How do you solve #1/4n+12>=3/4n-4# and graph the solution on a number line?

2 Answers
Oct 13, 2017

Answer:

#nle32#

Explanation:

add 4 to both sides
#1/4n+16le3/4n#
subtract #1/4n# from both sides
#16le1/2n#
divide #16# by #1/2#
it looks like this
#16/1 * 2/1#
your answer is #32#
so your final equation is #nle32#
on a number line, put a closed circle on 32 and draw the line going towards the negatives indefinitely.

Here is the graph
enter image source here

Oct 13, 2017

Answer:

#n<=32#

Explanation:

#1/4 n + 12 >= 3/4 n - 4#
Let's start by subtracting #color(red)(3/4 n)# from both sides
#1/4 n + 12 - color(red)(3/4 n) >=cancel (3/4 n) - 4 cancelcolor(red)(-3/4 n)#
#(-1)/2 n + 12 >= -4#
Then, we can subtract #color(green)(12)# from both sides
#(-1)/2 n + cancel12 - cancelcolor(green)(12) >= -4 - color(green)(12)#
#(-1)/2 n >= -16#
In order to find #n#, we need to multiply both sides by #color(orange)(2/(-1))#
#color(orange)((2/-1))((-1)/2 n) >= color(orange)(2/-1)(-16)#
#cancel((-2)/-2 n) >= ((-32)/-1)#
#n <=32#

enter image source here