How do you solve #abs(7x3)<2#?
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}).
By signing up, you agree to our Terms of Service and Privacy Policy
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
 Out of 7 lottery tickets 3 are prizewinning tickets. If someone buys 4 tickets what is the probability of winning at least two prizes?
 How do you solve the inequality #x < 2x^3 < 8x^3#?
 How do you solve #11w+99<77#?
 How do you solve #abs(x^2  9 )= x^2  9#?
 How do you solve #2x +(2) + (6) = 4#?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7