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Solve |ax + b| < c or |ax + b| > c with automatic compound inequality conversion
Inequality: |1x + 3| < 5
Absolute value inequalities involve the absolute value of an expression compared to a number. The key is understanding how they convert to compound inequalities.
Absolute value measures distance from zero. |x| < 5 means x is within 5 units of zero (between -5 and 5). |x| > 5 means x is more than 5 units from zero (less than -5 or greater than 5).
Because x must be within c units of zero, meaning it's between -c and c. Both conditions (greater than -c AND less than c) must be true simultaneously.
Because x must be more than c units from zero, which happens in two separate regions: either far to the left (less than -c) OR far to the right (greater than c).
Absolute values are always non-negative. So |x| < -3 has no solution (absolute value can't be negative), while |x| > -3 is always true (all real numbers).
Convert to -5 < 2x + 3 < 5, then solve the compound inequality by subtracting 3 from all parts and dividing by 2.
No, they have either no solution, infinitely many solutions (a range), or all real numbers. They never have a single solution because absolute value creates ranges, not points.
≤ includes the boundary (closed circle/bracket), while < excludes it (open circle/parenthesis). |x| ≤ 5 means -5 ≤ x ≤ 5, while |x| < 5 means -5 < x < 5.
Pick test values from each region (inside and outside your solution) and substitute into the original inequality. Values in your solution should make it true.
Yes, but you must first isolate the absolute value expression on one side before converting to a compound inequality.