How do you write the inequality and solve given "twice a number increased by 3 is less than the number decreased by 4"?

2 Answers
Jun 2, 2017

Answer:

See a solution process below:

Explanation:

To write the inequality we will start by defining "a number" as #n#.

Then, "twice a number" can be written as #2n#.

"increased by 3" becomes: #2n + 3#

"is less than" turns it into" #2n + 3 <#

"the number decreased by 4" is: #n - 4#

Putting this together gives the inequality:

#2n + 3 < n - 4#

To solve, first subtract #color(red)(3)# and #color(blue)(n)# from each side of the inequality to solve for #n# while keeping the inequality balanced:

#-color(blue)(n) + 2n + 3 - color(red)(3) < -color(blue)(n) + n - 4 - color(red)(3)#

#-color(blue)(1n) + 2n + 0 < 0 - 7#

#(-color(blue)(1) + 2)n < -7#

#1n < -7#

#n < -7#

Answer:

Any number less than #-7# will make this inequality true.

#x < -7#

Explanation:

Let the number be #x#

Write the left side as maths first:

#color(red)("Twice a number ") color(blue)(" increased by 3 ") color(magenta)("is less than") #

#color(red)(2x) color(white)(wwwwwwwwwwwww)color(blue)(+ 3 ) color(white)(wwwwwww)color(magenta)( <)#

Now do the same for the right side..

#color(red)(2x)" " color(blue)(+ 3 ) " "color(magenta)( <)" "color(green)("the number decreased by 4"#

#color(red)(2x)" " color(blue)(+ 3 ) " "color(magenta)( <)" "color(green)(x-4)#

We have the inequality, now solve it:

#2x +3 < x-4#

#2x-x < -4 -3#

#x < -7#