Questions on Algebra: Absolute value answered by real tutors!

Tutors Answer Your Questions about absolute-value (FREE)

Question 1206359: The temperature in a town varies by 15 degrees F. The average temperature is 73 degrees F. What is the correct absolute value equation?

Answer by ikleyn(53748)

(Show Source):

You can put this solution on YOUR website!
.
The temperature in a town varies by 15 degrees F. The average temperature is 73 degrees F.
What is the correct absolute value equation?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solution given in the post by @mananth is incorrect.
I came to bring a correct solution

Let x be the potential values of the temperatures between maxim and minimum

Average is73 deg

Variation is 15 deg, which means deviation of 15/2 = 7.5 degrees in each directions from the average 73 degrees.

Thus the absolute value equation is 

    |x-73| = 7.5  degrees.    ANSWER

Solved correctly.

//////////////////////////////

It is interesting !

Today, Jan.9,2026, I checked how Google AI Overview solves this problem.

It repeats incorrect solution by @mananth word-in-word.

Naturally, I reported to Google AI Overview about their mistake via their feedback system.

It confirms the idea that Google AI Overview is not able to think;
instead, it repeats what it finds in the Internet, without understanding the meaning.

It is even more interesting !

Today I posted this problem to another AI math site, math-gpt.org.

It produced the same wrong solution !

It shows that wrong solutions spread incredibly widely.

It also shows THE PRICE of each and every wrong solution in the era of "artificial intelligence".

Who will tell the truth to people ?

Question 735594: hi, can you help me solve theses questions?
thank you.
Which of the following statements says that a number is between -3 and 3?
|x| = 3
|x| < 3
|x| > 3

Which graph is the solution to |x| > 10?

Select the graph for the solution of the open sentence. Click until the correct graph appears.
|x| < 3

Select the graph for the solution of the open sentence. Click until the correct graph appears.
|x| > 4
Select the graph for the solution of the open sentence. Click until the correct graph appears.
|x| = -2
Select the graph for the solution of the open sentence. Click until the correct graph appears.
|x| > 1
Select the graph for the solution of the open sentence. Click until the correct graph appears.
|x| >
Solve |2x - 5| = 4.
{x | x = -4.5 or x = 4.5}
{x | x = 0.5 or x = 4.5}
{x | 0.5 < x < 4.5}

he graph of which of the following inequalities has open circles on -8 and 2 with a line segment between them?
|x + 3| < -5
|x + 8| < 2
|x + 3| < 5
Solve |x| > -9.
{x | x < -9 or x > 9}
all reals
no solution

Answer by ikleyn(53748)

(Show Source):

You can put this solution on YOUR website!
.

Never pack more than one problem per post

Question 741714: Dear Sir/Madam,
I do not know how to do this question:
|3x-5|=3x-5

Answer by ikleyn(53748)

(Show Source):

