Solving Inequalities with Two Absolute Values by Testing Points
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- Опубліковано 5 жов 2024
- This may not be an SAT/ACT inequalities question, but it's more difficult than those seen on SAT or ACT. So, if you understand the two techniques demonstrated in this video about solving inequalities, you will become a lot better on your SAT or ACT preparation!
We are solving the inequality |x+3|+|x-1| less than or equal to 8 by breaking into cases and testing points.
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why do we need to draw the number line if u already got x=3,=5? cant u just automatically make the link to [-5,3]
Because the solution could been (-inf, -5] or [3, inf), so we need to test points and check.
How would you do it if it were subtraction of absolutes instead of addition
when doing the different cases, why does it have to be an equation (=)? is it not possible to retain the signs and solve it like a regular inequality?
If we have 0 instead of 8 and it is just |x+3| + |x-1|
Correct! In that case, we only consider the equation. And no need to test points since |x+3| and |x-1| must be both zero at the same time in order for the equation to be true. Can we get a solution then?
If on R.H.S has > 8 then what will be the cases to solve the question
Good question! The process is similar, but we want the sum of the absolute values to be greater than 8.
in VIET NAM , we have x+3+x-1 = 8 and -8 . that enought
Ok!
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try and solve something like |x+1|>|x+3|
If we move |x+3| to the left side, it is similar to the question in this video.
does it always work?
Yeah, for this type.
Nice
Thank you!
Number line before -5 y isn't it in solution
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