Main content

## Algebra (all content)

### Course: Algebra (all content) > Unit 8

Lesson 2: Solving absolute value equations- Intro to absolute value equations and graphs
- Worked example: absolute value equation with two solutions
- Worked example: absolute value equations with one solution
- Worked example: absolute value equations with no solution
- Solve absolute value equations

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Worked example: absolute value equations with no solution

Solving the equation 4|x+10|+4 = 6|x+10|+10 to find that it has no possible solution. Created by Sal Khan.

## Want to join the conversation?

- Just out of curiosity, is there something similar to imaginary number
**i**and square root, what makes absolute value of something to equal to a negative number?(12 votes)- I would think not, since the absolute value function is specifically designed to ensure a positive output. Square roots only have positive outputs as a side effect of the squaring laws, so I would think i is a unique case compared to absolute value.(5 votes)

- I have a question, how would I solve a question like

| x - 1 | + |2 - x| = 14

Is it possible to solve this equation algebraically or do I solve this graphically.

PleaseHelp.(8 votes)- The proper way of doing the problem is to consider the points at which each of the terms whose absolute value is taken changes sign

So, equate each term to 0, e.g in this question x-1=0 implies the abs x-1 term changes sign at 1 and equating 2-x=0, implies it changes sign at 2

These two points divide the number line into 3 areas: x<1; 1<x<2 and x>2

also note that the x-1 term is positive for x>1 and negative for x<1

the reverse is true for 2-x, which is positive for x<2 and is negative for x>2

for the three areas formulate the equations keeping in mind the sign of the terms and you'll be good to go.

e.g for x<1; -(x-1)+(2-x)=14

for 1<x<2: (x-1)+(2-x)= 14

for x>2: (x-1)+(-(2-x)) = 14

Solve each of the three equations and find the value of x valid in each region. Hope , it helps.(6 votes)

- What if the absolute value is not same on both sides?

example: 2|x - 8| + 6 = 3|x + 9| - 8

Can I solve it?(3 votes) - Did any one notice how Sal said "There is absolutely no solution". Caught the pun?(2 votes)
- ok, here's a problem I got wrong in the practice

5|x+4|+8=2|x+4|+5

Subtract 2|x+4| from both sides:

5|x+4|+8

−2|x+4|=

3|x+4|+8=|5

Subtract 8 from both sides:

3|x+4|+8=5

-8=-8

Divide both sides by 3:

3|x+4|=−3

Simplify:

|x+4|=−1

The absolute value cannot be negative. Therefore, there is no solution.

Because it always says when you take the absolute symbols off, you get a negative number as well as a positive number the solutions I came up with were

-3 and -5.

All throughout the practice there are solutions with fractions where both fractions are negative.

What's the difference? Why is this one unsolvable where problems with 2 negative fractions for answers are considered solvable?(3 votes)- 2 negative fractions make up a positive number, just think about the x-(-a)= x+a(0 votes)

- Does the absolute value equation have to be |x+a| = 0 to get the solution?(1 vote)
- If you have: |x+a|=0, you get one solution.

If you have: |x+a|=a positive number, you usually get 2 solutions.

If you have: |x+a|=a negative number, then it has no solution.

Hope this helps.(3 votes)

- So if the absolute value equals a positive number you can change it to a negative number to get an alternative answer, but if it's negative you can't change it to a positive?(1 vote)
- Absolute value is distance from 0. You can' that've a negative distance, so we put "no solution".(3 votes)

- I am so confused for this equation. I have 3|x+1|+7=6|x+1|+6.

I watched all the videos but I am still very confused. Can anyone help?(1 vote)- 1) Notice that the absolute values on both sides match. This means they can be combined just like 3y and 6y can be combined. So, start by subtracting 3|x+1| from both sides:

3|x+1|-3|x+1|+7 = 6|x+1|+6-3|x+1|

7 = 3|x+1|+6

2) Subtract 6 from both sides

7-6 = 3|x+1|+6-6

1 = 3|x+1|

3) Split the equation into 2:

1 = 3(x+1) and 1 = -3(x+1)

4) Solve each equation to get the solutions.

