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Solve absolute Value Inequalities

Solve absolute Value Inequalities. Isolate the absolute value on one side of the equation. Write the problem as a compound inequality. LESS THAN: If absolute value is <a, then write the problem as a compound inequality -a< || < a GREATER THAN:

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Solve absolute Value Inequalities

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  1. Solve absolute Value Inequalities Isolate the absolute value on one side of the equation. Write the problem as a compound inequality. LESS THAN: If absolute value is <a, then write the problem as a compound inequality -a< || < a GREATER THAN: If absolute value is > a, then write the problem as 2 inequality problems: || < -a and || > a Solve for x. Simplify. Check it in the ORIGINAL Equation!! Note : When you have an “x” on both sides of the “=“, you may get an extraneous solution.

  2. Absolute Value Inequalities Examples |2x-3|<5 -5 < 2x-3 <5 -2 < 2x <8 -1 < x < 4 Write your answer using proper interval notation and graph It on a number line. (-1,4)

  3. Absolute Value Inequalities Example #2 |2x-1| >5 |2x-1| < -5 2x < -4 X < -2 | 2x-1| > 5 2x -1 > 5 2x > 6 X > 3 Write your answer using proper interval notation and graph it on a number line. (-∞, -2) U ( 3, ∞)

  4. Absolute value inequality Word Problems A manufacturer has a .6 oz tolerance for a bottle of salad dressing advertised as for 16 oz. Write and solve an absolute value inequality that describes the acceptable volumes for “16 oz” bottles HINT: If you want WITHIN the range use <, If you want OUTSIDE the range use > . | actual amount – ideal amount| < tolerance | x -16|< .6 -.6< x-16< .6 15.4 < X < 16.6 This means that you could actually buy a 15.4 to 16.6 oz salad dressing that is labeled and sold as 16 oz.

  5. Absolute Value inequality Word Problem A city ordinance states that pools must be enclosed by a fence that is between 3 to 6 feet high. Write an absolute value inequality describing fences that don’t meet this ordinance. How do you find the tolerance? Tolerance = (high end) – (average of low and high ends) | actual height – average height| > tolerance | h – [(3+6)/2] > 6-4.5 |h – 4.5| > 1.5

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