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Finding Absolute Value Inequality Statements

From mathematics, 3 months ago

Finding Absolute Value Inequality Statements

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Slide 1: Finding absolute value inequality statements Copyright ©2007-2008 Home Tuition Services

Slide 2: We can show that if a≥0, −a ≤ x ≤ a ⇔ ∣ x ∣ ≤ a x ≤ −a or x ≥ a ⇔ ∣ x ∣ ≥ a .

Slide 3: Example 1 Find the absolute value inequality statement corresponding to x ≥ −3 or x ≤ 3 . ⇒ x ≥ −3 or x ≤ 3 ⇒ −3 ≤ x ≤ 3 We write in standard≤ form. ⇒ ∣ x ∣≤ 3 The statement is ∣ x ∣ ≤ 3

Slide 4: Example 2 Find the absolute value inequality statement corresponding to x ≤ 3 or x ≥ −3 . This is the same as Example 1 with the inequality statements in reverse order. Don't confuse this with statements x ≤ −3 or x ≥ 3 that lead to a standard ≥ absolute value inequality statement.

Slide 5: ⇒ x ≤ 3 or x ≥ −3 ⇒ −3 ≤ x ≤ 3 We write in standard ≤ form. ⇒ ∣ x ∣≤ 3 The statement is ∣ x ∣ ≤ 3

Slide 6: Example 3 Find the absolute value inequality statement corresponding to x  −5 or x  5 . ⇒ x  −5 or x  5 Standard  form. ⇒ ∣ x ∣ 5 The statement is ∣ x ∣  5

Slide 7: Example 4 Find the absolute value inequality statement corresponding to −1  x  6 .

Slide 8: ⇒ −1  x  6 5 5 5 ⇒ −1−  x−  6− 2 2 2 7 5 7 ⇒ −  x−  2 2 2 ∣ ∣ 5 We note the inequality is in a form that 7 ⇒ x−  2 we2can convert to standard  form.

Slide 9: ⇒ −1  x  6 Average −1 and 6 . 5 5 5 ⇒ −1−  x−  6− −16 5 = 2 2 2 2 2 7 5 7 ⇒ −  x−  2 2 2 ∣ ∣ 5 7 ⇒ x−  2 2 5 Subtract from all three sides. 2

Slide 10: ⇒ −1  x  6 5 5 5 ⇒ −1−  x−  6− 2 2 2 7 5 7 ⇒ −  x−  2 2 2 Standard  form. ∣ ∣ 5 7 ⇒ x−  2 2 ∣ ∣ 5 7 The statement is x− 2 2

Slide 11: Example 5 Find the absolute value inequality statement corresponding to −5  x  3 .

Slide 12: ⇒ −5  x  3 ⇒ −5−−1  x−−1  3−−1 ⇒ −4  x1  4 ⇒ ∣ x1 ∣  4 We note the inequality is in a form that we can convert to standard  form.

Slide 13: ⇒ −5  x  3 Average −5 and 3 . ⇒ −5−−1  x−−1  3−−1 −53 = −1 ⇒ −4  x1  4 2 ⇒ ∣ x1 ∣  4 Subtract −1 from all three sides.

Slide 14: ⇒ −5  x  3 ⇒ −5−−1  x−−1  3−−1 ⇒ −4  x1  4 Standard  form. ⇒ ∣ x1 ∣  4 The statement is ∣ x1 ∣  4

Slide 15: Example 6 Find the absolute value inequality statement corresponding to x  3 or x  5 .

Slide 16: ⇒ x  3 or x  5 ⇒ x−4  3−4 or x−4  5−4 ⇒ x−4  −1 or x−4  1 ⇒ ∣ x−4 ∣  1 We note the inequality is in a form that we can convert to standard  form.

Slide 17: ⇒ x  3 or x  5 Average 3 and 5 . ⇒ x−4  3−4 or x−4  5−4 35 =4 ⇒ x−4  −1 or x−4  1 2 ⇒ ∣ x−4 ∣  1 Subtract 4 from all sides.

Slide 18: ⇒ x  3 or x  5 ⇒ x−4  3−4 or x−4  5−4 ⇒ x−4  −1 or x−4  1 Standard  form. ⇒ ∣ x−4 ∣  1 The statement is ∣ x−4 ∣  1

Slide 19: Example 7 Find the absolute value inequality statement with integer coefficients corresponding to −1  x  6 .

