How do you solve #4abs[x+1]-2<10#?

Answer 1

See a solution process below:

First, add #color(red)(2)# to each side of the inequality to isolate the absolute value term while keeping the inequality balanced:
#4abs(x + 1) - 2 + color(red)(2) < 10 + color(red)(2)#
#4abs(x + 1) - 0 < 12#
#4abs(x + 1) < 12#
Now, divide each side of the inequality by #color(red)(4)# to isolate the absolute value function while keeping the inequality balanced:
#(4abs(x + 1))/color(red)(4) < 12/color(red)(4)#
#(color(red)(cancel(color(black)(4)))abs(x + 1))/cancel(color(red)(4)) < 3#
#abs(x + 1) < 3#

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-3 < x + 1 < 3#
Now, subtract #color(red)(1)# from each segment of the system of inequalities to solve for #x# while keeping the system balanced:
#-3 - color(red)(1) < x + 1 - color(red)(1) < 3 - color(red)(1)#
#-4 < x + 0 < 2#
#-4 < x < 2#

Or

#x > -4# and #x < 2#

Or, in interval notation:

#(-4, 2)#
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Answer 2
To solve \(4| x + 1 | - 2 < 10\), we first add 2 to both sides of the inequality to isolate the absolute value term. This gives us \(4| x + 1 | < 12\). Then, we divide both sides by 4 to solve for the absolute value term. This gives us \(| x + 1 | < 3\). Now, we consider two cases: 1. If \(x + 1 \geq 0\), then \(| x + 1 | = x + 1\). 2. If \(x + 1 < 0\), then \(| x + 1 | = - (x + 1)\). For case 1, when \(x + 1 \geq 0\), we have \(x + 1 < 3\), which gives \(x < 2\). For case 2, when \(x + 1 < 0\), we have \(-(x + 1) < 3\), which simplifies to \(x + 1 > -3\), giving \(x > -4\). Therefore, the solution to \(4| x + 1 | - 2 < 10\) is \(-4 < x < 2\).
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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.

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.

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.

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.

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