How do you solve the equation $3x+y=6$ for $x$ ?

Hint: We need to solve this equation in terms of y. As it is given to us a normal equation, therefore we try to make the value of x in terms of y. So, we start solving this problem. We will subtract y on both sides of the equation. Then we know, the same terms with opposite signs cancel out, thus we get a new equation, and then divide both sides of the new equation by 3. After the necessary calculation, we split the denominator on the right-hand side of the equation with respect to subtraction and make needful calculations to get the value of x in terms of y.

Complete answer:
According to the question, it is given to a normal equation $3x+y=6$. We have to solve this equation for x and get the value in terms of y.
The equation is given to us: $3x+y=6$ ------------- (1)
Firstly, we will subtract $y$ on both sides in the equation (1), we get
$\Rightarrow 3x+y-y=6-y$
As we know, the same terms with opposite signs cancel out, therefore we get
$\Rightarrow 3x=6-y$
Now, we will divide 3 into both sides of the above equation, we get
$\Rightarrow \dfrac{3x}{3}=\dfrac{6-y}{3}$
We know in the division, the same terms will cancel out to 1,thus in LHS we apply the division rule, we get
$\Rightarrow x=\dfrac{6-y}{3}$
Now, we will split the denominator with respect to subtraction on the right-hand side of the above equation, we get
$\Rightarrow x=\dfrac{6}{3}-\dfrac{y}{3}$
After necessary calculations, we get
$\Rightarrow x=2-\dfrac{y}{3}$
Therefore, we see that for the equation $3x+y=6$ value of x is equal to $2-\dfrac{y}{3}$ that is in terms of y, which is the required answer.

Note: Make all the necessary calculations and do a step-by-step calculation, to avoid errors. In this question, we have to solve for x, thus we get the value of x in terms of y. Now, if you substitute any value of y in the required answer, you will get an integer value as your value of x.