Algebra (all content)
- 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
Solving the equation 8|x+7|+4 = -6|x+7|+6 which has two possible solutions. Created by Sal Khan.
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- how does the absolute value thing separate the equation into 2 equations(8 votes)
- 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(8 votes)
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.(9 votes)
- |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.(5 votes)
- 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(8 votes)
- whats the difference between |-4|+3 and |-4+3|(5 votes)
- At4:15, how did you get 49/7 ? How come 7 is equal to 49/7 ?(4 votes)
- Wait, there's 2 solutions?(3 votes)
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!(3 votes)
- so...when you add |x+7| to both sides, why don't you get |x+14| ?(2 votes)
- Anything in the brackets stay the same while your are organizing the problem(absolute value on the left, lone numbers on the right)(3 votes)
- I'm not quite understanding why
14|x+7| and not 14|x+14|.
Do you not add the 7's because that is a property of absolute value?(2 votes)
- 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(3 votes)
We're asked to solve for x. Let me just rewrite this equation so that the absolute values really pop out. So this is 8 times the absolute value of x plus 7 plus 4-- in that same color-- is equal to negative 6 times the absolute value of x plus 7 plus 6. Now the key here-- at first it looks kind of daunting. It's this complex equation. You have these absolute values in it. But the way to think about this is if you could solve for the absolute value expression, you could then-- it then turns into a much simpler problem, then you can take it from there. So you could almost treat this expression-- the absolute value of x plus 7, you can just treat it as a variable, and then once you solve for that, it becomes a simpler absolute value problem. So let's try to do that. Let's try to solve for not x first. We're just going to solve for the absolute value of x plus 7. You'll see what I mean. So I want to get all of the absolute values of x plus 7 on the left-hand side, so I want to get rid of this one on the right-hand side. Easiest way to get rid of it is to add 6 times the absolute value of x plus 7 to the right-hand side. We can't, of course, only do that to the right-hand side. If these two things are equal and we are being told that they are, then if you add something on this side, the only way that the equality will hold is if you still do it on the left-hand side. So let's do that, so plus 6 times the absolute value of x plus 7. And I want to get all of these constant terms on to the right-hand side. So I want to get rid of this positive 4. Easiest way is to subtract 4 right over there, but if we do it on the left-hand side, we have to do it on the right-hand side as well. And so what does this get us? So our left-hand side, if I have 8 of something-- and in this case the something is absolute values of x plus 7's-- but if I have 8 of something and I add 6 of that same something, I now have 14 of that something. So that's going to be 14 absolute values of x plus 7, 14 times the absolute value of x plus 7. The 4 and the negative 4 cancel out, and that was intentional. The negative 6 and the 6 x plus 7's cancel out, or absolute values of x plus 7's cancel out, and that was intentional. And then we're left with 6 minus 4, which is just 2. So that's going to be equal to 2. Now just as promised, we want to solve for the absolute value of x plus 7, so let's divide both sides by 14 to get rid of that coefficient there, or that factor, or whatever you want to call it, the thing that's multiplying the absolute value of x plus 7. So we'll divide both sides by 14, and we are left with the absolute value of x plus 7 is equal to 2/14. They're both divisible by 2, so this is the same thing as 1/7. So just as promised, we've now solved for the absolute value of x plus 7, but we really need to solve for x. So how can we reason through this? So if I take the absolute value of something and I got you 1/7, there's two possible things that I took the absolute value of. I could have taken the absolute value of positive 1/7, or I could've taken the absolute value of negative 1/7. So this thing that we're taking the absolute value of-- so x plus 7-- could be equal to positive 1/7, or x plus 7 could be equal to negative 1/7. And just think about that for a second. If this thing right over here were equal to 1/7, you take its absolute value, it'd be 1/7. If this thing was negative 1/7, you take its absolute value, it would be positive 1/7. So that's how we got this. So now let's just solve for x. So if we subtract 7 from both sides for this left-hand equation, we get x is equal to 1/7 minus-- and 7 we can rewrite as 49/7, which is equal to negative 48/7. So that's one possibility for x. And then the other possibility we would get x is equal to-- so we have negative 1/7 minus 49/7. We're just subtracting 7 from both sides. That's what 49/7 is. And then this gets us to negative 50/7. So the two solutions to this what we thought was a complicated equation are negative 48/7 and negative 50/7.