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

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.
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