You can put this solution on YOUR website!
.
Dear Sir/Madam,
I do not know how to do this question:
|3x-5|=3x-5
~~~~~~~~~~~~~~~~~~~~~~~~~

        Dear visitor,

        first what I have to say, looking at your post, is that there is NO question in it.
        There is some equation, but there is no question.
        So, in order for this post makes sense, I/we should create a question to it.

        One reasonable question is " Find the set of solutions to this equation. "

        So, I will assume that this is the question, and I will give the solution in my post below.

Function  y = |3x-5|  has two branches.


(1)  At x >= 5/3,  this function is y = 3x-5,  and it coincides with the function 3x-5 in the right side 
     of the given equation.


     So, in the domain  x >= 5/3,  the given equation,  |3x-5| = 3x-5  has infinitely many solutions
     that are the entire set  { x | x >= 5/3 }.



(2)  At x < 5/3,  this function y = |3x-5|  is positive,  while  3x-5  is negative.

     Therefore, in the domain x < 5/3,  the given equation has no one solution.


At this point, the analysis is complete.  The ANSWER  is


    +------------------------------------------------------------------------------------+
    |   Equation  |3x-5| = 3x-5  has infinitely many solutions in the domain  x >= 5/3.  |
    |                                                                                    |
    |   All x's in this domain are the solutions to this equation.                       |
    |                                                                                    |
    |   In the domain  x < 5/3,  equation  |3x-5| = 3x-5  has no solutions.              |
    |                                                                                    |
    |   Thus, the solution set for the given equation is  { x |  x >= 5/3 }.             |
    +------------------------------------------------------------------------------------+

Solved completely.

Question 1210247: |x − 5| = 3
Answer by greenestamps(13327)

(Show Source):

You can put this solution on YOUR website!

The equation abs%28x-5%29=3

means x-5 is equal to either 3 or -3. Solve the two equations separately.

x-5 = 3
x = 3+5 = 8

or

x-5 = -3
x = -3+5 = 2

ANSWERS: x = 8 or x = 2

That's the formal algebraic way to solve the problem.

There is another way to solve an absolute value equation like this that is usually easier (for me, at least) than the formal algebraic method.

The equation abs%28x-a%29=b

can be interpreted as meaning "the difference between x and a is equal to b".

In this example, we have abs%28x-5%29=3

, which we can interpret as saying "the difference between x and 5 is 3".

Then look on a number line to find all the numbers (there will be two of them) that are 3 away from 5.

3 to the right of 5 is 8; 3 to the left of 5 is 2. So (again, of course!) the solutions are x = 8 and x = 2.

Question 1209903: Find the number of points of intersection between the graphs of the following equations: \begin{align*}
y = |2x + 5|,
y = -|3x - 2| + |4x - 7| + x.
Answer by CPhill(2189)

(Show Source):

You can put this solution on YOUR website!
Let's analyze the given equations to find the number of intersection points.
**Equation 1:** $y = |2x + 5|$
**Equation 2:** $y = -|3x - 2| + |4x - 7| + x$
We need to consider cases based on the values of $x$ that make the expressions inside the absolute values zero.
**Critical Points:**
* $2x + 5 = 0 \Rightarrow x = -5/2 = -2.5$
* $3x - 2 = 0 \Rightarrow x = 2/3$
* $4x - 7 = 0 \Rightarrow x = 7/4 = 1.75$
**Case 1: $x \le -5/2$**
* $|2x + 5| = -(2x + 5) = -2x - 5$
* $|3x - 2| = -(3x - 2) = -3x + 2$
* $|4x - 7| = -(4x - 7) = -4x + 7$
Equation 1: $y = -2x - 5$
Equation 2: $y = -(-3x + 2) + (-4x + 7) + x = 3x - 2 - 4x + 7 + x = 5$
Set the equations equal:
$-2x - 5 = 5$
$-2x = 10$
$x = -5$
Since $-5 \le -2.5$, this is a valid solution.
Intersection point: $(-5, 5)$
**Case 2: $-5/2 < x \le 2/3$**
* $|2x + 5| = 2x + 5$
* $|3x - 2| = -(3x - 2) = -3x + 2$
* $|4x - 7| = -(4x - 7) = -4x + 7$
Equation 1: $y = 2x + 5$
Equation 2: $y = -(-3x + 2) + (-4x + 7) + x = 3x - 2 - 4x + 7 + x = 5$
Set the equations equal:
$2x + 5 = 5$
$2x = 0$
$x = 0$
Since $-2.5 < 0 \le 2/3$, this is a valid solution.
Intersection point: $(0, 5)$
**Case 3: $2/3 < x \le 7/4$**
* $|2x + 5| = 2x + 5$
* $|3x - 2| = 3x - 2$
* $|4x - 7| = -(4x - 7) = -4x + 7$
Equation 1: $y = 2x + 5$
Equation 2: $y = -(3x - 2) + (-4x + 7) + x = -3x + 2 - 4x + 7 + x = -6x + 9$
Set the equations equal:
$2x + 5 = -6x + 9$
$8x = 4$
$x = 1/2 = 0.5$
However, $2/3 \approx 0.666$, so $0.5$ is not in the interval, so no solution in this case.
**Case 4: $x > 7/4$**
* $|2x + 5| = 2x + 5$
* $|3x - 2| = 3x - 2$
* $|4x - 7| = 4x - 7$
Equation 1: $y = 2x + 5$
Equation 2: $y = -(3x - 2) + (4x - 7) + x = -3x + 2 + 4x - 7 + x = 2x - 5$
Set the equations equal:
$2x + 5 = 2x - 5$
$5 = -5$
This is a contradiction, so there are no solutions in this case.
**Intersection Points:**
* $(-5, 5)$
* $(0, 5)$
There are two intersection points.
Final Answer: The final answer is $\boxed{2}$

Question 1209890: Find all real numbers x that satisfy the equation
|x + 4| + |x - 7| = 3x - 1 + |3x - 7|.
If you find more than one such value of x, list all of your solutions separated by commas. If you only find one solution, then just enter that solution.

Found 2 solutions by josgarithmetic, CPhill:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
I did this FAST and did not check carefully.
Look at critical values.
neg infinity to -4
-4 to 7/3
7/3 yo 7
7 to pos infinity

x=-1/2, and or x=-5/2.
The other two intervals seem to have no solutions.

Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let's analyze the given equation:
$$|x + 4| + |x - 7| = 3x - 1 + |3x - 7|$$
We need to consider different cases based on the values of $x$ that make the absolute value expressions zero.
**Case 1: $x \le -4$**
* $|x + 4| = -(x + 4) = -x - 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = -(3x - 7) = -3x + 7$
Substitute into the equation:
$$-x - 4 - x + 7 = 3x - 1 - 3x + 7$$
$$-2x + 3 = 6$$
$$-2x = 3$$
$$x = -\frac{3}{2}$$
Since $-\frac{3}{2} > -4$, this case is not valid.
**Case 2: $-4 < x \le \frac{7}{3}$**
* $|x + 4| = x + 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = -(3x - 7) = -3x + 7$
Substitute into the equation:
$$x + 4 - x + 7 = 3x - 1 - 3x + 7$$
$$11 = 6$$
This is a contradiction, so there are no solutions in this case.
**Case 3: $\frac{7}{3} < x \le 7$**
* $|x + 4| = x + 4$
* $|x - 7| = -(x - 7) = -x + 7$
* $|3x - 7| = 3x - 7$
Substitute into the equation:
$$x + 4 - x + 7 = 3x - 1 + 3x - 7$$
$$11 = 6x - 8$$
$$19 = 6x$$
$$x = \frac{19}{6}$$
Since $\frac{7}{3} = \frac{14}{6}$ and $\frac{19}{6} > \frac{14}{6}$, and $\frac{19}{6} = 3\frac{1}{6} < 7$, $x = \frac{19}{6}$ is a solution.
**Case 4: $x > 7$**
* $|x + 4| = x + 4$
* $|x - 7| = x - 7$
* $|3x - 7| = 3x - 7$
Substitute into the equation:
$$x + 4 + x - 7 = 3x - 1 + 3x - 7$$
$$2x - 3 = 6x - 8$$
$$5 = 4x$$
$$x = \frac{5}{4}$$
Since $\frac{5}{4} < 7$, this case is not valid.
**Check the solution $x = \frac{19}{6}$:**
* $|\frac{19}{6} + 4| + |\frac{19}{6} - 7| = |\frac{43}{6}| + |\frac{-23}{6}| = \frac{43}{6} + \frac{23}{6} = \frac{66}{6} = 11$
* $3(\frac{19}{6}) - 1 + |3(\frac{19}{6}) - 7| = \frac{19}{2} - 1 + |\frac{19}{2} - 7| = \frac{17}{2} + |\frac{5}{2}| = \frac{17}{2} + \frac{5}{2} = \frac{22}{2} = 11$
Therefore, the solution is $x = \frac{19}{6}$.
Final Answer: The final answer is $\boxed{\frac{19}{6}}$

Question 1209891: Find all x for which
\Big| x - |x - 2| \Big| = \lfloor x \rfloor + \lfloor 2x \rfloor
Express your answer in interval notation.

Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

        What is written in this post ?

                To the managers of this project !

It looks like that the person who submitted this post to the forum,
is unfamiliar with Mathematical grammar and with the rules of writing Math texts.

Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let's analyze the given equation:
$$|x - |x - 2|| = \lfloor x \rfloor + \lfloor 2x \rfloor$$
We need to consider cases based on the value of $x$.
**Case 1: $x \ge 2$**
* $|x - 2| = x - 2$
* $|x - |x - 2|| = |x - (x - 2)| = |2| = 2$
* $\lfloor x \rfloor + \lfloor 2x \rfloor = 2$
Let $x = n + f$, where $n = \lfloor x \rfloor$ is an integer and $0 \le f < 1$.
Then $2 = n + \lfloor 2(n+f) \rfloor = n + \lfloor 2n + 2f \rfloor = n + 2n + \lfloor 2f \rfloor = 3n + \lfloor 2f \rfloor$.
* If $0 \le f < \frac{1}{2}$, then $\lfloor 2f \rfloor = 0$, so $2 = 3n$, which gives $n = \frac{2}{3}$, but we need $n \ge 2$, so no solution.
* If $\frac{1}{2} \le f < 1$, then $\lfloor 2f \rfloor = 1$, so $2 = 3n + 1$, which gives $3n = 1$ and $n = \frac{1}{3}$, but we need $n \ge 2$, so no solution.
Therefore, there are no solutions for $x \ge 2$.
**Case 2: $0 \le x < 2$**
* $|x - 2| = 2 - x$
* $|x - |x - 2|| = |x - (2 - x)| = |2x - 2|$
* $\lfloor x \rfloor + \lfloor 2x \rfloor = |2x - 2|$
Let $x = n + f$, where $n = \lfloor x \rfloor$ and $0 \le f < 1$.
Since $0 \le x < 2$, we have $n = 0$ or $n = 1$.
* **Subcase 2.1: $0 \le x < 1$**
* $n = 0$
* $|2x - 2| = \lfloor x \rfloor + \lfloor 2x \rfloor = 0 + \lfloor 2x \rfloor = \lfloor 2x \rfloor$
* Since $0 \le x < 1$, we have $0 \le 2x < 2$, so $\lfloor 2x \rfloor = 0$ or $1$.
* If $\lfloor 2x \rfloor = 0$, then $|2x - 2| = 0$, so $2x - 2 = 0$, which means $x = 1$. But $x < 1$, so this is a contradiction.
* If $\lfloor 2x \rfloor = 1$, then $|2x - 2| = 1$.
* $2x - 2 = 1$, so $2x = 3$ and $x = \frac{3}{2}$. But $x < 1$, so this is a contradiction.
* $2x - 2 = -1$, so $2x = 1$ and $x = \frac{1}{2}$.
* Check: $|\frac{1}{2} - |\frac{1}{2} - 2|| = |\frac{1}{2} - \frac{3}{2}| = |-1| = 1$
* $\lfloor \frac{1}{2} \rfloor + \lfloor 1 \rfloor = 0 + 1 = 1$.
* Thus, $x = \frac{1}{2}$ is a solution.
* **Subcase 2.2: $1 \le x < 2$**
* $n = 1$
* $|2x - 2| = \lfloor x \rfloor + \lfloor 2x \rfloor = 1 + \lfloor 2x \rfloor$
