How do you write an inequality and solve given "three fourths of a number decreased by nine is at least forty two"?

2 Answers
Feb 26, 2017

Answer:

The inequality is: #3/4n - 9 >= 42#

The solution is: #n >= 68#

Explanation:

First, let's call "a number" - #n#

"three fourths of a number" then can be written as:

#3/4n#

This, "decreased by nine" then can be written as:

#3/4n - 9#

"is at least" is the same as #>=# so we can now write:

#3/4n - 9 >= #

and what it is "at least" is "forty two" so:

#3/4n - 9 >= 42#

To solve this we first add #color(red)(9)# to each side of the inequality to isolate the #n# term while keeping the inequality balanced:

#3/4n - 9 + color(red)(9) >= 42 + color(red)(9)#

#3/4n - 0 >= 51#

#3/4n >= 51#

Now, we 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 3/4n >= color(red)(4)/color(blue)(3) xx 51#

#cancel(color(red)(4))/cancel(color(blue)(3)) xx color(blue)(cancel(color(black)(3)))/color(red)(cancel(color(black)(4)))n >= color(red)(4) xx 17#

#n >= 68#

Feb 26, 2017

Answer:

Without any punctuation there are different interpretations which lead to different solutions:

#3/4 x -9 >= 42" or "3/4(x-9) >=42#

#x >=56" or "x >=65#

Explanation:

Let's use some punctuation so show exactly what is meant:
note how the placement of a comma gives a different meaning.

#color(blue)("three fourths of a number"), color(red)(" decreased by 9") color(limegreen)(" is at least 42")#

Let the number be #x#

#color(blue)(3/4x) color(red)(-9) color(limegreen)( >=42)#

Solving gives:

#3/4x >= 51#

#4/3 xx 3/4x >= 51 xx 4/3#

#x >=56#

However, if we interpret it differently we could have:

#color(blue)("three fourths of,") color(red)(" a number decreased by 9,") color(limegreen)(" is at least 42")#

#color(blue)(3/4) color(red)((x-9)) color(limegreen)( >= 42#

Solving gives:

#4/3 xx 3/4(x-9) >= 42 xx 4/3#

#x-9 >= 56#

#x >=65#