate of one bending of a triangle is twice the measure of the smallest angle. The mensurate of the tertiary bending is three times the measure out of the smallest angle. Find the measures of all three angles.

$30\text{°},60\text{°},90\text{°}$

The length of a rectangle is v feet more than 4 times the width. The perimeter is 60 feet. Find the dimensions of the rectangle.

2 planes leave Dallas at the same time. One heads east at a speed of 428 miles per 60 minutes. The other plane heads west at a speed of 382 miles per hour. How many hours will it have them to be 2,025 miles autonomously?

$2.5$ hours

Leon drove from his house in Cincinnati to his sister's house in Cleveland, a distance of 252 miles. It took him $4\frac{1}{2}$ hours. For the first half 60 minutes, he had heavy traffic, and the rest of the time his speed was v miles per 60 minutes less than twice his speed in heavy traffic. What was his speed in heavy traffic?

Sara has a upkeep of ?one,000 for costumes for the xviii members of her musical theater group. What is the maximum she can spend for each costume?

At nearly ?55.56 per costume.

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Source: https://pressbooks.bccampus.ca/algebraintermediate/chapter/solve-absolute-value-inequalities/

Bergman Thertow

What Is Another Way to Write the Absolute Value Inequality P 12

Solve Absolute Value Inequalities

Learning Objectives

Solve Absolute Value Equations

Solve Absolute Value Inequalities with "Less Than"

Solve Absolute Value Inequalities with "Greater Than"

Solve Applications with Absolute Value

Key Concepts

Section Exercises

Practice Makes Perfect

Writing Exercises

Self Bank check

Affiliate Review Exercises

Employ a General Strategy to Solve Linear Equations

Use a Problem-Solving Strategy

Solve a formula for a Specific Variable

Solve Mixture and Uniform Move Applications

Solve Linear Inequalities

Solve Chemical compound Inequalities

Solve Absolute Value Inequalities

Exercise Examination

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