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Set 1.2
Questions 34, 36, 38
David Cohen
Rewrite without using absolute value notation.
See attachment.
Question 34
Since x > 4, the value inside is positive for both absolute value expressions added.
| x - 3 | + | x - 4|
x - 3 + x - 4
2x - 7
Question 36
Since x = 4, | x - 3 | is positive and | x - 4 | becomes 0.
We have x - 3 + 0 = x - 3.
Question 38
Since x < -3, | x + 1 | is negative.
We also know that | x + 3 | is negative.
We have -(x + 1) - 4(x + 3).
-x - 1 - 4x - 12
-5x - 13
You say?
Questions 34, 36, 38
David Cohen
Rewrite without using absolute value notation.
See attachment.
Question 34
Since x > 4, the value inside is positive for both absolute value expressions added.
| x - 3 | + | x - 4|
x - 3 + x - 4
2x - 7
Question 36
Since x = 4, | x - 3 | is positive and | x - 4 | becomes 0.
We have x - 3 + 0 = x - 3.
Question 38
Since x < -3, | x + 1 | is negative.
We also know that | x + 3 | is negative.
We have -(x + 1) - 4(x + 3).
-x - 1 - 4x - 12
-5x - 13
You say?
