Question

How do you solve \[3+7c=2c-3\]?

Answer 1

Hint: The given equation is a linear equation in one variable. As we know that to solve these types of equations means finding the value of the variable that satisfies the equation. To do this we need to take the variable terms to one side of the equation and the constant terms to the other side. The given equation has one variable term and a constant on each of its sides.

Complete step by step solution:
We are given the equation \[3+7c=2c-3\], we have to solve it. The highest power of the variable of the equation is 1, so the degree of the equation is also one. Hence, it is a linear equation. As we know to solve a linear equation, we have to take all the variable terms to one side of the equation and leave constants to the other side.
\[3+7c=2c-3\]
Subtracting 3 from both sides of the above expression, we get
\[\begin{align}
  & \Rightarrow 3+7c-3=2c-3-3 \\
 & \Rightarrow 7c=2c-6 \\
\end{align}\]
Subtracting \[2c\] from both sides of above equation, we get
\[\Rightarrow 5c=-6\]
Dividing by 5 to both sides of the above equation and cancelling out common factor, we get
\[\Rightarrow c=\dfrac{-6}{5}\]
Hence, the solution of the given equation is \[c=\dfrac{-6}{5}\].

Note: We can check if the answer is correct or not by substituting the value in the given equation. From the given equation, we get the left-hand side as \[3+7c\], and right-hand side as \[2c-3\]. Substituting \[c=\dfrac{-6}{5}\] in both sides of the equation, we get LHS as \[3+7\left( \dfrac{-6}{5} \right)=\dfrac{-27}{5}\], and RHS as \[2\left( \dfrac{-6}{5} \right)-3=\dfrac{-27}{5}\]. As \[LHS=RHS\], the solution is correct.