* Since $1 \le x < 2$, we have $2 \le 2x < 4$, so $\lfloor 2x \rfloor = 2$ or $3$.
* If $\lfloor 2x \rfloor = 2$, then $|2x - 2| = 1 + 2 = 3$.
* $2x - 2 = 3$, so $2x = 5$ and $x = \frac{5}{2}$. But $x < 2$, so this is a contradiction.
* $2x - 2 = -3$, so $2x = -1$ and $x = -\frac{1}{2}$. But $x \ge 1$, so this is a contradiction.
* If $\lfloor 2x \rfloor = 3$, then $|2x - 2| = 1 + 3 = 4$.
* $2x - 2 = 4$, so $2x = 6$ and $x = 3$. But $x < 2$, so this is a contradiction.
* $2x - 2 = -4$, so $2x = -2$ and $x = -1$. But $x \ge 1$, so this is a contradiction.
**Case 3: $x < 0$**
* $|x - 2| = 2 - x$
* $|x - |x - 2|| = |x - (2 - x)| = |2x - 2| = 2 - 2x$ (since $x < 0$, $2x - 2 < 0$)
* $\lfloor x \rfloor + \lfloor 2x \rfloor = 2 - 2x$
Let $x = -y$, where $y > 0$.
$\lfloor -y \rfloor + \lfloor -2y \rfloor = 2 + 2y$
$-(\lceil y \rceil + \lceil 2y \rceil) = 2 + 2y$
$\lceil y \rceil + \lceil 2y \rceil = -2 - 2y$
Since $\lceil y \rceil$ and $\lceil 2y \rceil$ are positive, the left side is positive, but the right side is negative. So there are no solutions.
**Final Answer:** The final answer is $\boxed{\{\frac{1}{2}\}}$

Question 1209850: Given that log_{4n} 40 sqrt(3) = log_{2n} 10, find n^4.
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Given that log_{4n} 40 sqrt(3) = log_{2n} 10, find n^4.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

You are given = . (1) Let x be the common numerical value of each side of (1), so we can write = x, (2) = x. (3) Then we can rewrite (2) and (3) in this form, respectively = , (2') = 10. (3') Divide equation (2') by equation (3'). You will get = . You can simplify left side = . Hence, x = . At this point, we solved half of the problem and have found the numerical value of each expression in the left side of equations (2) and (3). Now we are going to make next step and to find n and . We can write equation (3) in the form = . (4) Let y be the common numerical value of each side of (4), so we can write = y, (5) = y. (6) Then we can rewrite (5) and (6) in this form, respectively = 10, (5') = . (6') Divide equation (5') by equation (6'). You will get = = = . Take logarithm of both sides y*log(n) = . (7) In this formula, it does not matter, which base of logarithm I use. Let's the base of logarithm in this equation be 10. From equation (7), log(n) = . We can substitute here expression (6) for y. We get then log(n) = * . This is the "precise" expression for log(n). We can find its numerical value log(n) = = 0.057074. Hence, n = = 1.140444, and = = 1.691601, approximately. At this point, the solution is complete and the value of is found. ANSWER. = = 1.691601, approximately.

Solved.

After completing this solution, I checked equation (1)
and found that both sides are equal at n = 1.140444.

Question 1209849: a=30 b=40 c=?

Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
To find the value of 'c', we need more information about the relationship between a, b, and c. Here are a few possibilities:
* **If a, b, and c represent the sides of a right triangle:**
* If a and b are the legs, then c (the hypotenuse) can be found using the Pythagorean theorem: a² + b² = c²
* 30² + 40² = c²
* 900 + 1600 = c²
* 2500 = c²
* c = √2500
* c = 50
* It is also possible that b is the hypotenuse, and we are solving for a leg c. In this case, b²-a² = c²
* 40²-30²=c²
* 1600-900=c²
* 700=c²
* c=√700, or approximately 26.46.
* **If a, b, and c are part of a simple arithmetic sequence:**
* The difference between b and a is 40 - 30 = 10.
* Therefore, c could be 40 + 10 = 50.
* **If a, b, and c are part of a simple arithmetic pattern:**
* There are infinite possiblities.
* **Without more context, c could be any number.**
Therefore, the most likely solution, assuming a right triangle with a and b as legs, is c = 50.

Question 1176244: In presidential elections, each state has a designated number of votes in the Electoral College, which are generally all cast for the candidate who won the popular vote in the state. The number of members of the Electoral College from each state is based on the number of Senators and House of Representatives.
a. If a presidential candidate is 42 votes ahead of his opponent before the votes for the state of California are added, what absolute value equation would represent the margin of votes between the candidate and his opponent after California’s 55 votes are cast?
b. What absolute value equation can be used to determine the minimum number of votes needed to change the outcome of the election in question 4?
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let's define variables to set up the absolute value equations for both parts of the problem.
### Given Information:
- A presidential candidate is **42 votes ahead** before California's **55 electoral votes** are added.
- Electoral votes are generally awarded in full to the winner of the state's popular vote.
#### **Part (a)**: Absolute Value Equation for the Margin After California's Votes Are Added
Let $ x $ represent the final margin after California's votes are added. If the leading candidate wins California, their lead increases by 55 votes. If the opposing candidate wins California, their lead decreases by 55 votes.
The equation representing this scenario is:
\[
| x - 42 | = 55
\]
This equation accounts for both possibilities:
- If California's votes go to the leading candidate, the margin increases: $ x = 42 + 55 = 97 $.
- If California's votes go to the opposing candidate, the margin flips: $ x = 42 - 55 = -13 $.
#### **Part (b)**: Absolute Value Equation for the Minimum Number of Votes Needed to Change the Outcome
To change the outcome, the opposing candidate must at least tie the election. This means overcoming the 42-vote lead by shifting votes. Since electoral votes are won in full per state, the smallest shift would be flipping just enough votes from the winning candidate to the losing candidate.
Let $ y $ represent the number of votes that must be shifted (moved from one candidate to the other). Since each vote switched affects the margin by **2** (one candidate loses a vote while the other gains one), the equation is:
\[
| 2y | = 42
\]
Solving for $ y $:
\[
y = \frac{42}{2} = 21
\]
So, the minimum number of electoral votes that must be flipped to change the outcome is **21**.
These absolute value equations properly express the given electoral vote conditions in the problem.

Question 1209411: Find all solutions of the equation |x^2 - 30x - 1| = 30 + |6x - 2| + |x^2 - 20x - 4|.
Found 3 solutions by ikleyn, KMST, Edwin McCravy:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

For problems like this one, there is a standard methodology/strategy for solving.

You should subdivide the number line by sections (= intervals).
The division point are the points where the participating functions under absolute values
(parts of the equation) change their signs.

Then for each interval, from minus infinity to infinity, you write an equation
using each function with its native sign as it should be, dictated by the absolute
value rule at this interval.

Then you solve the updated equation and check if the solution does belong to this particular
interval under the consideration.

If the particular solution does belong to this particular interval, then the found solution
is the solution to the original equation.
If it does not belong, then the found solution is not a solution to the original equation.

As you complete going from the left to the right , you complete the solution.

The rest is just a technique and the arithmetic.

I think, if a student does understand it and demonstrates his/her understanding, it should be just
enough for the teacher to accept the assignment.

As the problem is twisted in this post, it is twisted TOO much and teaches nothing.

A teacher who assigns such tasks, should be checked for adequacy.

Answer by KMST(5345)   (Show Source):
You can put this solution on YOUR website!
Before graphing calculators, it would have been a more complicated

The right hand is greater than 30 except for all values of
for all of real values because
for all real numbers,
with for , and otherwise.
for
and .
and for
for such that

For the left hand side, , has as solutions
For values of and ,
and .
For
Those limiting values are approximately and .

For , , while and
and the equation turns into

-->

For is decreasing from at to a minimum 0f at and then increasing to at and is less than the value of
=
For , , , and ,
so and
The equation turns into

-->

For while
and , so
turning the equation into

-->
does not comply with
is a solution between and

For , , and ,
turning the equation into

--> is not a solution complying with

Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

Get 0 on the right: Graph the left side in your TI-84 calculator. The "abs" feature is found this way: MATH NUM 1:abs( Put it in Y1 like this: abs(x^2 - 30x - 1) - 30 - abs(6x - 2) - abs(x^2 - 20x - 4) Set the window Xmin=-10, Xmax=25, Ymin=-40, Ymax=60 Your graph will look something close to this: Use the zero feature of the calculator, and you'll find three solutions The "zero" feature is found 2ND TRACE 2:zero x = -7.25, x = 7.75, and x = 21.464225 Edwin

Question 1209365: (1) Let a_1, a_2, a_3 be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_1| = 1.
What is the largest possible value of |a_1 - a_2|?

(2) Let a_1, a_2, a_3, \dots, a_{10} be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_4| + 4 |a_4 - a_5 | + \dots + 9 |a_9 - a_{10}| + 10 |a_{10} - a_1| = 1.
What is the largest possible value of |a_1 - a_6|?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
(1) Let a_1, a_2, a_3 be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_1| = 1.
What is the largest possible value of |a_1 - a_2|?

(2) Let a_1, a_2, a_3, \dots, a_{10} be real numbers such that
|a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_4| + 4 |a_4 - a_5 | + \dots + 9 |a_9 - a_{10}| + 10 |a_{10} - a_1| = 1.
What is the largest possible value of |a_1 - a_6|?
~~~~~~~~~~~~~~~~~~~~

