Question

How do you solve $\dfrac{1}{6} = \dfrac{5}{{12}} + p$?

Answer 1

Hint: We will first of all, bring all the variables to the one side and the constants on the other side. After that, we will just take the least common multiple and solve the subtraction on constants side and thus have the value of p.

Complete step by step solution:
We are given that we are required to solve $\dfrac{1}{6} = \dfrac{5}{{12}} + p$.
Taking p from addition in the left hand side to subtraction in the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{1}{6} - p = \dfrac{5}{{12}}$
Taking $\dfrac{1}{6}$ from addition in the in the left hand side to subtraction in the right hand side, we will then obtain the following equation with us:-$ \Rightarrow p = \dfrac{1}{6} - \dfrac{5}{{12}}$
$ \Rightarrow - p = \dfrac{5}{{12}} - \dfrac{1}{6}$
Multiplying the whole equation by – 1, we will then obtain the following equation with us:-

Taking the least common multiple of the denominators on the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow p = \dfrac{{2 - 5}}{{12}}$
Simplifying the numerator on the right hand side by calculations, we will then obtain the following equation with us:-
$ \Rightarrow p = \dfrac{{ - 3}}{{12}}$
We can write the above fraction as follows:-
$ \Rightarrow p = \dfrac{{ - 1 \times 3}}{{3 \times 4}}$
Crossing – off the 3 from both the numerator and denominator on the right hand side, we will then obtain the following equation with us:-

$ \Rightarrow p = \dfrac{{ - 3}}{4}$

Thus, we have the required answer.

Note:
The students must notice that we could cross – off the 3 in the last second step of the solution mentioned above because we know that 3 can never be equal to 0. We can never cross – off any variable which can never be equal to 0.
Now, the students must also notice that we need as many equations as many numbers of variables we have so as to find the value of each of the variables. Here, we had one variable p and we had one equation only, so we could find its value very easily.