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

Question

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

Genesis Allen · Accepted Answer

#nle32#

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

Eva Adamson · Answer

#n<=32#

#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#

Abigail Allen · Answer

undefined To solve the inequality $\frac{1}{4}n + 12 \geq \frac{3}{4}n - 4$, first, eliminate the fractions by multiplying both sides by the least common denominator, which is 4:

$4\left(\frac{1}{4}n + 12\right) \geq 4\left(\frac{3}{4}n - 4\right)$

This simplifies to:

$n + 48 \geq 3n - 16$

Next, subtract $n$ from both sides:

$48 \geq 2n - 16$

Now, add 16 to both sides:

$64 \geq 2n$

Finally, divide both sides by 2:

$32 \geq n$

So, the solution to the inequality is $n \leq 32$.

To graph this solution on a number line, draw an arrow pointing to the left from the number 32, indicating all values of $n$ that are less than or equal to 32.

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

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