Creating an absolute value equation from an inequallity

Click For Summary
SUMMARY

The discussion focuses on converting the inequality $$-6 \leq x \leq 14$$ into an absolute value equation. The key transformation involves recognizing that $$|y| \leq c$$ can be expressed as $$-c \leq y \leq c$$. By subtracting 4 from the original inequality, the equation $$|x-4| \leq 10$$ is derived, confirming that the absolute value representation accurately reflects the original range of x. Additionally, the midpoint and radius method is highlighted as a systematic approach for such conversions.

PREREQUISITES
  • Understanding of absolute value equations
  • Familiarity with inequalities
  • Basic algebraic manipulation skills
  • Knowledge of midpoint and radius concepts in mathematics
NEXT STEPS
  • Study the properties of absolute value functions
  • Learn how to convert different types of inequalities into absolute value equations
  • Explore the concept of midpoints and radii in mathematical contexts
  • Practice solving absolute value inequalities with various examples
USEFUL FOR

Students, educators, and anyone interested in mastering the conversion of inequalities to absolute value equations in algebra.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
if you are given $$-6 \leq x \leq 14$$

from this how do you create an abs equation like $$|x-4| \leq 10$$

k
 
Mathematics news on Phys.org
Re: creating an abs equation

karush said:
if you are given $$-6 \leq x \leq 14$$

from this how do you create an abs equation like $$|x-4| \leq 10$$

k

Note that we can express the absolute value in terms of an inequality: $|y|\leq c \iff -c \leq y \leq c$.

Now, note that if we subtract 4 from each piece of $-6\leq x\leq 14$, we get $-10\leq x-4 \leq 10$. Thus, by what I mentioned in the first line, this means that $|x-4|\leq 10$.

Does this clarify things?
 
Re: creating an abs equation

karush said:
if you are given $$-6 \leq x \leq 14$$
from this how do you create an abs equation like $$|x-4| \leq 10$$

a \le x \le b converts to \left| {x - \frac{{a + b}}{2}} \right| \le \frac{{b - a}}{2}.

Think of the mid-point of [a,b] as well as the radius.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K