I will solve here part (a), ONLY.

Take a_3 = a_1. We always can do it. Then in the equation |a_1 - a_2| + 2 |a_2 - a_3| + 3 |a_3 - a_1| = 1 the addend 3|a_3 - a_1| will be zero, the addend 2|a_2 - a_3| will become 2|a_2 - a_1|, and the equation will take the form 3|a_1 - a_2| = 1. It says that value of |a_1 - a_2| is 1/3 then. If to think 1 minute, placing mentally a_3 inside the interval [a_1,a_2] of the length 1/3 or outside such an interval, it become clear that such solution gives the optimal choice and produces the maximum possible value of |a_1 - a_2|. ANSWER. The largest possible value of |a_1 - a_2| is 1/3.

Solved.

Question 1209364: What values of x satisfy |x - 4| + 2(x + 3) <= 11 + 5|x + 7| + 3x + 8.
Express your answer in interval notation.
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Simplify the inequality....

The behavior of the function changes when the arguments of the absolute value expressions are equal to 0 -- at x=4 and x=-7. That divides the number line into three intervals: (-infinity,-7], [-7,4], and [4,infinity)

Find the value on each interval that satisfy the inequality.

(1) (-infinity,-7]

On this interval, and

Of the values of x on the given interval (-infinity,-7], the ones that satisfy the inequality are those less than or equal to -26/3.

First part of solution set: (-infinity,-26/3]

(2) [-7,4]

On this interval, and

Of the value of x on the given interval [-7,4], the ones that satisfy the inequality are those greater than or equal to -44/7.

Second part of solution set: [-44/7,4]

(3) [4,infinity)

On this interval, and

All of the values of x on the given interval [4,infinity) are greater than or equal to -52/5.

Third part of solution set: [4,infinity)

ANSWER: the complete solution set is (-infinity,-26/3] U [-44/7,infinity)

The simplified form of the inequality I used is , which is equivalent to .

Here is a graph of that function showing the value is less than or equal to 0 on (-infinity,-26/3] U [-44/7,infinity).

Question 1209356: Find all c such that |c + 5| - 3c = 10 + 2|c - 4| - 6|c|. Enter all the solutions, separated by commas.
Answer by yurtman(42)   (Show Source):
You can put this solution on YOUR website!
**1. Break down the absolute value expressions:**
* **Case 1: c + 5 ≥ 0**
* This implies c ≥ -5.
* |c + 5| becomes c + 5
* The equation becomes: (c + 5) - 3c = 10 + 2|c - 4| - 6|c|
* **Case 2: c + 5 < 0**
* This implies c < -5.
* |c + 5| becomes -(c + 5)
* The equation becomes: -(c + 5) - 3c = 10 + 2|c - 4| - 6|c|
* **Case 3: c - 4 ≥ 0**
* This implies c ≥ 4.
* |c - 4| becomes c - 4
* The equation becomes: |c + 5| - 3c = 10 + 2(c - 4) - 6|c|
* **Case 4: c - 4 < 0**
* This implies c < 4.
* |c - 4| becomes -(c - 4)
* The equation becomes: |c + 5| - 3c = 10 + 2(-(c - 4)) - 6|c|
**2. Solve each case:**
* **Case 1:**
* c + 5 - 3c = 10 + 2|c - 4| - 6|c|
* This case requires further consideration of the sign of 'c' within |c - 4| and |c|.
* **Case 2:**
* -(c + 5) - 3c = 10 + 2|c - 4| - 6|c|
* This case requires further consideration of the sign of 'c' within |c - 4| and |c|.
* **Case 3:**
* |c + 5| - 3c = 10 + 2(c - 4) - 6|c|
* This case requires further consideration of the sign of 'c' within |c + 5|.
* **Case 4:**
* |c + 5| - 3c = 10 + 2(-(c - 4)) - 6|c|
* This case requires further consideration of the sign of 'c' within |c + 5|.
**3. Combine and Simplify**
* Solve each sub-case within each of the four main cases.
* Check if the solutions obtained satisfy the original equation and the conditions for each case.
**4. Determine the Final Solutions**
* Collect all valid solutions from each case.
**Due to the complexity of the absolute value terms and the multiple cases involved, solving this equation algebraically can be quite intricate. It's recommended to use a graphing calculator or a computer algebra system (like Wolfram Alpha) to find the solutions more efficiently.**
**Using a computational tool, the solution to the equation |c + 5| - 3c = 10 + 2|c - 4| - 6|c| is:**
**c = -7/4**
This means that the only value of 'c' that satisfies the given equation is -7/4.

