How do you solve #30>1/2n#?

1 Answer
Feb 6, 2017

Answer:

See the entire solution process below:

Explanation:

Multiply each side of the inequality by #color(red)(2)# to solve for #n# while keeping the inequality balanced:

#color(red)(2) xx 30 > color(red)(2) xx 1/2n#

#60 > cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))n#

#60 > n#

To solve in terms of #n# we need to reverse or "flip" the inequality:

#n < 60#