考虑边界条件： x-1 >=0 (x>=1) 2x+1 >= 0 (x>=-1/2) 所以， if x < -1/2 then |x-1| + |2x+1| 变成 -（x-1) - (2x + 1) if -1/2 <= x < 1 |x-1| + |2x+1| 变成 -（x-1) + (2x + 1) if x >= 1 |x-1| + |2x+1| 变成 （x-1) + (2x + 1)
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No offense. In general, we break the | | sign case by case... For example, Assume x-1>=0, 2x+1>=0, then we can break it to x-1+2x+1=12 and get x=4, which satisfies x-1>=0 and 2x+1>=0.. So x=4 is one solution. Assume x-1>=0 and 2x+1<0, then we can break it into x-1-(2x+1)=12 and get x=-14, which does not satisfies the two conditions. Basically, we break the equation that contains the | | sign into four groups of equations and inequality.... Two of four have solutions and the other two do not.... |
Thank both of you for your help. In general, how many solutions exist for this type of equation? |ax+b|+|cx+d|+...+|mx+n|=constant. (It is better to use a1x..., but I cannot write 1 as subscript) If the answer is bigger than 2, can you show me an example? |
solve each ax+b = 0, and get a sequece of numbers x1 |
It is easy to find a case of no solution, such as |x|=-1 or |x-1|+|x|=0. Could show an example for infinitely many solution and 3 solutions Thank you a lot. It seems so easy for you. But I just could not find them. |
Infinitely many: |x+1|+|x-1|=2. Obviously, [-1, 1] is the solution set. Three solutions: Well, I do not have one example. Actually I do not think we can have one. Cause f(x)=|ax+b|+|cx+d|+...+|mx+n| is convex..... Therefore, if there are only a finite number of solutions, then there are 2, 1, or zero solutions. Need to check again though. |
it can only be 0, 1, 2, or infinite. And congratulations! |
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