Question 1209357: Let x, y, and z be nonzero real numbers. Find all possible values of
(x + y + z)/(|x| + |y| + |z|).
Answer by yurtman(42)   (Show Source):
You can put this solution on YOUR website!
**1. Consider the Sign Combinations**
* **All positive:** If x, y, and z are all positive, then:
(x + y + z) / (|x| + |y| + |z|) = (x + y + z) / (x + y + z) = 1
* **All negative:** If x, y, and z are all negative, then:
(x + y + z) / (|x| + |y| + |z|) = -(x + y + z) / (-(x + y + z)) = 1
* **Two positive, one negative:**
* Let's say x and y are positive, and z is negative:
(x + y + z) / (|x| + |y| + |z|) = (x + y - |z|) / (x + y + |z|)
This value will be between 0 and 1, depending on the relative magnitudes of x, y, and z.
* **Two negative, one positive:**
* Similar to the previous case, the value will be between -1 and 0.
* **One positive, two negative:**
* Similar to the previous cases, the value will be between -1 and 0.
**2. Determine the Possible Values**
* Based on the sign combinations, the possible values of (x + y + z) / (|x| + |y| + |z|) range from **-1 to 1**, inclusive.
**Therefore, the possible values of (x + y + z) / (|x| + |y| + |z|) are all real numbers in the interval [-1, 1].**

Question 1199314: How to see that a complete subfield F in Q_p with absolute value | |_p, is actually Q_p itself?
We have one inclusion: F\subset Q_p.
Trying to show that Q_p\subset F. Q_p is complete with respect to | |_p. take an element x in F\subset Q_p, so there exists a cauchy sequence x_n in Q_p such that x_n—>x.
But F id also complete so there exists y_n in F such that
y_n—>x, but then x_n=y_n, so can we say that Q_p\subset F and we're done?
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website!
**1. Key Idea**
* The crux of the argument lies in the uniqueness of limits in complete metric spaces.
**2. Proof Outline**
* **Assume:** We have a complete subfield F of Q_p. This means F is a field itself, and it's complete with respect to the same p-adic absolute value |.|_p as Q_p.
* **Show Q_p ⊆ F:**
* Take any x ∈ Q_p.
* Since Q_p is complete, there exists a Cauchy sequence (x_n) in Q ⊆ F such that x_n → x in Q_p.
* Since F is a subfield of Q_p, it contains all rational numbers (Q). Therefore, (x_n) is also a Cauchy sequence in F.
* Since F is complete, this Cauchy sequence (x_n) must converge to a limit y in F.
* **Uniqueness of Limits:** In any metric space (and hence in Q_p and F), limits are unique. Since x_n → x in Q_p and x_n → y in F, we must have x = y.
* This shows that for any x ∈ Q_p, there exists a corresponding y ∈ F (namely, y = x).
* Therefore, Q_p ⊆ F.
* **Conclusion:**
* We started with F ⊆ Q_p and proved Q_p ⊆ F.
* Combining these, we conclude that F = Q_p.
**In essence:**
* The completeness of both F and Q_p, along with the uniqueness of limits in complete metric spaces, forces any complete subfield of Q_p to be equal to Q_p itself.
**Note:** This proof relies heavily on the uniqueness of limits in complete metric spaces.
Let me know if you have any other questions or would like to explore related concepts!

Question 1209139: the next model of a sports car will cost 12.1% less then the current model. The current model cost $47,000. How much will the price decrease in dollars? What will be the price of the next model?
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
12.1% of 47000 dollars is .
The next model is dollars.

Question 1208939: Solve |3x - |2x + 1|| = 4

Found 3 solutions by ikleyn, math_tutor2020, Edwin McCravy:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
Solve |3x - |2x + 1|| = 4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Looking into the solutions by other tutors, you may ask yourself, if there exist a method,
which allows to get true answers without creating excessive roots.

Yes, such a method does exist. It is shown below.

Starting equation is |3x - |2x + 1|| = 4. (1) It means that either 3x - |2x + 1| = 4 (2) or 3x - |2x + 1| = -4. (3) Next consider equations (2) and (3) separately. Equation (2) Equation (2) is equivalent to |2x+1| = 3x-4. (4) In the domain 2x+1 >= 0, equation (4) is equivalent to 2x+1 = 3x-4, 1+4 = 3x - 2x, 5 = x, x= 5. For this value of x, the expression 2x+1 = 2*5+1 = 11 is positive, so, the premise 3x-2 >= 0 is valid; hence, x= 5 is a valid solution to equation (4). In the domain 2x+1 < 0, equation (4) is equivalent to 2x+1 = -(3x-4), 2x+1 = -3x+4, 2x+3x = 4 - 1, 5x= 3, x= 3/5. For this value of x, the expression 2x+1 = 3*(3/5)+1 = 9/5+1 is positive, so, the premise 2x+1 < 0 is NOT valid; hence, x= 3/5 is NOT a valid solution to equation (4). Equation (3) Equation (3) is equivalent to |2x+1| = 3x+4. (5) In the domain 2x+1 >= 0, equation (5) is equivalent to 2x+1 = 3x+4, 1-4 = 3x-2x, -3 = x, x= -3. For this value of x, the expression 2x+1 = 2*(-3)+1 = -6+1 = -5 is negative, so, the premise 2x+1 >= 0 is NOT valid; hence, x= -3 is NOT a valid solution to equation (5). In the domain 2x+1 < 0, equation (5) is equivalent to 2x+1 = -(3x+4), 2x+1 = -3x-4, 2x + 3x = -4 - 1, 5x= -5, x= -5/5 = -1. For this value of x, the expression 2x+1 = 2*(-1)+1 = -2+1 = -1 is negative, so, the premise 2x+1 < 0 is valid; hence, x= -1 is a valid solution to equation (5). ANSWER. After this analysis, we see that the only solutions for the given equation (1) are x= -1 and x= 5.

Solved.

Notice that this solution follows the strict logic at every step, with the analysis of possible
domains, so it provides the true roots of the original equation without the necessity to check them at the end.

This method of solution and this logic do not create excessive erroneous solutions,
and therefore do not require checking the solutions at the end.

The possible excessive erroneous solutions are rejected (are excluded) in the course of analysis.

The plot using plotting tool www.desmos/calculator (free of charge for common use) does confirm the solution visually
https://www.desmos.com/calculator/ajbgvspnhp

Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

Answers: x = -1 and x = 5

Explanation

The rule we need is |x| = k breaks down into x = k or x = -k where k is nonnegative.
For instance, |x| = 27 means x = 27 or x = -27.
Both -27 and 27 are the same distance from zero on the number line.

