Question

How do you solve $\dfrac{1}{6}x + \dfrac{2}{3}x = 1$?

Answer 1

Hint: We will first of all, multiply both the sides of the given equation $\dfrac{1}{6}x + \dfrac{2}{3}x = 1$ by 6 and then simplify the left hand side by clubbing the like terms and thus simplify further.

Complete step by step answer:
We are given that we are required to solve $\dfrac{1}{6}x + \dfrac{2}{3}x = 1$.
Let us, first of all, multiply both (left hand side as well as right hand side) of the above equation by 6, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{6}{6}x + \dfrac{{2 \times 6}}{3}x = 6$
Simplifying the calculations on the left hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow x + 4x = 6$
Simplifying the left hand side of the above equation by clubbing both the terms with x, we will then obtain the following equation with us:-
$ \Rightarrow 5x = 6$
Taking 5 from multiplication in the left hand side to division in the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow x = \dfrac{6}{5}$
Thus, we have the required answer with us.

Note:
The students must note that we need as many equations as many numbers of unknown variables we have so that we have a value for all of them individually. Like, in the above question, we were given one equation and we had only one unknown variable x, therefore, we could find its value.
The students must also note that while clubbing the like terms, we used the following underlying procedure for it:-
$ \Rightarrow x + 4x = 6$
The above equation can be written as follows:-
$ \Rightarrow (1 + 4)x = 6$
Simplifying the left hand side of the above equation by calculating the required addition, we will then obtain the following equation with us:-
$ \Rightarrow 5x = 6$
Thus, we have reached out to the required result.
The students must also note that we could divide both the sides of the equation by 5 in the last step, because we know that 5 can never be equal to 0. We can never multiply or divide a quantity by 0.