a very TRICKY double absolute value equation

This equation has two absolute values. It seems like it will give use 4 answers at first but two of them are actually extraneous! So we must be careful when we solve this kinda double absolute value equation. Be sure to try the two practice problems at the end of the video.

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"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Free feel to leave your questions in the comment!
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#justalgebra

[awildscrub]

Case 1: x>=0
Case 2: -3/2<=x<0 however the answer we got is outside this domain
Case 3 is impossible so we don't need to consider that
Case 4: x<-3/2

[marcushendriksen]

By the definition of absolute value, |x| = x if x >= 0 and -x if x < 0; similarly, |2x+3| = 2x+3 if x >= -3/2 and -2x-3 if x < -3/2 (you can work that out by setting the inner expression to 0 and solving for x. Since it's a linear function with a positive slope, all x values to the left of the root will produce a negative value).

That gives 3 cases to consider: x being bigger than 0, x being smaller than -3/2 and x being between -3/2 and 0. For each case, use the appropriate rendition of each absolute value term and solve for x. You'll find that the third case gives you a nonsensical answer of x = 1, so it's really just two cases (x = 1 is a nonsensical answer because in the third case, x is constrained by -3/2 =< x < 0)

[kobethebeefinmathworld]

Suppose x>=0, then 2x + 3>=0 + 3= 3> 0, so in fact we just really need 3 cases.

[Si.top]

With absolue values, you need to study the sign of the functions involved :)

[CombustibleLmon]

This is so much easier just finding out where the 0's of each moduli and studying the case for intervals between these values

[bonnienesbitt]

Look at the intended audience of this channel. While most of the alternative methods proposed below are more efficient, will they work for the intended audience? I am a retired high school mathematics teacher. Getting started is the most difficult part for the inexperienced math student. The method used here: Set up all the cases and then check for extraneous answers works not only for this problem but for many other types. As the student builds mathematical knowledge and confidence, then some of the methods proposed below would be appropriate.

[ZipplyZane]

The way I did it was to find the critical points first--where the sign of x changes when taking the absolute value. Those are x=0 and 2x+3=0→x=-3/2. Then I checked the cases where x < -3/2, -3/2≤x<0, and x≥0.

Case 1 you must negate both values, and you get x = -7/3, which is within the domain. Case 2 you negate only |x|, and get x=1, which is not in the domain And in case 3, you negate none of them, and get x = 1/3, which is in the domain.

[robertlunderwood]

My method involves trying to figure out what f(x) = |x| + |2x+3| looks like. Fortunately, everything is linear. f(x) = 3x+3 for x >= 0, x+3 if -3/2 <= x < 0, and -3x-3 if x < -3/2. Setting each of these equal to 4 yields x = 1/3, x =1, and x = -7/3 respectively. x = 1 can be rejected for the middle portion as that's outside the domain, leaving us with the two solutions.

[shmuelzehavi]

If x ≥0 than: 2x+3 >0 and therefore Case 3 is logically impossible and should not be analyzed at first place. Besides, nice explanation.

[זאבגלברד]

To further elaborate on your explanation we have |2x+3| + |x| = 4 and we have the x< - 3/2 and we have till x=0 and we have x>0 . For each of these 3 "areas" we have an equation without the | | ....

[yoyoezzijr]

Case 3 is impossible. x < -3/2 and x > 0 at the same time is impossible. So, case 1, 2, 4 are the only cases, and in case 2 where x € [-3/2, 0] the x solution is outside the interval so we can reject that solution

[ExtinctTrain]

My thought for this was only solve for case 1 and 4 because x is in both parts of the absolute value, and one part of x doesn't make sense to be negative while the other part stays positive. This is what I mean. |-1(x+2x+3)| and |+1(x+2x+3)|. Just solve for that. I know I didn't provide it, but the original equation can be rewritten as |x+2x+3|. Then apply the same idea to the other side. |-1|4 and |+1|4. The other 2 cases don't make sense to spend time solving because they don't follow the rule (I forget the name). This is only tricky because the absolute value isn't on both sides of the equation. However, do not combine the like terms in the rewritten equation before distributing the +1 or -1. Doing so will come up with an incorrect answer, and original equation must be followed to verify and come up with a correct answer.

[michalchik]

It seems to me like the x value will always both have to be positive or both have to be negative since they represent the same X

[jeffthevomitguy]

Real big brains realize the second absolute value is around the plus sign. 🧠

[VictorGoddamnSullivan]

This video helped me the most, thank you!!!!

[realshadowtaka]

shortly x + 2x+3 = ±4
Would work in every scenario, right?

[Katiex]

thank you you are the only one that explains it well ❤

[JDC]

I'll start with the critical values where the absolute values are 0, which is x = 0, and 2x + 3 = 0, or x = -3/2

1. |x| - |2x + 3| = 4, critical values x = 0, -3/2. If x = 0, LHS = -3 != 4, if x = -3/2, LHS = 3/2 != 4, so the critical values are not solutions


x < -3/2
-x - (-2x - 3) = 4
-x + 2x + 3 = 4
x = 1, no as it is not less than -3/2

-3/2 < x < 0
-x - 2x - 3 = 4
-3x = 7
x = -7/3, no as it is not on interval above

x > 0

x - 2x - 3 = 4
-x = 7
x = -7, no as it is not greater than 0

No solutions at all.

2. |2x + 3| - |x| = 4, critical values x = 0, -3/2, If x = 0, LHS = 3 != 4, If x = -3/2, LHS = -3/2 != 4, so the critical values are not solutions


x < -3/2
-2x - 3 + x = 4
-x = 7,
x = -7, ok

-3/2 < x < 0
2x + 3 + x = 4
3x = 1
x = 1/3, no as it does not lie in above interval

x > 0
2x + 3 - x = 4
x = 1, ok
x = -7, 1 are the solutions

[cornerstone]

Use graph and critical values . Much more easier . Idk why no body is talking about graphical method

[SUPERDUPERPETS]

Dang the way u made it look so simple 😅☺️