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How do you solve #abs(7x-3)<2#?

Answer 1

To solve the inequality (|7x - 3| < 2), we need to consider two cases: when the expression inside the absolute value is positive and when it's negative.

When (7x - 3) is positive ((7x - 3 > 0)), the inequality simplifies to (7x - 3 < 2).
When (7x - 3) is negative ((7x - 3 < 0)), the inequality becomes (-(7x - 3) < 2).

Solving each case separately:

For (7x - 3 < 2): (7x - 3 < 2)
(7x < 5)
(x < \frac{5}{7})
For (-(7x - 3) < 2): (-(7x - 3) < 2)
(-7x + 3 < 2)
(-7x < -1)
(x > \frac{1}{7})

Combining the solutions: (x) must be greater than (\frac{1}{7}) and less than (\frac{5}{7}), so the solution set is (\frac{1}{7} < x < \frac{5}{7}).

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Answer 2

Avery Alexander

#1/7< x < 5/7#

#"absolute value functions of the type "|x| < a#

#"always have solutions of the form " -a < x < a #

#rArr-2 < 7x-3 < 2#

#"add 3 to all 3 intervals"#

#-2+3 < 7x < 2+3#

#rArr1 < 7x < 5#

#"divide all 3 intervals by 7"#

#rArr1/7 < x < 5/7" is the solution"#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #abs(7x-3)&lt;2#?

How do you solve #abs(7x-3)<2#?