How do you solve #abs(10+8x)-2>16#?

2 Answers

Answer:

#color(blue)(ul(bar(abs(color(black)(x>1, x<-7/2))))#

Explanation:

Let's first isolate the absolute value operation:

#abs(10+8x)-2>16#

#abs(10+8x) > 18#

When solving absolute value questions, we have to consider the positive and negative value of what is sitting inside the absolute value sign. (For instance, when #absx=3#, #x# can be 3. It can also b #-3#. So we need to consider both cases:

Positive

#10+8x>18#

#8x>8#

#x>1#

Negative

#-10-8x>18#

#-8x>28#

#x<-28/8=-14/4=-7/2#

Put together, we have:

#color(blue)(ul(bar(abs(color(black)(x>1, x<-7/2))))#

Mar 4, 2017

Answer:

See the entire solution process below:

Explanation:

First, add #color(red)(2)# to each side of the equation to isolate the absolute value expression while keeping the equation balanced:

#abs(10 + 8x) - 2+ color(red)(2) > 16 + color(red)(2)#

#abs(10 + 8x) - 0 > 18#

#abs(10 + 8x) > 18#

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

#-18 > 10 + 8x > 18#

Next, subtract #color(red)(10)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-18 - color(red)(10) > 10 + 8x - color(red)(10) > 18 - color(red)(10)#

#-28 > 10 - color(red)(10) + 8x > 8#

#-28 > 0 + 8x > 8#

#-28 > 8x > 8#

Now, divide each segment of the system by #color(red)(8)# to solve for #x# while keeping the system balanced:

#-28/color(red)(8) > (8x)/color(red)(8) > 8/color(red)(8)#

#-(4 xx 7)/color(red)(4 xx 2) > (color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) > 1#

#-(color(red)(cancel(color(black)(4))) xx 7)/color(red)(cancel(4) xx 2) > x > 1#

#-7/2 > x > 1#