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Solve |x| = 5

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Solve |x| = 5

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Solve |x| = 5 x = 5

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Solve |x| = 5 x = 5

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Solve |x| = 5 x = 5 or x = –5

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Solve |x| = 5 |x| = a

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Solve |x| = 5 |x| = a x = a “or” x = –a

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Solve |x| > 5

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Solve |x| > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5

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Solve |x| > 5 x > 5 or x < – 5

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Solve |x| > 5 x > 5 or x < – 5

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Solve |x| > 5 x > 5 or x < – 5

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Solve |x| > 5 |x| > a

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Solve |x| > 5 |x| > a x > a “or” x < –a

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Solve |x| 5

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Solve |x|

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Solve |x| x 5

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Solve |x| x 5

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Solve |x| x 5 and x –5

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Solve |x| x 5 and x –5

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Solve |x| x 5 and x –5

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Solve |x| x –5 and x 5

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Solve |x| x –5 and x 5

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Solve |x| x –5 and x 5 –5 x 5

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Solve |x| |x| < a

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Solve |x| |x| < a x –a “and” x a

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|ax + b| = c

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ax + b = c or ax + b = –c

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|ax + b| = c ax + b = c or ax + b = –c |ax + b| > c

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|ax + b| = c ax + b = c or ax + b = –c |ax + b| > c ax + b > c or ax + b < –c

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|ax + b| = c ax + b = c or ax + b = –c |ax + b| > c ax + b > c or ax + b < –c |ax + b| < c

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|ax + b| = c ax + b = c or ax + b = –c |ax + b| > c ax + b > c or ax + b < –c |ax + b| < c ax + b –c

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Solve |x – 4| = 2

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x – 4 = 2 or x – 4 = –2

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Solve |x – 4| = 2 x – 4 = 2 or x – 4 = –2 x = 6 or x = 2

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Solve |6x + 3| = 15

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6x + 3 = 15 or 6x + 3 = –15 6x = 12 or 6x = –18 x = 2 or x = – 3

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Solve |5 – 4x| + 3 = 4

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|5 – 4x| = 1

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Solve |5 – 4x| + 3 = 4 |5 – 4x| = 1 5 – 4x = 1 or 5 – 4x = –1

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Solve |5 – 4x| + 3 = 4 |5 – 4x| = 1 5 – 4x = 1 or 5 – 4x = –1 –4x = –4 or –4x = –6

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Solve |5 – 4x| + 3 = 4 |5 – 4x| = 1 5 – 4x = 1 or 5 – 4x = –1 –4x = –4 or –4x = –6 x = 1 or x =

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Solve |2x + 5| 9

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2x + 5 9 and 2x + 5 –9

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Solve |2x + 5| 9 2x + 5 9 and 2x + 5 –9 2x 4 and 2x –14

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Solve |2x + 5| 9 2x + 5 9 and 2x + 5 –9 2x 4 and 2x –14 x 2 and x –7

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Solve |2x + 5| 9 2x + 5 9 and 2x + 5 –9 2x 4 and 2x –14 x 2 and x –7

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Solve |2x + 5| 9 2x + 5 9 and 2x + 5 –9 2x 4 and 2x –14 x 2 and x –7

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Solve |4x – 3| + 7 > 20

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|4x – 3| > 13

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Solve |4x – 3| + 7 > 20 |4x – 3| > 13 4x – 3 > 13 or 4x – 3 < –13

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Solve |4x – 3| + 7 > 20 |4x – 3| > 13 4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10

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Solve |4x – 3| + 7 > 20 |4x – 3| > 13 4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10 x > 4 or x < –

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Solve |4x – 3| + 7 > 20 |4x – 3| > 13 4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10 x > 4 or x < –

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Solve |4x – 3| + 7 > 20 |4x – 3| > 13 4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10 x > 4 or x < –

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A soda manufacturer has a tolerance of 0.25 ounce for a can of soda that is supposed to weigh 12 ounces. Write and solve an absolute value inequality that describes the acceptable weights for a “12 ounce” soda.

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|actual weight – ideal weight| tolerance

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A soda manufacturer has a tolerance of 0.25 ounce for a can of soda that is supposed to weigh 12 ounces. Write and solve an absolute value inequality that describes the acceptable weights for a “12 ounce” soda. |actual weight – ideal weight| tolerance |x – 12| 0.25

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A soda manufacturer has a tolerance of 0.25 ounce for a can of soda that is supposed to weigh 12 ounces. Write and solve an absolute value inequality that describes the acceptable weights for a “12 ounce” soda. |actual weight – ideal weight| tolerance |x – 12| 0.25 x – 12 0.25 and x – 12 –0.25

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A soda manufacturer has a tolerance of 0.25 ounce for a can of soda that is supposed to weigh 12 ounces. Write and solve an absolute value inequality that describes the acceptable weights for a “12 ounce” soda. |actual weight – ideal weight| tolerance |x – 12| 0.25 x – 12 0.25 and x – 12 –0.25 x and x 11.75

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A soda manufacturer has a tolerance of 0.25 ounce for a can of soda that is supposed to weigh 12 ounces. Write and solve an absolute value inequality that describes the acceptable weights for a “12 ounce” soda. |actual weight – ideal weight| tolerance |x – 12| 0.25 x – 12 0.25 and x – 12 –0.25 x and x 11.75

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] Is x = –4 a solution to |3x + 8| = 20? 2] |5x – 4| = 16 3] |10 + 3x| – 1 17 4] Solve & graph the solution on the # line: |4x + 11| 23

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] Is x = –4 a solution to |3x + 8| = 20? 2] |5x – 4| = 16 3] |10 + 3x| – 1 17 4] Solve & graph the solution on the # line: |4x + 11| 23 x = 4 or x = – NO ; 4 20 x – x or x –

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