How do you solve for x in \[ax+bx-c=0\]?

Hint: In this problem, we have to solve the linear equation \[ax+bx-c=0\], solve for x is nothing but we have to find the value of x. In the given equation, we can take the constant term c to the right-hand side or we can add c on both sides, then in the left-hand side we have x terms in which common x in both the terms can be taken out, then we will get the addition of two terms. Now we can divide the right-hand side and the left-hand side by the similar terms to be cancelled, to get the value of x.

Complete step by step answer:
We know that the given linear equation is,
\[ax+bx-c=0\].
Here we can add c on both sides, we get
\[\begin{align}
  & \Rightarrow ax+bx-c+c=0+c \\
 & \Rightarrow ax+bx=c \\
\end{align}\]
Now we can take the x outside in the left-hand side as it is the common term, the equation becomes,
\[\Rightarrow x\left( a+b \right)=c\]
We can divide by \[\left( a+b \right)\]on both the left-hand side and the right-hand side, we get
\[\Rightarrow \dfrac{x\left( a+b \right)}{\left( a+b \right)}=\dfrac{c}{\left( a+b \right)}\]
We can cancel the similar terms to get the value of x.
\[\Rightarrow x=\dfrac{c}{a+b}\]

Therefore, the value of x is \[\dfrac{c}{a+b}\]

Note: Students make mistakes in adding or subtracting some terms to be cancelled in the given equation. We have to know that instead of adding or subtracting to cancel the terms, we can also take it to the right-hand side by changing the symbol. We also have to know that solving for x is nothing but finding the value of x.