Solutions for all numbers x ∈ R for the following equation?

The particular questions are:

x+3+5+4x=16

1 Answer
Mar 9, 2018

Answers are x=85andx=245

Explanation:

We have two modulus monomials added to be equal 16.

It means that for every single monomial we will have two options :
when the expression inside is positive and when it is negative.

It means that overall we will have four different cases:

  1. When x+3>0and5+4x>0
    so in this case, x has to be :x>3andx>54

What this means is that x should be x > -5/4

when you solve the equation for this these conditions, you get
x+3+5+4x=16 where x=5/8 , which agrees with your condition that x has to be greater than 54.

You do in all cases the same process.

  1. (The second case ) you have x+3>0and5+4x<0

x>3andx<54, so x should be between -3 and -5/4

3<x<54

when you solve x+3(5+4x)=16 you get that x = -6

6 is not between 3and54 , so in the second case there is no solution

Two more cases you do on the same way.
You will get:
3.x+3<0and5+4x<0
x=245

4.x+3<0and5+4x>0
no solution

So the possible solutions are only x=58andx=245

This can be done using graphical method also, but I prefer this one.