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

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

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