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

Lesson 2: Solving absolute value equations

# Worked example: absolute value equation with two solutions

Solving the equation 8|x+7|+4 = -6|x+7|+6 which has two possible solutions. Created by Sal Khan.

## Want to join the conversation?

• how does the absolute value thing separate the equation into 2 equations
• One of the equation's answer is a positive while the other equation's answer is a negative.
• What happens when the absolute values on either side of the equation are not equal to each other, such as
(Im using \'s for absolute value signs)
6 \x+9\ +7 = -4 \x+2\ +3
• Jspofforth,
6 |x+9| +7 = -4 |x+2| +3 has no solution.
If you graph the y = 6 |x+9| +7 and y= -4 |x+2| +3 you see they have no points in common.https://www.khanacademy.org/cs/y-6x97-and-y-4x23/5898680397725696

Algebraically, you can find the four solutions to
6*( ±(x+9)) +7 = -4*(±(x+2) +3
which expands into four separate equations.
6 (x+9) +7 = -4 (x+2) +3
6 (-x-9) +7 = -4 (x+2) +3
6 (x+9) +7 = -4 (-x-2) +3
6 (-x-9) +7 = -4 (-x-2) +3
If the answer you get for any of the four equations gives you a positive number for (x+9) and for (x+2) that is a valid answer.
But if the answer for the equation when put into (x+2) or (x+9) is negative in either case, the answer fails.

The answer I get for the first equation is x=-6.6
But -6.6+2 is negative so this answer fails.
The answer I get for the second equation is x=-21
But -21+2 and -21+9 are both negative so this answer fails.
The answer I get for the third equation is x=-25
But -25+2 and -25+9 are both negative so this answer fails.
The answer I get for the first equation is x=-5.8
But -5.8+2 is negative so this answer fails.

So no answer works and the equation has no solution.

I hope that helps make it click for you.
• whats the difference between |-4|+3 and |-4+3|
• The |-4| = 4, so |-4| + 3= 4 +3 = 7
While |-4 +3| = |-1| = 1
• |c-3| = |c-7|

How can we solve this question? Answer is 5, but how? Please answer. Here 'c' is the 'y-intercept' of a graph.
• We can apply casework for the possible values that c can attain:
Note, this identity for absolute values is used: for |x-k|, if x<k, then |x-k|=k-x and for x>k, then |x-k|=x-k

If c<3, then |c-3|=3-c and |c-7|=7-c
3-c=7-c, which results with no solution
2) if c is on the interval 3<c<7 and including 3 and 7, then |c-3|=c-3 and |c-7|=7-c
c-3=7-c, 2c=10, c=5 is a solution
Finally, we consider c>7
|c-3|=c-3 and |c-7|=c-7
c-3=c-7, which has no solution
Thus the only solution is c=5
• Wait, so how did we get 49/7?? How is 1/7-7 equivalent to 49/7??
• Wait, there's 2 solutions?
• Yes!
Suppose I said |x|=10.
There are 2 solutions, 10 and -10.
Why -10? Because |-10| = 10.
If this seems weird now, you will get used to it with time.
Just keep studying!
• so...when you add |x+7| to both sides, why don't you get |x+14| ?
• Anything in the brackets stay the same while your are organizing the problem(absolute value on the left, lone numbers on the right)
• At , how did you get 49/7 ? How come 7 is equal to 49/7 ?
• Because 49/7=7 and because 7 squared is 49.
• I'm not quite understanding why
8|x+7|
+ 6|x+7|
-------------
14|x+7| and not 14|x+14|.
Do you not add the 7's because that is a property of absolute value?
• Imagine this as the distributive property: 14 |x+7| = (6+8) |x+7| =6 |x+7| + 8 |x+7|
(1 vote)
• At the beginning of the problem, couldn't you just made it as 8*x+7+4= -6*x+7+6?
(1 vote)
• No, you can't for 2 reasons.
1) You completely ignored the absolute values. They have meaning. You can not ignore them. The absolute values are what cause this equation to have 2 solutions, not one. Your version creates one solution.
2) The absolute values (besides being the distance from zero) act as grouping symbols. Once you get to the point that you can actually drop the absolute values, if there is a number in front, that number must be distributed. Example: 14 |x+7| = 2 becomes 14(x+7) = 2 and 14(x+7) = -2.
The 14 must be distributed and you get: 14x+98 = 2 and 14x+98 = -2

I suggest you start with the 1st video in this section. Pay attention to the basic concepts on why these equations have 2 solutions and when you can drop the absolute value symbols. Here is the link: https://www.khanacademy.org/math/algebra/absolute-value-equations-functions/absolute-value-equations/v/absolute-value-equations