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

Lesson 2: Solving absolute value equations

# 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?
• 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.
• 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.
• 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.
• What if the absolute value is not same on both sides?
example: 2|x - 8| + 6 = 3|x + 9| - 8
Can I solve it?
• Did any one notice how Sal said "There is absolutely no solution". Caught the pun?
• 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?
• 2 negative fractions make up a positive number, just think about the x-(-a)= x+a
• 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.
• 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".
• 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.