# 2.7 Linear inequalities and absolute value inequalities  (Page 4/11)

 Page 4 / 11
$\begin{array}{c}-200\le x-600\le 200\\ -200+600\le x-600+600\le 200+600\\ 400\le x\le 800\end{array}$

This means our returns would be between $400 and$800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

## Absolute value inequalities

For an algebraic expression X, and $\text{\hspace{0.17em}}k>0,$ an absolute value inequality is an inequality of the form

These statements also apply to $\text{\hspace{0.17em}}|X|\le k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}|X|\ge k.$

## Determining a number within a prescribed distance

Describe all values $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ within a distance of 4 from the number 5.

We want the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 5 to be less than or equal to 4. We can draw a number line, such as in [link] , to represent the condition to be satisfied.

The distance from $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to 5 can be represented using an absolute value symbol, $\text{\hspace{0.17em}}|x-5|.\text{\hspace{0.17em}}$ Write the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that satisfy the condition as an absolute value inequality.

$|x-5|\le 4$

We need to write two inequalities as there are always two solutions to an absolute value equation.

$\begin{array}{lll}x-5\le 4\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & x-5\ge -4\hfill \\ \phantom{\rule{1.8em}{0ex}}x\le 9\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}x\ge 1\hfill \end{array}$

If the solution set is $\text{\hspace{0.17em}}x\le 9\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x\ge 1,$ then the solution set is an interval including all real numbers between and including 1 and 9.

So $\text{\hspace{0.17em}}|x-5|\le 4\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}\left[1,9\right]\text{\hspace{0.17em}}$ in interval notation.

Describe all x- values within a distance of 3 from the number 2.

$|x-2|\le 3$

## Solving an absolute value inequality

Solve $|x-1|\le 3$ .

$\begin{array}{l}|x-1|\le 3\hfill \\ \hfill \\ -3\le x-1\le 3\hfill \\ \hfill \\ -2\le x\le 4\hfill \\ \hfill \\ \left[-2,4\right]\hfill \end{array}$

## Using a graphical approach to solve absolute value inequalities

Given the equation $y=-\frac{1}{2}|4x-5|+3,$ determine the x -values for which the y -values are negative.

We are trying to determine where $\text{\hspace{0.17em}}y<0,$ which is when $\text{\hspace{0.17em}}-\frac{1}{2}|4x-5|+3<0.\text{\hspace{0.17em}}$ We begin by isolating the absolute value.

Next, we solve for the equality $|4x-5|=6.$

$\begin{array}{lll}4x-5=6\hfill & \hfill & 4x-5=-6\hfill \\ \phantom{\rule{1.9em}{0ex}}4x=11\hfill & \phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\hfill & \phantom{\rule{1.9em}{0ex}}4x=-1\hfill \\ \phantom{\rule{2em}{0ex}}x=\frac{11}{4}\hfill & \hfill & \phantom{\rule{2em}{0ex}}x=-\frac{1}{4}\hfill \end{array}$

Now, we can examine the graph to observe where the y- values are negative. We observe where the branches are below the x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $\text{\hspace{0.17em}}x=-\frac{1}{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=\frac{11}{4},$ and that the graph opens downward. See [link] .

Solve $\text{\hspace{0.17em}}-2|k-4|\le -6.$

$k\le 1\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}k\ge 7;$ in interval notation, this would be $\text{\hspace{0.17em}}\left(-\infty ,1\right]\cup \left[7,\infty \right).$

Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.

## Key concepts

• Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See [link] and [link] .
• Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See [link] , [link] , [link] , and [link] .
• Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See [link] and [link] .
• Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See [link] and [link] .
• Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See [link] .

The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)