Solve the following absolute value inequalities:
|x+1|<=|x-3|
I promise I'm not posting my whole homework assignment. These are questions that I skipped and have no clue how to do out of a huge packet.
The key to absolute value problems is to remember the definition of |n|.
|n| = n if n >= 0
|n| = -n if n < 0
So, here we have
|x+1|<=|x-3|
if (x-3) >= 0, |x-3| = x-3
Also, since that means x > 3, x+1 >= 0, so |x+1| = x+1 and we have
x+1 <= x-3
1 < -3
not possible
If (x-3) < 0, (that is, x<3) we have
|x+1| <= -(x-3)
Now, if x >= -1, x+1 >= 0, and we have
x+1 <= -x+3
2x <= 2
x <= 1
But, we required above that x >= -1, so we get a solution set -1 <= x <= 1
If (x-3) < 0 and (x+1) < 0 we have
-(x+1) <= -(x-3)
-x-1 <= -x+3
-1 < 3
So x < -1 is also a solution.
Combining the solution sets, we see that
x <= 1 is the complete solution set
To see how this works graphically, visit
http://rechneronline.de/function-graphs/
and enter
abs(x+1)
abs(x-3)
as your two functions, and set the domain for x from -5 to 5, and the range y from 0 to 5
This is an excellent place to try out viewing function graphs to confirm your algebra.
No problem! I can help you solve the given absolute value inequality: |x + 1| <= |x - 3|.
To solve it, we'll need to consider different cases based on the signs of the expressions inside the absolute value bars. There are three possibilities:
Case 1: x + 1 and x - 3 are both positive.
When both expressions inside the absolute value bars are positive, the original inequality can be written without the absolute value signs:
x + 1 <= x - 3
To solve this equation, we can simplify it by subtracting x from both sides:
1 <= -3
However, this is not possible since 1 is not less than or equal to -3. Therefore, there are no solutions for this case.
Case 2: x + 1 is positive, while x - 3 is negative.
In this case, the original inequality becomes:
x + 1 <= -(x - 3)
To solve this, we need to distribute the negative sign inside the parenthesis:
x + 1 <= -x + 3
Next, we can add x to both sides:
2x + 1 <= 3
Now, subtract 1 from both sides:
2x <= 2
Divide both sides by 2:
x <= 1
So, any value of x that is less than or equal to 1 is a solution for this case.
Case 3: x + 1 is negative, while x - 3 is positive.
In this case, the original inequality becomes:
-(x + 1) <= x - 3
To solve this, we need to distribute the negative sign:
-x - 1 <= x - 3
Next, add x to both sides:
-1 <= 2x - 3
Now, add 3 to both sides:
2 <= 2x
Divide both sides by 2:
1 <= x
So, any value of x that is greater than or equal to 1 is a solution for this case.
Combining the solutions from both cases, we have:
x <= 1 or x >= 1
Therefore, the solution to the absolute value inequality |x + 1| <= |x - 3| is x <= 1 or x >= 1.