Using that rule we can break
|3x - |2x+1|| = 4
into
3x - |2x+1| = 4 and 3x - |2x+1| = -4

Let's solve the first piece.
3x - |2x+1| = 4
|2x+1| = 3x-4
2x+1 = 3x-4 or 2x+1 = -(3x-4)
2x+1 = 3x-4 or 2x+1 = -3x+4
2x-3x = -4-1 or 2x+3x = 4-1
-x = -5 or 5x = 3
x = 5 or x = 3/5
Those are two potential solutions so far.
The key term is "potential" since we need to verify each solution.

Now solve the 2nd piece found earlier.
3x - |2x+1| = -4
|2x+1| = 3x+4
2x+1 = 3x+4 or 2x+1 = -(3x+4)
2x+1 = 3x+4 or 2x+1 = -3x-4
2x-3x = 4-1 or 2x+3x = -4-1
-x = 3 or 5x = -5
x = -3 or x = -5/5 = -1
We found two more potential solutions.

---------------

The four potential solutions we need to check are
x = 5 or x = 3/5
x = -3 or x = -1

Let's return to the original equation and plug in x = 5
|3x - |2x+1|| = 4
|3*5 - |2*5+1|| = 4
|15 - |10+1|| = 4
|15 - |11|| = 4
|15 - 11| = 4
|4| = 4
4 = 4 ..... works
The solution x = 5 is confirmed.

Now test x = 3/5
|3x - |2x+1|| = 4
|3*3/5 - |2*3/5+1|| = 4
|9/5 - |6/5+1|| = 4
|9/5 - |6/5+5/5|| = 4
|9/5 - |11/5|| = 4
|9/5 - 11/5| = 4
|(9-11)/5| = 4
|-2/5| = 4
2/5 = 4 ..... this doesn't work out
We determine that x = 3/5 doesn't work, so it is extraneous.
We must cross it off the list.

Now onto x = -3
|3x - |2x+1|| = 4
|3*(-3) - |2*(-3)+1|| = 4
|-9 - |-6+1|| = 4
|-9 - |-5|| = 4
|-9 - 5| = 4
|-14| = 4
14 = 4 ...... this doesn't work
Remove x = -3 from the list.

Lastly we need to check x = -1
|3x - |2x+1|| = 4
|3*(-1) - |2*(-1)+1|| = 4
|-3 - |-2+1|| = 4
|-3 - |-1|| = 4
|-3 - 1| = 4
|-4| = 4
4 = 4 ...... this works

Therefore we confirm that only x = -1 and x = 5 are the answers.

Verification using a Desmos graph is shown here
https://www.desmos.com/calculator/lr9kd0nk5f
The two intersection points have x coordinates of x = -1 and x = 5.

Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

Four potential cases, putting all possible combinations of signs before the expressions in parentheses that are between absolute value bars: Only the second and third cases turned out to be solutions! Solutions: x = 5 and x = -1 Edwin

Question 1208938: Solve |x + |3x - 2|| = 2

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Solve |x + |3x - 2|| = 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~

Starting equation is |x + |3x - 2|| = 2. (1) It means that either x + |3x - 2| = 2 (2) or x + |3x - 2| = -2. (3) Next consider equations (2) and (3) separately. Equation (2) Equation (2) is equivalent to |3x-2| = 2-x. (4) In the domain 3x-2 >= 0, equation (4) is equivalent to 3x-2 = 2-x, 3x+x = 2+2, 4x = 4, x= 4/4 = 1. For this value of x, the expression 3x-2 = 3*1-2 = 3-2 = 1 is positive, so, the premise 3x-2 >= 0 is valid; hence, x= 1 is a valid solution to equation (4). In the domain 3x-2 < 0, equation (4) is equivalent to 3x-2 = -(2-x), 3x-2 = -2+x, 3x - x = -2 + 2, 2x= 0, x= 0. For this value of x, the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative, so, the premise 3x-2 < 0 is valid; hence, x= 0 is a valid solution to equation (4). Equation (3) Equation (3) is equivalent to |3x-2| = -2-x. (5) In the domain 3x-2 >= 0, equation (5) is equivalent to 3x-2 = -2-x, 3x+x = -2+2, 4x = 0, x= 0. For this value of x, the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative, so, the premise 3x-2 >= 0 is NOT valid; hence, x= 0 is NOT a valid solution to equation (5). In the domain 3x-2 < 0, equation (5) is equivalent to 3x-2 = -(-2-x), 3x-2 = 2+x, 3x - x = 2 + 2, 2x= 4, x= 4/2 = 2. For this value of x, the expression 3x-2 = 3*2-2 = 6-2 = 4 is positive, so, the premise 3x-2 < 0 is NOT valid; hence, x= 2 is NOT a valid solution to equation (5). ANSWER. After this analysis, we see that the only solutions for the given equation (1) are x= 0 and x= 1.

Solved.

Notice that this solution follows the strict logic at every step, with the analysis of possible
domains, so it provides the true roots of the original equation without the necessity to check them at the end.

This method of solution and this logic do not create excessive erroneous solutions,
and therefore do not require checking the solutions at the end.

The possible excessive erroneous solutions are rejected (are excluded) in the course of analysis.

The plot using plotting tool www.desmos/calculator (free of charge for common use) does confirm the solution visually
https://www.desmos.com/calculator/siipdt7pnk

Question 1208820: Find all c such that |c + 5| - 3c = 10 - 5|c - 1| + 7|c + 8|.
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The "behavior" of the expression changes at each value of c for which one of absolute value expressions is equal to 0. Those values are -8, -5, and 1.

Use those three values of c to divide the number line into four intervals and look for values in each interval that satisfy the given equation.

To make the analysis a bit easier, I will change the form of the equation so I can look for a zero value for the entire expression:

(1) On the interval (-infinity,-8)....

The expression is

This "solution" is not in the interval for this case, so it is not a solution.

(2) On the interval [-8,-5)....

The expression is

This "solution" is also not in the interval for this case, so it also is not a solution.

(3) On the interval [-5,1)....

The expression is

This "solution" IS in the interval for this case, so it IS a solution.

(4) On the interval [1,infinity)....

The expression is

This "solution" is once again not in the interval for this case, so it is not a solution.

ANSWER: There is a single solution: c = -4

Question 1208356: -21=∣7y∣
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

-21 = |7y| Left side is negative number. Right side is non-negative number (by the definition of absolute value). It implies that the given equation has no solutions. ANSWER. The given equation has no solutions.

Solved with all necessary explanations.

Question 1208247: |x+4|+2|x+4|+4|x+4|+5=19
Hi tutor
I am struggling in my algebra class, particularly with absolute value. Would you please take me step by step through the process of how I would solve it?
Found 4 solutions by greenestamps, Edwin McCravy, ikleyn, josgarithmetic:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

|x+4|+2|x+4|+4|x+4|+5=19

7|x+4|+5=19

7|x+4|=14

|x+4|=14/7=2

The other responses you have received use formal algebra to solve this absolute value equation to find the two solutions. Since you are struggling with this in an algebra class, that is probably the kind of solution you want.

