How do you solve #3x-8>=1/2 (10x+7)#?

1 Answer
Jul 29, 2017

Answer:

See a solution process below:

Explanation:

First, expand the term on the right side of the inequality by multiplying each term within the parenthesis by the term outside the parenthesis:

#3x - 8 >= color(red)(1/2)(10x + 7)#

#3x - 8 >= (color(red)(1/2) xx 10x) + (color(red)(1/2) xx 7)#

#3x - 8 >= 10/2x + 7/2#

#3x - 8 >= 5x + 7/2#

Next, subtract #color(red)(3x)# and #color(blue)(7/2)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#-color(red)(3x) + 3x - 8 - color(blue)(7/2) >= -color(red)(3x) + 5x + 7/2 - color(blue)(7/2)#

#0 - (2/2 xx 8) - color(blue)(7/2) >= (-color(red)(3) + 5)x + 0#

#-16/2 - color(blue)(7/2) >= 2x#

#-23/2 >= 2x#

Now. multiply each side of the inequality by #color(red)(1/2)# to solve for #x# while keeping the inequality balanced:

#color(red)(1/2) + -23/2 >= color(red)(1/2) xx 2x#

#-23/4 >= color(red)(1/color(black)(cancel(color(red)(2)))) xx color(red)(cancel(color(black)(2)))x#

#-23/4 >= x#

To state the solution in terms of #x# we can reverse or "flip" the entire inequality:

#x <= -23/4#