Hope this helps.(3 votes)

- So when you solve an absolute value equation, and you first get a negative it's no solution ,but if you solve it and you get a positive solution you can solve for two solutions making the other negative?

why??(1 vote)- Hi! :) Absolute value is simply the distance that number is from zero (0). Think of each number as a certain number of "jumps" from zero. Because it is a distance from zero, you can never have the answer of an absolute value equation equal to a negative number. For example, |5| = 5, |-6| = 6....... Hope this helps! :-)(2 votes)

- Do you always have to subtract something from the right-hand side or does it also work if you do it on the left hand side?(1 vote)
- You can do opposites from either side of the equation, so yes it works from both sides. You can also flip equations around from one side to the other.(2 votes)

## Video transcript

So we're asked to
solve for x, and we have this equation with
absolute values in it. So it's 4 times
the absolute value of x plus 10 plus 4 is equal to
6 times the absolute value of x plus 10 plus 10. And at first, this
looks really daunting, but the key is to just solve for
this absolute value expression and then go from there. Let me just rewrite it so
that the absolute value expression really jumps out. So this is 4 times
the absolute value of x plus 10 plus 4 is equal to
6 times the absolute value of x plus 10 plus 10. So let's get all of the
absolute values of x plus 10 on the left-hand side. So I want to get
rid of the 6 times the absolute value of
x plus 10 on the right. Well, how would I do that? Well, I could subtract 6 times
the absolute value of x plus 10 from the right,
but we've already seen this multiple times. If these two things
are equal, and if I want to keep them
equal, if I subtract 6 from the right-hand side,
I've got to subtract-- or if I subtract 6 times the
absolute value of x plus 10 from the right-hand side, I
have to subtract the same thing from the left-hand side. So we're going to
have minus 6 times the absolute value of x plus 10. And likewise, I want to
get all my constant terms, I want to get this 4 out
of the left-hand side. So let me subtract
4 from the left, and then I have to also
do it on the right, otherwise my equality
wouldn't hold. And now let's see
what we end up with. So on the left-hand side,
the 4 minus 4, that's 0. You have 4 of something
minus 6 of something, that means you're going
to end up with negative 2 of that something. Negative 2 of the absolute
value of x plus 10. Remember, this might
seem a little confusing, but remember, if
you had 4 apples and you subtract 6 apples, you
now have negative 2 apples, I guess you owe
someone the apples. Same way, you have 4
of this expression, you take away 6 of
this expression, you now have negative
2 of this expression. Let me write it a
little bit neater. So it's negative 2 times the
absolute value of x plus 10 is equal to, well the whole
point of this, of the 6 times the absolute value of
x plus 10 minus 6 times the absolute value of x plus
10 is to make those cancel out, and then you have 10 minus
4, which is equal to 6. Now, we want to solve for the
absolute value of x plus 10. So let's get rid
of this negative 2, and we can do that by dividing
both sides by negative 2. You might realize,
everything we've done so far is just treating
this red expression as almost just like
a variable, and we're going to solve for
that red expression and then take it from there. So negative 2 divided
by negative 2 is 1. 6 divided by negative
2 is negative 3. So we get the absolute
value of x plus 10 is equal to negative 3. Now, this gets us to a
very interesting situation. You might say maybe this
could be the positive version or the negative, but
remember, absolute value is always non-negative. If you took the absolute
value of 0, you would get 0. But the absolute
value of anything else is going to be positive. So this thing right
over here is definitely going to be greater
than or equal to 0. Doesn't matter what
x you put in there, when you take its
absolute value, you're going to
get a value that's greater than or equal to 0. So there's no x that you
could find that's somehow-- you put it there, you add 10,
you take the absolute value of it, you're actually
getting a negative value. So this right over here
has absolutely no solution. And I'll put some exclamation
marks there for emphasis.