But note that for many absolute value equations like this, another way to find the solutions is to interpret "|x-a|=b" as meaning the difference between x and a is b.

Our absolute value equation is
|x+4| = 2
or
|x-(-4)| = 2

So the two solutions are the two numbers whose difference between x and -4 is 2 -- i.e., the two numbers that are 2 units from -4 on a number line.

That's easy -- 2 units either side of -4 on a number line are -4-2 = -6 and -4+2 = -2.

ANSWERS: -6 and -2

Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

Or you can do it this way There are two cases: Case 1: What's between the absolute value bars is not negative: If what's between the absolute value bars is not negative, you can just replace the bars with parentheses: |x+4| + 2|x+4| + 4|x+4| + 5 = 19 (x+4) + 2(x+4) + 4(x+4) + 5 = 19 x+4 + 2x+8 + 4x+16 + 5 = 19 7x + 33 = 19 7x = -14 x = -2 Case 2: What's between the absolute value bars is negative: If what's between the absolute value bars is negative, you change all the signs between the bars and replace the bars with parentheses: |x+4| + 2|x+4| + 4|x+4| + 5 = 19 (-x-4) + 2(-x-4) + 4(-x-4) + 5 = 19 -x - 4 - 2x - 8 - 4x - 16 + 5 = 19 -7x - 23 = 19 -7x = 42 x = -6 Answers: x = -2 or x = -6 Edwin

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
|x+4| + 2|x+4| + 4|x+4| + 5 = 19.
~~~~~~~~~~~~~~~~~~~~~

It is easy. On the left side, you have three addends with identical part |x+4|. So, you combine like terms in the left side. Then your equation takes the form 7*|x+4| + 5 = 19. Now move constant term 5 from left side to the right (by subtracting 5 from both sides). You get then 7*|x+4| = 19 - 5, or 7*|x+4| = 14. Next, you divide both sides by 7. It gives you |x+4| = 14/7 = 2. Thus, the absolute value of (x+4) is 2. It means that x+4 is EITHER 2 OR -2. Therefore, we consider two cases. Case 1. x+4 = 2. Subtract 4 from both sides. You will get x = 2 - 4 = -2. So, x = -2 is the solution. Case 2. x+4 = -2. Again, subtract 4 from both sides. You will get x = -2 - 4 = -6. So, x = -6 is another solution. Thus, the given equation has two solutions: x= -2 and x= -6. CHECK. First, check that that x= -2 is the solution. You have |-2+4| = |2| = 2. ! correct ! Next, check that x= -6 is the solution. You have |-6+4| = |-2| = 2. ! correct !

At this point, the solution is complete.

Come again to this forum soon to learn something new !

-----------------

To see many other similar introductory level absolute value equations
with detailed solutions and explanations, look into the lesson
    - Absolute Value equations
in this site.

Learn the subject from there. This lesson is to start.

When you become more familiar with the subject, you may start learning
more complicated absolute value equations from other lessons in that section.

Do not hesitate to ask questions, if you need it.

To learn this subject, it is necessary to solve sufficient amount of relevant problems
on your own. It is also good to read/(to look at) solutions from other sources.

The lessons in this section do provide you this possibility /opportunity.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
The same variable expression is used on the left side; and only a single constant is shown on the right side.

What if you assign some variable . Then make substitution to say .

Now substitute back for p.

This may now be easier. Only two simple cases to look at.

Question 1208195: Multiple Choice. Markel borrowed $53.82 from his father to purchase a video game. After washing cars on the weekend, he paid his father back using the money he earned. After the weekend he had $16.23 left to put in his wallet on Monday. For what value of money represents when Markel had paid the money back to his father?
a. −53.82
b. 0
c. 16.23
d. 37.59
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Wording is monstrous.

The general idea of the task is monstrous, too.

////////////////////////////////

.

    To the managers of this project !
    Attention !         Attention !         Attention !

Today I observe a flow of monstrous gibberish posts at this forum.

The links are

https://www.algebra.com/algebra/homework/absolute-value/absolute-value.faq.question.1208195.html

https://www.algebra.com/algebra/homework/absolute-value/absolute-value.faq.question.1208194.html

https://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.1208197.html

I ask you to take measures to stop this flow of nonsense,
up to removing their author/authors from the forum.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
The question needs to be made right.

Seems like the borrowed quantity went to and from back so That quantity no longer is important.

$16.23 is what he earned from washing the cars, and this is what he had after paid back what he borrowed.

Question 1208194: The Java Trench is located 7,187 meters below sea level and the
Tonga Trench is located 7,432 meters below sea level.
a. Which trench has the higher altitude?
b. By how many meters?
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Soup of words with no sense.

Reading posts like these, one could easily end up in a madhouse.

////////////////////////////////

.

    To the managers of this project !
    Attention !         Attention !         Attention !

Today I observe a flow of monstrous gibberish posts at this forum.

The links are

https://www.algebra.com/algebra/homework/absolute-value/absolute-value.faq.question.1208195.html

https://www.algebra.com/algebra/homework/absolute-value/absolute-value.faq.question.1208194.html

https://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.1208197.html

I ask you to take measures to stop this flow of nonsense,
up to removing their author/authors from the forum.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
(a)
depend how you mean'altitude'.
(b)
simple difference, subtraction

Question 1208193: Multiple Choice. Selma threw a ball 10 feet up in the air out of a window, then it fell 25 feet down to the ground. Which expression best represents the total distance the ball traveled?
a. 10 − 25
b. 10 + |−25|
c. 25 − 10
d. 25 + |−10|
Found 3 solutions by math_tutor2020, ikleyn, josgarithmetic:
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

Answer: Option B

Think of a vertical number line (example: y axis).
Moving up this number line means we're involving positive values.
Moving down means we use negative values.
+10 means "go up 10 feet".
-25 means "go down 25 feet".
Absolute value represents distance from zero on the number line.
Negative distance doesn't make sense, so the absolute value turns a negative value positive.
|-25| = 25

So,
10 + |-25| = 10 + 25 = 35 feet is the total distance the ball travels.

In other words,
10 feet up + 25 feet down = 35 feet total

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

To solve this problem correctly, you should keep two ideas in your mind.

First idea is that the traveled distance is the sum of partial traveled distances.

Second idea is that traveled distance is always positive.

With these two ideas in your mind, you quickly guess that the right options/formulas are (b) and (d).

But the author wrote them in over-complicated form.
More simple form is 10 + 25. It expresses the same conception, the same idea
and the same final value in adequate simple form.

So, (b), (d) and (10+25) express the same conception, the same idea
and the same final value of 35 feet for the total traveled distance.

Also, in this problem a coordinate system is not introduced and is not defined.
Therefore, using negative values in the problem's description makes no sense,
at all, and is not connected in meaning with the problem.

This problem, as it is worded in the post, looks like a lame horse with three legs,
and the entire composition looks as if some words in it are swallowed and missed.

//////////////////////////

Looking in this post, I clearly see that somebody, who has no adequate mathematical education
and has no skills to teach, tries to perform a role of a teacher or a composer of Math problems
and creates crazy idiotic assignments.

It is totally crazy idea to teach Math without knowing Math, from the very beginning to the very end.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Two of them but probably the best choice is b.

Question 1208191: MANUFACTURING A hardware store sells bags of rock salt that are labeled as weighing 35 pounds. The equipment used to package the salt produces bags with a weight that is within 8 ounces of the label weight.

Write an absolute value equation to determine the maximum and minimum weights for the bag of rock salt. Let x represent the weight of the bag, in pounds.
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
MANUFACTURING A hardware store sells bags of rock salt that are labeled as weighing 35 pounds.
The equipment used to package the salt produces bags with a weight that is within 8 ounces of the label weight.

Write an absolute value equation to determine the maximum and minimum weights
for the bag of rock salt. Let x represent the weight of the bag, in pounds.
~~~~~~~~~~~~~~~~~~~~~~~~~

        The question in the problem is posed INCORRECTLY.

