Question

How do you solve for ‘y’ for the equation $ 2x-3y=6 $ ?

Answer 1

Hint: We need to find out the value of ‘y’ from the given linear equation in two variables. As here it is asked to define the value of ‘y’. so we will use some basic mathematical operation to have ‘y’ on the LHS side and other variables and constant on the RHS side.

Complete step by step answer:
Moving ahead with the question in step wise manner,
To solve for ‘y’ here means we need to find the value of ‘y’ in terms of other constants and variables given in the equation. So we have to have ‘y’ on one side and another variable and constant on the other side, let it be ‘y’ on the LHS side and another variable on the RHS side.
As the given equation is $ 2x-3y=6 $ . Here on the LHS side we have two variables i.e. ‘x’ and ‘y’ so to find the value of ‘y’ let us vanish the other variable ‘x’. So to vanish let us subtract $ 2x $ in both side of equation, i.e.
 $ 2x-2x-3y=6-2x $
So on solving we will get;
 $ -3y=6-2x $
Since now on the LHS side we have a negative of $ 3y $ so in order to make it positive let us multiply the equation by $ -1 $ on both sides. So we will get;
 $ -\left( -3y \right)=-\left( 6-2x \right) $
As we know that when $ '-' $ is multiplied by $ '-' $ or $ '+' $ is multiplied by $ '+' $ then it is $ '+' $ and if $ '-' $ is multiplied with $ '+' $ then it is $ '-' $ . So applying the same rule in above equation we will get;
 $ \begin{align}
  & -\left( -3y \right)=-\left( 6-2x \right) \\
 & 3y=-6+2x \\
\end{align} $
Now on the LHS side we have $ 3y $ and we want only the value of ‘y’. So let us divide the equation by ‘3’ in order to get a simple ‘y’. So by diving the equation by ‘3’ we will get;
 $ \dfrac{3y}{3}=\dfrac{-6+2x}{3} $
On simplifying we will get;
 $ \begin{align}
  & y=\dfrac{-6}{3}+\dfrac{2x}{3} \\
 & y=-2+\dfrac{2x}{3} \\
\end{align} $
So we got ‘y’ equal to $ y=-2+\dfrac{2x}{3} $ .
Hence the answer is $ y=-2+\dfrac{2x}{3} $ .

Note: If we are asked to solve for any variable whether it is a linear variable equation or quadratic, cubic and so on, in all we have to pick out the variable for which we need to solve and find its relation with other variables and constants present in the equation.