Solve the linear equation:
\[\dfrac{x}{2}-\dfrac{1}{5}=\dfrac{x}{3}+\dfrac{1}{4}\]

Hint: In this question we can see that we are given one linear equation with only one variable. Now to solve this question one needs to start by taking the LCM of both the left hand side and the right hand side of the equation to simplify it and make it easier to solve. And then by cross multiplying it will become a simplified linear equation. And then after performing arithmetic operations on it like addition, subtraction multiplication and division you can find the answer of x.

Complete answer:Now here in this question we can see that we are given a linear equation which is;
\[\dfrac{x}{2}-\dfrac{1}{5}=\dfrac{x}{3}+\dfrac{1}{4}\]
Now taking the LCM for the left hand side of the equation we get;
\[\dfrac{5x-2}{10}=\dfrac{x}{3}+\dfrac{1}{4}\]
After this we need to take the LCM of the RHS of the equation
\[\dfrac{5x-2}{10}=\dfrac{4x+3}{12}\]

Now to solve this further what we can do is cross multiplying both sides of the equation therefore simplifying it
\[12\left( 5x-2 \right)=10\left( 4x+3 \right)\]
Opening the brackets on both sides and multiplying
\[60x-24=40x+30\]
Now taking the x terms together on left hand side and the constant terms together on right hand side
\[60x-40x=30+24\]
Now subtracting and adding respectively
\[20x=54\]
Dividing both sides by 20
\[x=\dfrac{54}{20}\]

Simplifying we get the value of x which is
\[x=\dfrac{27}{10}\]
Therefore for \[x = 2\] matrix \[A\] is a skew symmetric matrix .

Note:
Student can verify that their answer is correct in questions like this by putting the value of x and checking if LHS is equal to RHS and if it is then you can be verified that your answer is correct
\[\dfrac{27}{20}-\dfrac{1}{5}=\dfrac{23}{30}+\dfrac{1}{4}\]
Taking LCM
\[\dfrac{27-4}{20}=\dfrac{54+15}{60}\]
\[\dfrac{23}{20}=\dfrac{69}{60}\]
\[\dfrac{23}{20}=\dfrac{23}{20}\]
Therefore LHS is equal to RHS which means our answer is correct