# 2.7 Linear inequalities and absolute value inequalities  (Page 3/11)

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Solve the inequality and write the answer in interval notation: $\text{\hspace{0.17em}}-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x.$

$\left[-\frac{3}{14},\infty \right)$

## Understanding compound inequalities

A compound inequality    includes two inequalities in one statement. A statement such as $\text{\hspace{0.17em}}4 means $\text{\hspace{0.17em}}4 and $\text{\hspace{0.17em}}x\le 6.\text{\hspace{0.17em}}$ There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.

## Solving a compound inequality

Solve the compound inequality: $\text{\hspace{0.17em}}3\le 2x+2<6.$

The first method is to write two separate inequalities: $\text{\hspace{0.17em}}3\le 2x+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2x+2<6.\text{\hspace{0.17em}}$ We solve them independently.

$\begin{array}{lll}3\le 2x+2\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & 2x+2<6\hfill \\ 1\le 2x\hfill & \hfill & \phantom{\rule{2em}{0ex}}2x<4\hfill \\ \frac{1}{2}\le x\hfill & \hfill & \phantom{\rule{2em}{0ex}}x<2\hfill \end{array}$

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

$\frac{1}{2}\le x<2$

In interval notation, the solution is written as $\text{\hspace{0.17em}}\left[\frac{1}{2},2\right).$

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

We get the same solution: $\text{\hspace{0.17em}}\left[\frac{1}{2},2\right).$

Solve the compound inequality: $\text{\hspace{0.17em}}4<2x-8\le 10.$

$6

## Solving a compound inequality with the variable in all three parts

Solve the compound inequality with variables in all three parts: $\text{\hspace{0.17em}}3+x>7x-2>5x-10.$

Let's try the first method. Write two inequalities :

$\begin{array}{lll}3+x>7x-2\hfill & \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\hfill & 7x-2>5x-10\hfill \\ \phantom{\rule{1.2em}{0ex}}3>6x-2\hfill & \hfill & 2x-2>-10\hfill \\ \phantom{\rule{1.2em}{0ex}}5>6x\hfill & \hfill & \phantom{\rule{1.8em}{0ex}}2x>-8\hfill \\ \phantom{\rule{1.2em}{0ex}}\frac{5}{6}>x\hfill & \hfill & \phantom{\rule{2.5em}{0ex}}x>-4\hfill \\ \phantom{\rule{1.3em}{0ex}}x<\frac{5}{6}\hfill & \hfill & \phantom{\rule{1.5em}{0ex}}-4

The solution set is $\text{\hspace{0.17em}}-4 or in interval notation $\text{\hspace{0.17em}}\left(-4,\frac{5}{6}\right).\text{\hspace{0.17em}}$ Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See [link] .

Solve the compound inequality: $\text{\hspace{0.17em}}3y<4-5y<5+3y.$

$\left(-\frac{1}{8},\frac{1}{2}\right)$

## Solving absolute value inequalities

As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at $\text{\hspace{0.17em}}\left(-x,0\right)\text{\hspace{0.17em}}$ has an absolute value of $\text{\hspace{0.17em}}x,$ as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.

An absolute value inequality is an equation of the form

Where A , and sometimes B , represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all $\text{\hspace{0.17em}}x$ - values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of$600. We can solve algebraically for the set of x- values such that the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 600 is less than 200. We represent the distance between $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and 600 as $\text{\hspace{0.17em}}|x-600|,$ and therefore, $\text{\hspace{0.17em}}|x-600|\le 200\text{\hspace{0.17em}}$ or

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1