Question

How do you solve and write the following in interval notation:
$- 6x + 2 < \- 3x - 12$?

Answer 1

Here we are asked to solve the given inequality and write that in the interval notation. First, we will solve the given inequality by separating the $x$ terms on one side and the constant term on another side. After some simplification we will end up by getting a lower or upper bound of $x$ then we will find the other bound with the possibilities.

Complete step-by-step solution:
It is given that$- 6x + 2 < \- 3x - 12$. We aim to solve this inequality and then we will write it in interval notation.
We will first solve this inequality. Consider the given expression$- 6x + 2 < \- 3x - 12$.
Let us add a term $3x$ on both sides of the inequality
$\left( { - 6x + 3x} \right) + 2 < \left( { - 3x + 3x} \right) - 12$
Now let us group like terms together.
$\left( { - 6x + 3x} \right) + 2 < \left( { - 3x + 3x} \right) - 12$
On simplifying this we get
$\left( { - 3x} \right) + 2 < \left( 0 \right) - 12$
Let us rewrite the above inequality.
$- 3x + 2 < \- 12$
Now let us subtract two from both sides of the above inequality.
$- 3x + 2 - 2 < \- 12 - 2$
On simplifying the above inequality, we get
$\dfrac{{ - 3x}}{{ - 1}} < \dfrac{{ - 14}}{{ - 1}}$
Now let us divide the above by $- 3$ we get
$\dfrac{{ - 3x}}{{ - 3}} < \dfrac{{ - 14}}{{ - 3}}$
On simplifying this we get
$x < \dfrac{{14}}{3}$
Thus, we have obtained the solution $x < \dfrac{{14}}{3}$. Now we have to express this solution in the interval form. From the solution, we found that the upper bound is $\dfrac{{14}}{3}$. Here we don’t have any lower boundary thus the value $x$ goes indefinitely on the left side on the number line that is $- \infty$. Thus, the lower bound is$- \infty$.
Therefore, the interval notation of the solution $x < \dfrac{{14}}{3}$ is $\left( { - \infty ,\dfrac{{14}}{3}} \right)$.

Note: In algebraic equations like terms are the terms that are having the same unknown variable, addition, and subtraction of algebraic expression are done by grouping the like terms. Then in the interval notation, we have used the curved braces not the square braces because the value $x$ is less than $\dfrac{{14}}{3}$and not less than or equal to $\dfrac{{14}}{3}$ also, it has indefinite value on the left side thus we cannot use square brackets.