Question

How do you solve and write the following in interval notation: $4x < 2x+1\le 3x+2$?

Answer 1

We start solving the problem by considering one of the inequalities from the given $4x < 2x+1\le 3x+2$ and then find the values of x which satisfies the assumed inequality. We then consider the other inequality given $4x < 2x+1\le 3x+2$ and then find the values of x which satisfies the assumed inequality. We then find the common values from values obtained from both the considered inequalities and then write them in the interval notation.

Complete step by step solution:
According to the problem, we are asked to solve and write the following in interval notation: $4x < 2x+1\le 3x+2$.
Let us consider $4x < 2x+1$.
$\Rightarrow 2x < 1$.
$\Rightarrow x < \dfrac{1}{2}$ ---(1).
Now, let us consider $2x+1\le 3x+2$.
$\Rightarrow -1\le x$.
$\Rightarrow x\ge -1$ ---(2).
From equations (1) and (2), we can see that the values of x lie in the given interval: $-1\le x < \dfrac{1}{2}$. Now, let us write this obtained solution in the interval notation. We know that the interval notation of $a\le x < b$ is represented as $\left[ a,b \right)$. Using this result, we get the interval notation as $\left[ -1,\dfrac{1}{2} \right)$.
So, we have found the solution set (i.e., the values of x satisfying) of given inequality $4x < 2x+1\le 3x+2$ as $\left[ -1,\dfrac{1}{2} \right)$.
$\therefore$ The solution set (i.e., the values of x satisfying) of given inequality $4x < 2x+1\le 3x+2$ is $\left[ -1,\dfrac{1}{2} \right)$.

Note: Whenever we get this type of problem, we first try to consider two or more inequalities from the given inequality and then find the common interval of values of x which is the solution set. We should not make calculation mistakes while solving this problem. We should report the obtained answer as $\left[ -1,\dfrac{1}{2} \right]$ because it included the value $x=\dfrac{1}{2}$, which is the common mistake done by students. Similarly, we can expect problems to find the solution set of the given inequality $2x+5\le 5-3x < 2+2x$.