Help with complex inequalities

Question

Hello I am confused about how to open modulus of complex abs value inequalities. The below example is provided in the GMAT CLUB Math Book, but I would love an explanation of how they solve for each possible value of X. To provide additional clarity, I am referring to how you figure out how to setup the equation under each condition.

|x+3|-|4-x|=|8+x|

4 conditions:
1) x x=-1 (reject solution)
2) -8 x=-15 (reject solution)
3) -3 x=9 (reject solution)
4) x>=4, (x+3)-(4-x)=(8+x)--> x=-1

Answer = 0

|x+3|-|4-x|=|8+x|

4 conditions:
1) x x=-1 (reject solution)
2) -8 x=-15 (reject solution)
3) -3 x=9 (reject solution)
4) x>=4, (x+3)-(4-x)=(8+x)--> x=-1

Answer = 0

AlexanderSupertramp · Answer

Well i am not very clear as to what was the original question asked in your book, but basically the absolute value function behaves like this with respect to positive and negative values:

|x-a| = x-a if the value of x-a is positive.
|x-a| = -(x-a) if the value of x-a is negative.

Lets see this theory for condition 1 from above :

|x+3|-|4-x|=|8+x|

Condition 1 : x < -8.

Now lets say x = -9 which is less than -8, then you know that x+3 will have a negative value -6.4-x will have a positive value 13 and 8+x will have a negative value -1.

So applying the rules i mentioned above to this equation we get: -(x+3) -(4-x) = -(8 +x) => x = -1.

You can try the same for the other options.

Hope this helps.

Bunuel · Answer

This question is discussed here: x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html Hope it helps. In case of any further questions please post in that thread.

Help with complex inequalities

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