How do you solve "quotient of three times a number and 4 is at least -16" and graph the solution on a number line?

2 Answers
Nov 7, 2017

Answer:

See a solution process below:

Explanation:

Let's call "a number": #x#

"The quotient" is the result of division.

In this problem, the numerator is: #3x#

The denominator is: #4#

So we can write:

#(3x)/4#

"is at least" means this is an inequality and specifically a #>=# inequality.

So, we can continue to write:

#(3x)/4 >=#

And we can finish the inequality as:

#(3x)/4 >= -16#

To solve this, multiply each side of the inequality by #color(red)(4)/color(blue)(3)# to solve for #n# while keeping the inequality balanced:

#color(red)(4)/color(blue)(3) xx (3x)/4 >= color(red)(4)/color(blue)(3) xx -16#

#cancel(color(red)(4))/cancel(color(blue)(3)) xx (color(blue)(cancel(color(black)(3)))x)/color(red)(cancel(color(black)(4))) >= -64/3#

#x >= -64/3#

To graph this we will draw a vertical line at #-64/3# on the horizontal axis.

The line will be a solid line because the inequality operator contains an "or equal to" clause.

We will shade to the right side of the line because the inequality operator also contains a "greater than" clause:

graph{x>=-64/3 [-30, 30, -15, 15]}

Nov 7, 2017

Answer:

#x>=-21 1/3#
To graph this on a number line, you would make a solid dot on the point #(-21 1/3)#, with the line moving to the right (#rarr#)

Explanation:

First, let's analyze what each word means.

"quotient (#-:#) of three times a number (#3x#) and four (#+4#) is at least -16 (#>=-16#)"

Now take out the numbers.

#3x-:4>=-16#

Now to find the possibilities of #x#, balance the inequality.

#3x-:4>=-16# Multiply both sides by 4.
#3x>=-64# Divide both sides by 3.
#x>=-21 1/3#

To graph this on a number line, you would make a solid dot on the point #(-21 1/3)#, with the line moving to the right (#rarr#)