Find the sum of all real numbers x such that (absolute value(x – 2009)) + (absolute value(x – 2010)) = 3.
Given:
(absolute value(x – 2009)) + (absolute value(x – 2010)) = 3
or
|x-2009|+|x-2010|=3
There are 4 cases:
1. x≥2009 and x≥2010 => x≥2010
2. x≤2009 and x≤2010 => x≤2009
3. x≥2009 and x≤2010 => 2009≤x≤2010
4. x≤2009 and x≥2010
Case 4 is clearly not possible, so it will be rejected.
Case 1: x≥2010
x-2010+x-2009=3
2x=3+2010+2009
x=2022/2=2011
Case 2: x≤2009
2009-x+2010-x=3
2x=2009+2010-3
x=2016/2=2008
Case 3:2009≤x≤2010
2010-x+x-2009=3
2010-2009=3 => a contradiction
So there is no solution whereby
2009≤x≤2010
Can you take it from here?
umm no?
u lost me
what was the answer u got??
There are two possible answers that have been calculated and posted. If you reread the post carefully, you will find where they are.
In fact, it is even more important to understand what was written, and the logic behind the calculations.
If you do not follow the logic, make sure you post what you do not understand. I do not want you to sit at the exam hall and stare at your teacher.
I just don't understand the math you're doing
im in 11th grade taking pre-calc and i don't think we've ever done anything quite like that. Thank you for your help
To find the sum of all real numbers x that satisfy the given equation, we need to solve the equation. Let's break down the equation step by step:
1. Start with the equation: |x – 2009| + |x – 2010| = 3
2. Remove the absolute value signs by considering both positive and negative cases:
a) For x - 2009 ≥ 0 and x - 2010 ≥ 0:
In this case, the equation becomes: (x - 2009) + (x - 2010) = 3
Simplifying, we get: 2x - 4019 = 3
Solving for x, we have: 2x = 4022
x = 2011
b) For x - 2009 ≥ 0 and x - 2010 < 0:
In this case, the equation becomes: (x - 2009) - (x - 2010) = 3
Simplifying, we get: x - 2009 - x + 2010 = 3
Combining like terms, we have: 1 = 3
This equation has no solution.
c) For x - 2009 < 0 and x - 2010 ≥ 0:
In this case, the equation becomes: -(x - 2009) + (x - 2010) = 3
Simplifying, we get: -x + 2009 + x - 2010 = 3
Combining like terms, we have: -1 = 3
This equation has no solution.
d) For x - 2009 < 0 and x - 2010 < 0:
In this case, the equation becomes: -(x - 2009) - (x - 2010) = 3
Simplifying, we get: -x + 2009 - x + 2010 = 3
Combining like terms, we have: 4029 = 3
This equation has no solution.
3. From the above solutions, we found that x = 2011 is the only solution that satisfies the equation.
Therefore, the sum of all real numbers x that satisfy the given equation is 2011.