# 2.7 Linear inequalities and absolute value inequalities  (Page 2/11)

 Page 2 / 11

## Using interval notation to express all real numbers less than or equal to a Or greater than or equal to b

Write the interval expressing all real numbers less than or equal to $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ or greater than or equal to $\text{\hspace{0.17em}}1.$

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at $\text{\hspace{0.17em}}-\infty \text{\hspace{0.17em}}$ and ends at $\text{\hspace{0.17em}}-1,$ which is written as $\text{\hspace{0.17em}}\left(-\infty ,-1\right].$

The second interval must show all real numbers greater than or equal to $\text{\hspace{0.17em}}1,$ which is written as $\text{\hspace{0.17em}}\left[1,\infty \right).\text{\hspace{0.17em}}$ However, we want to combine these two sets. We accomplish this by inserting the union symbol, $\cup ,$ between the two intervals.

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

Express all real numbers less than $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ or greater than or equal to 3 in interval notation.

$\left(-\infty ,-2\right)\cup \left[3,\infty \right)$

## Using the properties of inequalities

When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol.

## Properties of inequalities

These properties also apply to $\text{\hspace{0.17em}}a\le b,$ $a>b,$ and $\text{\hspace{0.17em}}a\ge b.$

## Demonstrating the addition property

Illustrate the addition property for inequalities by solving each of the following:

• (a) $x-15<4$
• (b) $6\ge x-1$
• (c) $x+7>9$

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

Solve: $\text{\hspace{0.17em}}3x-2<1.$

$x<1$

## Demonstrating the multiplication property

Illustrate the multiplication property for inequalities by solving each of the following:

1. $3x<6$
2. $-2x-1\ge 5$
3. $5-x>10$

1. $\begin{array}{l}\phantom{\rule{1.5em}{0ex}}3x<6\hfill \\ \frac{1}{3}\left(3x\right)<\left(6\right)\frac{1}{3}\hfill \\ \phantom{\rule{2em}{0ex}}x<2\hfill \end{array}$

Solve: $\text{\hspace{0.17em}}4x+7\ge 2x-3.$

$x\ge -5$

## Solving inequalities in one variable algebraically

As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.

## Solving an inequality algebraically

Solve the inequality: $\text{\hspace{0.17em}}13-7x\ge 10x-4.$

Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the interval $\text{\hspace{0.17em}}\left(-\infty ,1\right],$ or all real numbers less than and including 1.

Solve the inequality and write the answer using interval notation: $\text{\hspace{0.17em}}-x+4<\frac{1}{2}x+1.$

$\left(2,\infty \right)$

## Solving an inequality with fractions

Solve the following inequality and write the answer in interval notation: $\text{\hspace{0.17em}}-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x.$

We begin solving in the same way we do when solving an equation.

The solution set is the interval $\text{\hspace{0.17em}}\left(-\infty ,\frac{15}{34}\right].$

#### Questions & Answers

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
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
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
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
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
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
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar