Question

How do you solve $\dfrac{x}{2}+\dfrac{x}{4}=5$

Answer 1

Hint: The above question is a simple problem of linear equations in one variable. So, we will simply first understand the concept and definition of linear equations in one variable and then we will solve the above question to get the value of ‘x’. We will first take x as common form the fractional terms and then we will find the LCM of 2, and 4 and then we will add them. Then we will simplify the obtained equation by using cross-multiplication and using basic algebraic operations. Then we will get the value of x.

Complete step-by-step solution:
We know that the linear equation in one variable is an equation which has maximum one variable in the equation which is of order 1. Also, we know that a linear equation in one variable has only one solution.
And, also standard form of linear equation in one variable is represented as:
ax + b = 0 where a, b is a real number and a is not equal to zero.
So, we can say that $\dfrac{x}{2}+\dfrac{x}{4}=5$ is a linear equation in one variable with variable x.
$\Rightarrow \dfrac{x}{2}+\dfrac{x}{4}=5$
Now, we will first take x as common from the given fractional terms.
$\Rightarrow x\left( \dfrac{1}{2}+\dfrac{1}{4} \right)=5$
Now, we will add the fractional terms by taking the LCM of 2 and 4.
LCM of 2 and 4 is 4
$\Rightarrow x\left( \dfrac{2+1}{4} \right)=5$
$\Rightarrow x\left( \dfrac{3}{4} \right)=5$
$\Rightarrow \dfrac{3x}{4}=5$
Now, we will use the cross-multiplication property and multiply both sides of the given equation by 4.
$\Rightarrow \dfrac{3x}{4}\times 4=5\times 4$
$\Rightarrow 3x=20$
So, after dividing both side by 3 we will get:
\[\begin{align}
  & \Rightarrow \dfrac{3x}{3}=\dfrac{20}{3} \\
 & \Rightarrow x=\dfrac{20}{3} \\
\end{align}\]
Now, when we put \[x=\dfrac{20}{3}\] in the equation given above, then we will get $5=5$ , which satisfies the equality.
Hence, \[x=\dfrac{20}{3}\] is our required answer.

Note: Students are required to recall the linear equations and basic properties of linear equations to solve these types of questions. We can verify the answer by putting the value in the given equation. On solving the equation if we get the values of LHS and RHS equal it means that the answer is correct. Students are also required to note that when we shift a variable or constant from one side of the equation to the other side of the equation then the sign of the variable gets reversed and magnitude remains the same.