How to explain?

y · 发表于 2005-2-21 20:40:19

I try to teach my daughter to solve some equations with absolute value signs, such as

|x-1| + |2x+1| = 12

We can break it into x-1+2x+1=12 and -(x-1)-(2x+1)=12.

How can I explain to her that she cannot use

x-1-(2x+1)=12 or -(x-1)+2x+1=12 to get the answers?

新用户 · 发表于 2005-2-21 21:07:43

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)

原贴：

文章来源: y 于 2005-2-21 12:40:19
标题：How to explain?

I try to teach my daughter to solve some equations with absolute value signs, such as

|x-1| + |2x+1| = 12

We can break it into x-1+2x+1=12 and -(x-1)-(2x+1)=12.

How can I explain to her that she cannot use

x-1-(2x+1)=12 or -(x-1)+2x+1=12 to get the answers?

Sorry · 发表于 2005-2-21 21:08:07

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....

y · 发表于 2005-2-22 23:18:06

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?

fzy · 发表于 2005-2-22 23:31:07

solve each ax+b = 0, and get a sequece of numbers x1

乱弹 · 发表于 2005-2-23 01:13:58

Or even infinitely many.

乱弹 · 发表于 2005-2-23 02:13:59

Or no solution

y · 发表于 2005-2-23 03:44:02

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.

乱弹 · 发表于 2005-2-23 05:00:20

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.

fzy · 发表于 2005-2-23 20:18:04

it can only be 0, 1, 2, or infinite. And congratulations!

		自动登录	找回密码
密码			立即注册

动态微博

How to explain?

回复：How to explain?

Your method seems wrong.

Here is a general question

回复：Here is a general question

Or even infinitely many. [:D)]

Or no solution

No solution

Infinitely many and three solutions

you are right