# 2.7 Linear inequalities and absolute value inequalities  (Page 5/11)

 Page 5 / 11

## Verbal

When solving an inequality, explain what happened from Step 1 to Step 2:

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

When solving an inequality, we arrive at:

$\begin{array}{l}x+2

Explain what our solution set is.

When writing our solution in interval notation, how do we represent all the real numbers?

$\left(-\infty ,\infty \right)$

When solving an inequality, we arrive at:

$\begin{array}{l}x+2>x+3\hfill \\ \phantom{\rule{1.2em}{0ex}}2>3\hfill \end{array}$

Explain what our solution set is.

Describe how to graph $\text{\hspace{0.17em}}y=|x-3|$

We start by finding the x -intercept, or where the function = 0. Once we have that point, which is $\text{\hspace{0.17em}}\left(3,0\right),$ we graph to the right the straight line graph $\text{\hspace{0.17em}}y=x-3,$ and then when we draw it to the left we plot positive y values, taking the absolute value of them.

## Algebraic

For the following exercises, solve the inequality. Write your final answer in interval notation.

$4x-7\le 9$

$3x+2\ge 7x-1$

$\left(-\infty ,\frac{3}{4}\right]$

$-2x+3>x-5$

$4\left(x+3\right)\ge 2x-1$

$\left[\frac{-13}{2},\infty \right)$

$-\frac{1}{2}x\le \frac{-5}{4}+\frac{2}{5}x$

$-5\left(x-1\right)+3>3x-4-4x$

$\left(-\infty ,3\right)$

$-3\left(2x+1\right)>-2\left(x+4\right)$

$\frac{x+3}{8}-\frac{x+5}{5}\ge \frac{3}{10}$

$\left(-\infty ,-\frac{37}{3}\right]$

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

For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation.

$|x+9|\ge -6$

All real numbers $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$

$|2x+3|<7$

$|3x-1|>11$

$\left(-\infty ,\frac{-10}{3}\right)\cup \left(4,\infty \right)$

$|2x+1|+1\le 6$

$|x-2|+4\ge 10$

$\left(-\infty ,-4\right]\cup \left[8,+\infty \right)$

$|-2x+7|\le 13$

$|x-7|<-4$

No solution

$|x-20|>-1$

$|\frac{x-3}{4}|<2$

$\left(-5,11\right)$

For the following exercises, describe all the x -values within or including a distance of the given values.

Distance of 5 units from the number 7

Distance of 3 units from the number 9

$\left[6,12\right]$

Distance of10 units from the number 4

Distance of 11 units from the number 1

$\left[-10,12\right]$

$-4<3x+2\le 18$

$3x+1>2x-5>x-7$

$3y<5-2y<7+y$

$x+7

## Graphical

For the following exercises, graph the function. Observe the points of intersection and shade the x -axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation.

$|x-1|>2$

$\left(-\infty ,-1\right)\cup \left(3,\infty \right)$

$|x+3|\ge 5$

$|x+7|\le 4$

$\left[-11,-3\right]$

$|x-2|<7$

$|x-2|<0$

It is never less than zero. No solution.

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y -values of the lines.

$x+3<3x-4$

$x-2>2x+1$

Where the blue line is above the orange line; point of intersection is $\text{\hspace{0.17em}}x=-3.$

$\left(-\infty ,-3\right)$

$x+1>x+4$

$\frac{1}{2}x+1>\frac{1}{2}x-5$

Where the blue line is above the orange line; always. All real numbers.

$\left(-\infty ,-\infty \right)$

$4x+1<\frac{1}{2}x+3$

## Numeric

For the following exercises, write the set in interval notation.

$\left\{x|-1

$\left(-1,3\right)$

$\left\{x|x\ge 7\right\}$

$\left\{x|x<4\right\}$

$\left(-\infty ,4\right)$

For the following exercises, write the interval in set-builder notation.

$\left(-\infty ,6\right)$

$\left\{x|x<6\right\}$

$\left(4,+\infty \right)$

$\left[-3,5\right)$

$\left\{x|-3\le x<5\right\}$

$\left[-4,1\right]\cup \left[9,\infty \right)$

For the following exercises, write the set of numbers represented on the number line in interval notation.

12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8