Question

How do you solve this inequality? $\dfrac{1}{2}x\le 6?$

Answer 1

Hint: Here, given that, the resultant may be equal to or less than $6,$ as $\left( \le \right)$ symbol is used. For solving the given equation, we have to replace $\left( \le \right)$ symbol by $\left( = \right)$ symbol.
Suppose, for value $'x'$ the equation gives resultant value $'x'$ the equation gives resultant as $6,$ then the actual value of solution will be $'x'$ or any number which is greater than $'x'$.

Complete step-by-step answer:
In inequality we use the following symbols.
$'\ge '$ it is known as greater than or equal to symbol and
$'\le '$ It is known as less than or equal to symbol.
As given is the question. The equation is,
$\Rightarrow$ $\dfrac{1}{2}x\le 6$
So, from above we can say that, maximum value should be $6.$
And the minimum value should be less than $6,$ as it is given in the question, that the resultant may be equal to $6$ or less than $6.$
By applying this concept, we have to solve this question.
So, as given in the question the equation.
$\Rightarrow$ $\dfrac{1}{2}x\le 6...(i)$
For solving the above equation we have to replace $\left( \le \right)$ symbol to $\left( = \right)$ symbol.
Then,
$\Rightarrow$ $\dfrac{1}{2}x=6...(ii)$
Now, we have to separate the like terms,
$\Rightarrow$ $x=\dfrac{6}{\dfrac{1}{2}}...(iii)$
$\Rightarrow$ $x=6\times 2$
$\Rightarrow$ $x=12$
As, here maximum value of $x$ for resultant $6$ is $12.$
So, $'x'$ may be equal to or less than $12.$

Thus, the value of $x\le 12$

Additional Information:
The given question we can solve in another way too.
The given equation is,
$\Rightarrow$ $\dfrac{1}{2}x\le 6...(i)$
Let, multiply equation $(i)$ by $2$ on both sides
$\Rightarrow$ $2\left( \dfrac{1}{2} \right)x\le 2\left( 6 \right)$
$\Rightarrow$ $\left( \dfrac{2}{2} \right)x\le 12$
$\Rightarrow$ $x\le 12$
Hence, as seen from above it is clear that the $x$ may be equal to or less than $12.$
Here we multiply the equation $(i)$ by $2$ on both sides for balancing the equation.
So that the $2$ is already given at one side of the equation. If we multiply it on that side. The $\left( 2\times \dfrac{1}{2} \right)$ gets canceled. But for balancing the equation we have to multiply $2$ on another side too.

Note:
As given in the question, resultant may be $6$ or less than $6.$ It means that, the minimum value of resultant is or less than $6$ and we have to find the maximum or equal value to resultant.
In equation $(iii)$
$\Rightarrow$ $x=\dfrac{6}{\dfrac{1}{2}}$
Here, $6$ divide by $\dfrac{1}{2}$ but for making the equation simple. We have to take the reciprocal of $\dfrac{1}{2}$. So then the $2$ comes at numerator and $1$ at denominator and then equation becomes
$\Rightarrow$ $x=6\times \dfrac{2}{1}$