Question

How do you solve $ - \dfrac{1}{3}y - 6 = - 11$?

Answer 1

Hint: In this question we have to solve the equation for $y$, the given equation is a linear equation as the degree of the highest exponent of $y$ is equal to 1. To solve the equation first take all $y$ terms of the equation to one side and all constants of the equation to the other side and solve for the required $y$.

Complete step by step solution:
Given equation is $ - \dfrac{1}{3}y - 6 = - 11$, and we have to solve for $y$,
Now add 6 to both sides of the equation, we get,
$ \Rightarrow - \dfrac{1}{3}y - 6 + 6 = - 11 + 6$,
Now simplifying by eliminating the like terms, we get,
$ \Rightarrow - \dfrac{1}{3}y = - 5$,
Now multiply -3 to both sides of the equation we get,
$ \Rightarrow - \dfrac{1}{3}y \times - 3 = - 5 \times - 3$,
Now simplifying by multiplying we get,
$ \Rightarrow y = 15$,
So the value of $y$ will be $15$, i.e., when we substitute the value of $y$ in the equation $ - \dfrac{1}{3}y - 6 = - 11$, then right hand side of the equation will be equal to left hand side of the equation, we get,
$ \Rightarrow - \dfrac{1}{3}y - 6 = - 11$,
Now substitute $x = 15$, we get,
$ \Rightarrow - \dfrac{1}{3}\left( {15} \right) - 6 = - 11$,
Now multiplying, we get,
$ \Rightarrow - 5 - 6 = - 11$ ,
Further simplifying we get,
$ \Rightarrow - 11 = - 11$,
Therefore here R.H.S=L.H.S.
So, the value of $y$ is $15$.

$\therefore $ The value of $y$ when the equation $ - \dfrac{1}{3}y - 6 = - 11$ is solved will be equal to $15$.

Note: A linear equation is an equation of a straight line having a maximum of one variable. The degree of the variable will be equal to 1. To solve any equation in one variable, pit all the variable terms on the left hand side and all the numerical values on the right hand side to make the calculation solved easily.