The modulus of a linear inequality refers to the inequality that results when the absolute value (or modulus) of a linear expression is involved. It essentially deals with the distance of that linear expression from zero.
Here's a breakdown to clarify further:
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Absolute Value/Modulus: The absolute value of a number x, denoted as |x|, is its distance from zero on the number line. It's always non-negative. For example, |3| = 3 and |-3| = 3.
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Linear Expression: A linear expression is an algebraic expression where the highest power of the variable is 1. Examples include 2x + 1, x - 5, and -3x.
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Linear Inequality: A linear inequality is a mathematical statement that compares two linear expressions using inequality symbols such as < (less than), > (greater than), ≤ (less than or equal to), or ≥ (greater than or equal to).
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Modulus of a Linear Inequality: This combines the concepts. It's an inequality where a linear expression is within absolute value bars. A general form is |ax + b| < c, |ax + b| > c, |ax + b| ≤ c, or |ax + b| ≥ c, where a, b, and c are constants, and x is the variable.
How to Solve Modulus Inequalities:
Solving modulus inequalities involves splitting them into two separate inequalities because the expression inside the absolute value can be either positive or negative.
Example:
Consider the inequality |x - 2| < 3.
This means the distance of (x - 2) from zero is less than 3. We can split this into two cases:
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x - 2 < 3: This represents the case where (x - 2) is positive or zero. Solving this gives x < 5.
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-(x - 2) < 3: This represents the case where (x - 2) is negative. Solving this gives -x + 2 < 3, which simplifies to -x < 1, and further to x > -1.
Therefore, the solution to |x - 2| < 3 is -1 < x < 5.
Another Example:
Consider the inequality |2x + 1| ≥ 5.
This means the distance of (2x + 1) from zero is greater than or equal to 5. We split this into two cases:
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2x + 1 ≥ 5: Solving this gives 2x ≥ 4, which simplifies to x ≥ 2.
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-(2x + 1) ≥ 5: Solving this gives -2x - 1 ≥ 5, which simplifies to -2x ≥ 6, and further to x ≤ -3.
Therefore, the solution to |2x + 1| ≥ 5 is x ≤ -3 or x ≥ 2.
In essence, the modulus of a linear inequality transforms a single inequality into a compound inequality (either a conjunction "and" or a disjunction "or"), reflecting the two possible signs of the expression inside the absolute value.