Slide 20: ⇒ −1  x  6 5 5 5 ⇒ −1−  x−  6− 2 2 2 7 5 7 ⇒ −  x−  2 2 2 ∣ ∣ 5 We note the inequality is in a form that 7 ⇒ x−  2 we2can convert to standard  form. ∣ ∣ 5 7 ⇒ 2× x−  2× 2 2 ∣  ∣5 ⇒ 2 x−  7 2 ⇒ ∣ 2 x−5 ∣  7

Slide 21: ⇒ −1  x  6 Average −1 and 6 . 5 5 5 ⇒ −1−  x−  6− −16 5 = 2 2 2 2 2 7 5 7 ⇒ −  x−  2 2 2 ∣ ∣ 5 7 ⇒ x−  2 2 5 Subtract 7 from all three sides. ∣ ∣ 5 2× x−  2×2 ⇒ 2 2 ∣  ∣5 ⇒ 2 x−  7 2 ⇒ ∣ 2 x−5 ∣  7

Slide 22: ⇒ −1  x  6 5 5 5 ⇒ −1−  x−  6− 2 2 2 7 5 7 ⇒ −  x−  2 2 2 Standard  form. ∣ ∣ 5 7 ⇒ x−  2 2 ∣ ∣ Multiply both sides by 5 7 ⇒ 2× x−  2× 2 , the Lowest Common 2 2 ∣  ∣5 Denominator. ⇒ 2 x−  7 2 ⇒ ∣ 2 x−5 ∣  7

Slide 23: ⇒ −1  x  6 5 5 5 ⇒ −1−  x−  6− 2 2 2 7 5 7 ⇒ −  x−  2 2 2 ∣ ∣ 5 7 ⇒ x−  2 2 ∣ ∣ Note that a×∣ b ∣=∣ ab ∣ 5 7 ⇒ 2× x−  2× 2 2 if a ≥ 0 . ∣  ∣5 ⇒ 2 x−  7 2 ⇒ ∣ 2 x−5 ∣  7 The statement is ∣ 2 x−5 ∣  7

Slide 24: Example 8 Find the absolute value inequality statement corresponding to x2  5 or x−1  4 .

Slide 25: ⇒ x2  5 or x−1  4 ⇒ x  3 or x  5 ⇒ x−4  3−4 or x−4  5−4 ⇒ x−4  −1 or x−4  1 Take care. Isolate x before continuing. ⇒ ∣ x−4 ∣  1

Slide 26: ⇒ x2  5 or x−1  4 ⇒ x  3 or x  5 ⇒ x−4  3−4 or x−4  5−4 ⇒ x−4  −1 or x−4  1 ⇒ ∣ x−4 ∣  1 We note the inequality is in a form that we can convert to standard  form.

Slide 27: ⇒ x2  5 or x−1  4 ⇒ x  3 or x  5 Average 3 and 5 . ⇒ x−4  3−4 or x−4  5−4 35 =4 ⇒ x−4  −1 or x−4  1 2 ⇒ ∣ x−4 ∣  1 Subtract 4 from all sides.

Slide 28: ⇒ x2  5 or x−1  4 ⇒ x  3 or x  5 ⇒ x−4  3−4 or x−4  5−4 ⇒ x−4  −1 or x−4  1 Standard  form. ⇒ ∣ x−4 ∣  1 The statement is ∣ x−4 ∣  1

Slide 29: Example 9 Find the absolute value inequality statement corresponding to 2 x−3  1 or x−2  4 .

Slide 30: ⇒ 2 x−3  1 or x−2  4 ⇒ 2 x  4 or x  6 ⇒ x  2 or x  6 ⇒ 2x 6 ⇒ 2−4 Isolate  6−4 continuing. x−4 x before Take care. ⇒ −2  x−4  2 ⇒ ∣ x−4 ∣  2

Slide 31: ⇒ 2 x−3  1 or x−2  4 ⇒ 2 x  4 or x  6 ⇒ x  2 or x  6 ⇒ 2x 6 We note we can write in this form ⇒ 2−4  x−4  6−4 in preparation for conversion to standard  form. ⇒ −2  x−4  2 ⇒ ∣ x−4 ∣  2

Slide 32: ⇒ 2 x−3  1 or x−2  4 ⇒ 2 x  4 or x  6 ⇒ x  2 or x  6 ⇒ 2x 6 Average 2 and 6 . ⇒ 2−4  x−4  6−4 26 =4 ⇒ −2  x−4  2 2 ⇒ ∣ x−4 ∣  2 Subtract 4 from all three sides.

Slide 33: ⇒ 2 x−3  1 or x−2  4 ⇒ 2 x  4 or x  6 ⇒ x  2 or x  6 ⇒ 2x 6 ⇒ 2−4  x−4  6−4 ⇒ −2  x−4  2 Standard  form. ⇒ ∣ x−4 ∣  2 The statement is ∣ x−4 ∣  2

Slide 34: Finding absolute value inequality statements Copyright ©2007-2008 Home Tuition Services