Question

How do you solve \[r - \dfrac{3}{{10}} = \dfrac{r}{{16}}\] ?

Answer 1

Hint:Here, we are given the linear equation with one variable. To solve this equation, we will first bring the terms with the variable to the left side and other terms to the right side. Then by taking LCM on the right side we will subtract the terms on the left side to make it one term with the variable. After that, we need to use the concept of multiplication or division in simple equations.

Complete step by step answer:
We are given \[r - \dfrac{3}{{10}} = \dfrac{r}{{16}}\]. Here, we have two terms with variables on both the sides. Therefore, we will first bring them together on the left hand side and move the constant term to the right hand side.
\[r - \dfrac{3}{{10}} = \dfrac{r}{{16}} \\
\Rightarrow r - \dfrac{r}{{16}} = \dfrac{3}{{10}} \\ \]
Now if we take LCM on the left side, it will be 16.
\[\dfrac{{16r - r}}{{16}} = \dfrac{3}{{10}} \\
\Rightarrow \dfrac{{15r}}{{16}} = \dfrac{3}{{10}} \\ \]
Now, to find the variable $r$, we need to remove the digit $\dfrac{{15}}{{16}}$ which is multiplied with it. This can be done by dividing the right hand side by $\dfrac{{15}}{{16}}$ which is nothing but multiplying by its reciprocal $\dfrac{{16}}{{15}}$.However, it is important to know that we need to perform any arithmetic operation on both the sides of the equation.Therefore, we will multiply both the sides of the equation by $\dfrac{{16}}{{15}}$.
\[\dfrac{{15r}}{{16}} \times \dfrac{{16}}{{15}} = \dfrac{3}{{10}} \times \dfrac{{16}}{{15}} \\
\therefore r = \dfrac{8}{{25}} \\ \]
Thus our final answer is $\dfrac{8}{{25}}$.

Note:It is important to keep in mind that while solving a simple equation, we need to think of the equation as a balance. Thus, if we do something to one side of the equation, we must do the same thing to the other side. Doing the same thing to both sides of the equation keeps the equation balanced. For example, we have multiplied both the sides by $\dfrac{{16}}{{15}}$ in this problem.