How do you solve for $x$ in$-ax+3b=5$?

Hint: To solve these questions, simply take all the constants on one side of the equation and the variables on the other side. The final solution for $x$ will be the expression on the right-hand side of the equality sign.

Complete step-by-step solution:
Linear equations can be defined as equations of the first order. The general form of a linear equation in one variable can be given as $ax+b=0$ , where $a$ is the coefficient of $x$ and $b$ is the constant. The solutions of linear equations give values, which if substituted back in the equation makes the equation true.
The given equation is$-ax+3b=5$.
To solve for $x$, we need to keep $x$ on the left-hand side of the equation and transpose all other constants and values to the right-hand side of the equation.
For the equation $-ax+3b=5$, first, subtract $3b\;$ from both the sides of the equation to get:
$-ax+3b-3b=5-3b\Rightarrow -ax=5-3b$ …$(i)$
Now divide both the sides of the equation $(i)$ by$-a$, to get:
$\Rightarrow \dfrac{-ax}{-a}=\dfrac{5-3b}{-a}\Rightarrow x=\dfrac{-\left( 5-3b \right)}{a}$
Hence, on solving the given equation step by step we get the value of $x$ as:
$\Rightarrow x=\dfrac{-\left( 5-3b \right)}{a}$
The above equation could have been further simplified if the values of $a$ and $b$ were given in the question.

Note: The linear equation $ax+b=0$ has a unique solution which can be given by $x=\dfrac{-b}{a}$, provided $a\ne 0$. However, if $a=0$, then there can be two cases. First case being that either $b$ also equals 0 and hence, every number is a solution and the second case being that $b\ne 0$ and therefore there exists no solution for the given linear equation. In the case where there exists no solution, the linear equation is said to be an inconsistent linear equation.