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How do you solve #|3- 5x | < 18#?

Answer 1

See a solution process below:

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.

#-18 < 3 - 5x < 18#

First, subtract #color(red)(3)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-18 - color(red)(3) < 3 - color(red)(3) - 5x < 18 - color(red)(3)#

#-21 < 0 - 5x < 15#

#-21 < -5x < 15#

Now, divide each segment by #color(blue)(-5)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

#(-21)/color(blue)(-5) color(red)(>) (-5x)/color(blue)(-5) color(red)(>) 15/color(blue)(-5)#

#21/5 color(red)(>) (color(blue)(cancel(color(black)(-5)))x)/cancel(color(blue)(-5)) color(red)(>) -3#

#21/5 color(red)(>) x color(red)(>) -3#

#x > -3#; #x < 21/5#

Or, in interval notation:

#(-3, 21/5)#

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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 #|3- 5x | &lt; 18#?

How do you solve #|3- 5x | < 18#?