How do you solve #2( 8x - 1) \geq 0#?

Question

1263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-27T14:31:05

See the entire solution process below:

First, divide each side of the inequality by #color(red)(2)# to eliminate the multiplier while keeping the inequality balanced:

#(2(8x - 1))/color(red)(2) >= 0/color(red)(2)#

#(color(red)(cancel(color(black)(2)))(8x - 1))/cancel(color(red)(2)) >= 0#

#8x - 1 >= 0#

Next, add #color(red)(1)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#8x - 1 + color(red)(1) >= 0 + color(red)(1)#

#8x - 0 >= 1#

#8x >= 1#

Now, divide each side of the inequalityby #color(red)(8)# to solve for #x# while keeping the inequality balanced:

#(8x)/color(red)(8) >= 1/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) >= 1/8#

#x >= 1/8#