|1-4k| ≥ -11

Solve

This is a trick question.

|u| is always positive, so

|1-4k| ≥ -11

for any value of k.

Damon's solution is correct, but it should read

k >= -2.5 OR k <= 3

So, any value of k will work.

To solve the absolute value inequality |1-4k| ≥ -11, we can use the definition of absolute value.

According to the definition, |a| is equal to a if a is greater than or equal to 0, and |a| is equal to -a if a is less than 0.

Considering the given inequality, we have two possibilities:

Case 1: 1-4k ≥ -11

If 1-4k ≥ -11, we can solve for k as follows:

1-4k ≥ -11
Add 4k to both sides:
1-4k+4k ≥ -11+4k
1 ≥ -11+4k
Add 11 to both sides:
1+11 ≥ -11+4k+11
12 ≥ 4k
Divide both sides by 4:
12/4 ≥ 4k/4
3 ≥ k

So, in this case, k can be any number less than or equal to 3.

Case 2: -(1-4k) ≥ -11

If -(1-4k) ≥ -11, we can solve for k as follows:

-(1-4k) ≥ -11
Multiply both sides by -1 (which reverses the inequality):
1-4k ≤ 11
Subtract 1 from both sides:
1-1-4k ≤ 11-1
-4k ≤ 10
Divide both sides by -4 (which reverses the inequality):
k ≥ 10/-4
k ≥ -5/2

So, in this case, k can be any number greater than or equal to -5/2.

Combining both cases, we find that k can be any number less than or equal to 3 or any number greater than or equal to -5/2.

To solve the inequality |1-4k| ≥ -11, we need to consider two cases:

Case 1: 1-4k is positive or zero.
In this case, the absolute value expression |1-4k| simplifies to 1-4k. Therefore, we can rewrite the inequality as 1-4k ≥ -11. To isolate k, we can subtract 1 from both sides: 1-4k-1 ≥ -11-1, which simplifies to -4k ≥ -12. Dividing both sides by -4, we get k ≤ 3.

Case 2: 1-4k is negative.
In this case, the absolute value expression |1-4k| simplifies to -(1-4k), which is equivalent to -1+4k. Therefore, we can rewrite the inequality as -1+4k ≥ -11. To isolate k, we can add 1 to both sides: -1+4k+1 ≥ -11+1, which simplifies to 4k ≥ -10. Dividing both sides by 4, we get k ≥ -2.5.

Combining the results from both cases, we have k ≤ 3 and k ≥ -2.5. To express the solution in a single interval notation, we can write it as:
[-2.5, 3]

1-4k >= -11 and -1+4k >=-11

-4k >= -12 _and ___4k>=-10
4 k <= 12 __and_____k>=-2.5
k <= 3
note
arrow
reverse
when
signs
changed

so in the end between -3and +3
-2.5 <= k <= 3