How do you solve #4x - 2> \frac { 1} { 4} ( 16x + 8)#?

1 Answer
smendyka
Jul 16, 2017

See a solution process below:

Explanation:

First, expand the terms in parenthesis on the right side of the inequality. Multiply each term within the parenthesis by the term outside the parenthesis:

#4x - 2 > color(red)(1/4)(16x + 8)#

#4x - 2 > (color(red)(1/4) xx 16x) + (color(red)(1/4) xx 8)#

#4x - 2 > 4x + 2#

Now, subtract #color(red)(4x)# from each side of the inequality:

#-color(red)(4x) + 4x - 2 > -color(red)(4x) + 4x + 2#

#0 - 2 > 0 + 2#

#-2 color(red)(cancel(color(black)(>))) 2#

Because this inequality is not true there is no solution to this problem. Or, the solution is the empty or null set: #{O/}#

However, if instead the inequality operator was reversed to:

#-2 < 2#

This is true for each and every value of #x#. Therefore the solution would be the set of all Real Numbers or: #{RR}#

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