Question

How do you solve the inequality $3x + 2 \leqslant 1$?

Answer 1

Hint: This problem deals with solving the linear inequation with one variable. A linear inequation is an inequation of a straight line, written in one variable. The only power of the variable is 1. Linear inequalities in one variable may take the form of $ax + b \leqslant 0$, and are usually solved for the variable $x$ using basic algebraic operations.

Complete step-by-step solution:
Given a linear inequation one variable which is considered as given below:
$ \Rightarrow 3x + 2 \leqslant 1$
Now rearrange the terms such that all the constants are on one side of an inequation and all the $x$ terms are on the other side of the inequation.
Now moving the term which is constant, 2, which is on the left hand side of the inequation to the right hand side of the inequation, as shown below:
$ \Rightarrow 3x \leqslant 1 - 2$
Now simplifying the above inequation, as the like terms are and the constants are grouped together.
On the right hand side, solving 1 and -2, :
$ \Rightarrow 3x \leqslant - 1$
Now dividing the above inequation with 3, so as to remove the coefficient of the $x$ term on the left hand side of the inequation, as shown below:
$\therefore x \leqslant \dfrac{{ - 1}}{3}$

The value of $x \leqslant \dfrac{{ - 1}}{3}$

Note: Please note that the linear inequalities in one variable which are expressed in the form of $ax + b \leqslant 0$, have only one solution. Where a and b are two integers, and x is a variable. This means that there will be no terms involving higher powers of x, not even the power of 2, which is ${x^2}$.