Solve |x| = 5 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 Solve |x| = 5.

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Presentation transcript:

Solve |x| = 5

Solve |x| = 5

Solve |x| = 5 x = 5

Solve |x| = 5 x = 5

Solve |x| = 5 x = 5 or x = –5

Solve |x| = 5 |x| = a

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

Solve |x| > 5

Solve |x| > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5

Solve |x| > 5 x > 5 or x < – 5

Solve |x| > 5 x > 5 or x < – 5

Solve |x| > 5 x > 5 or x < – 5

Solve |x| > 5 |x| > a

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

Solve |x|  5

Solve |x| 

Solve |x|  x  5

Solve |x|  x  5

Solve |x|  x  5 and x  –5

Solve |x|  x  5 and x  –5

Solve |x|  x  5 and x  –5

Solve |x|  x  –5 and x  5

Solve |x|  x  –5 and x  5

Solve |x|  x  –5 and x  5 –5  x  5

Solve |x|  |x| < a

Solve |x|  |x| < a x  –a “and” x  a

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

|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

Solve |x – 4| = 2

x – 4 = 2 or x – 4 = –2

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

Solve |6x + 3| = 15

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

Solve |5 – 4x| + 3 = 4

|5 – 4x| = 1

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

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

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

Solve |2x + 5|  9

2x + 5  9 and 2x + 5  –9

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

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

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

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

Solve |4x – 3| + 7 > 20

|4x – 3| > 13

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

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

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

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

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

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

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

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

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

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

] 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

] 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  –