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1.5 “absolute value equations & inequalities”. ex 1:. Solve |x| = 5. ex 1:. Solve |x| = 5. -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10. ex 1:. Solve |x| = 5. - PowerPoint PPT Presentation
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
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |x| = 5
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
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
x = 5 or 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
|x| = a
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |x| = 5
|x| = a
x = a “or” x = –a
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
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |x| > 5
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
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
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
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
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
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
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
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
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
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
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
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
x > 5 or 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
x > 5 or 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
x > 5 or 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
|x| > a
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |x| > 5
|x| > a
x > a “or” x < –a
Solve |x| 5
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
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x 5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x 5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x 5 and x –5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x 5 and x –5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x 5 and x –5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x –5 and x 5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x –5 and x 5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
x –5 and x 5
–5 x 5
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
|x| < a
Solve |x| 5
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
|x| < a
x > –a “and” x < a
|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|
> cax + b > c or ax + b < –c
|ax + b| = c
ax + b = c or ax + b = –c |ax + b|
> cax + b > c or ax + b < –c
|ax + b| < c
|ax + b| = c
ax + b = c or ax + b = –c |ax + b|
> cax + b > c or ax + b < –c
|ax + b| < c
ax + b < c and ax + b > –c
Solve |x – 4| = 2
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
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
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 = 3
2
Solve |2x + 5| 9
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
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |2x + 5| 9
2x + 5 9 and 2x + 5 –9 2x 4 and 2x –14 x 2 and x –7
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
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
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 < 52
–
Solve |4x – 3| + 7 > 20
|4x – 3| > 13
4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10x > 4 or x < 5
2–
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Solve |4x – 3| + 7 > 20
|4x – 3| > 13
4x – 3 > 13 or 4x – 3 < –13 4x > 16 or 4x < –10x > 4 or x < 5
2–
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
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.
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 –0.25x 12.25 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 –0.25x 12.25 and x 11.75
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
1] 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
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
1] 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 =
125
–
NO ; 4 20
x < 3 and x >
172
–
x or x
283
–83