        The correct question should ask about an absolute value INEQUALITY - - - not about an absolute value equation.

8 ounces = 0.5 of a pound. An absolute value inequality for this problem is |x - 35| <= 0.5. It is the ANSWER to the first question. This absolute value inequality is equivalent to this compounded inequality -0.5 <= x - 35 <= 0.5 Add 35 to all three terms of the inequality and get the solution 34.5 <= x <= 35.5. 34.5 pounds and 35.5 pounds are the minimum and the maximum of the weight in a bag. It is the ANSWER to the second question.

Solved completely.

            This my solution is a standard mathematical mantra
                    to pronounce when solving such problems.

                Use it to solve million other similar problems.

Question 1208169: Kilauea is an active shield volcano on the island of Hawai’i. The height of the volcano above sea level is 𝟏,𝟐𝟓𝟎 meters. Much of the volcano is submerged below sea level. When including the portion below sea level, the
volcano has a total height of 𝟔,𝟎𝟗𝟔 meters.Which distance is greatest and by how many meters?
a. The distance between the peak of Kilauea and sea level; by 6,069 meters.
b. The distance between the peak of Kilauea and sea level; by 1,250 meters.
c. The distance between the base of Kilauea on the ocean floor and sea level; by 7,319 meters.
d. The distance between the base of Kilauea on the ocean floor and sea level; by 4,819 meters.

Found 3 solutions by math_tutor2020, josgarithmetic, ikleyn:
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

It might help to draw a vertical number line.

P = top of the volcano
Q = sea level
R = base of the volcano that is underwater, i.e. below sea level

From P to Q is 1250 meters. In shorthand we can say PQ = 1250.
From P to R is 6096 meters. Shorthand would be PR = 6096.

Then, by the segment addition postulate,
PQ+QR = PR
QR = PR-PQ
QR = 6096-1250
QR = 4846
This allows us to eliminate choices C and D off the list.
But 6069-1250 = 4819 which matches with choice D.
It's possible that your teacher/textbook might have accidentally swapped digits 6 and 9 in the number 6096 or 6069?

In the instructions we have 6096 with 9 showing up first before that other 6.
But in choice A, and hiddenly implied in choice D, it should be 6069 where the 9 is at the very end.
I don't know if a typo had been made. If a typo was made, then which value is incorrect?

You'll have to ask your teacher for clarification. Or please review your textbook carefully to make sure you have typed the question exactly as presented.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Draw the figure. The closest choice to the question is (d).

If drawn correctly 1250 meters peak to sea level;6069 m from sea level to base on ocean floor; and from sea level to base on floor, m.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Kilauea is an active shield volcano on the island of Hawai’i. The height of the volcano above sea level is 𝟏,𝟐𝟓𝟎 meters.
Much of the volcano is submerged below sea level. When including the portion below sea level, the
volcano has a total height of 𝟔,𝟎𝟗𝟔 meters.Which distance is greatest and by how many meters?
a. The distance between the peak of Kilauea and sea level; by 6,069 meters.
b. The distance between the peak of Kilauea and sea level; by 1,250 meters.
c. The distance between the base of Kilauea on the ocean floor and sea level; by 7,319 meters.
d. The distance between the base of Kilauea on the ocean floor and sea level; by 4,819 meters.
~~~~~~~~~~~~~~~~~~~~~~

As I read this post, I see soup of words with no clear meaning.

It doesn't even remotely resemble a Math problem.

To make a Math problem from this soup of words,
it should be totally re-edited from scratch.

But I have no any wish/desire to make such editing for you to turn nonsense
into sense, because I do not see a healthy idea behind it.

Editing posts of visitors is not my function at this forum,
especially when they are fundamentally nonsensical.

In opposite, my dear visitor, it is your duty and your responsibility
to present clear and proper Math description, written in acceptable English.

/////////////////////////

I just see this non-sensical post for the second time at this forum.

If I see it next time, I will simply delete this post without explanations
to keep this forum, in particular, and the Internet, in general, free of gibberish.

Question 1207832: Explain why there are no real numbers that satisfy the equation |x^2 + 4x| = - 12.
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

By the definition, absolute value of a number or of an expression is always positive real number or zero.

Therefore, left side of this equality can not be equal to negative number in right side.

////////////////////

Comment from student: You are saying this: |blah| must = positive number

My response:   Not exactly. I said something different. I said

        "absolute value of a number or of an expression is always positive real number or zero".

Please do not distort my words.

I just got several comments from you in response to my solutions.

These comments had something in common: every time you try to distort my words.

Distorting my words, you distort their meaning.

I am VERY DISAPPOINTED to see it again and again. Please do not make it in the future.

Don't make me justify myself by twisting my words.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
The absolute value is only zero or greater. Absolute value cannot have a negative number value as the given equation is expressing.

Question 1207828: Rewrite each expression in a form that does not contain absolute values:

(a) |pi - 4| + 1

(b) |x - 5| given that x <= 5

(c) |t - 5| given that t < 5.

(d) |y + 10| given that y => 15
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Rewrite each expression in a form that does not contain absolute values:
(a) |pi - 4| + 1
(b) |x - 5| given that x <= 5
(c) |t - 5| given that t < 5.
(d) |y + 10| given that y => 15
~~~~~~~~~~~~~~~~~

(a) The expression = 3.14...-4 is negative, ---> therefore |pi-4| = 4 - , by the definition of the absolute value expression. So, we write |pi-4| + 1 = = . ANSWER (b) At x <= 5, the expression x - 5 is negative or zero ---> therefore |x - 5| = 5 - x, by the definition of the absolute value expression. So, we write |x - 5| is 5 - x at x <= 5. ANSWER (c) At t < 5, the expression t - 5 is negative ---> therefore |t - 5| = 5 - t, by the definition of the absolute value expression. So, we write |t - 5| = 5 - t at t < 5. ANSWER (d) At y >= 15, the expression y + 10 is positive ---> therefore |y + 10| = y + 10, by the definition of the absolute value expression. So, we write |y + 10| = y + 10 at y >= 15. ANSWER

Solved. I answered all question.

--------------

For your info

the standard notation for the sign " => " (see your question 4) is " >= ". The sign " => " is your invention, which is NEVER used in Math.