Question

How do you solve \[7v-\left( 6-2v \right)=12\]?

Answer 1

Hint: The degree of an equation is the highest power of the variable present in it. The degree of the equation decides whether the equation is linear, quadratic, cubic, etc. To solve a linear equation, we have to take all the variable terms to one side of the equation, and constant terms to another side of the equation.

Complete answer:
The given equation is \[7v-\left( 6-2v \right)=12\], the highest power to which variable is raised is one, hence the degree of the equation is 1. As the degree of the equation is 1, it is a linear equation. To solve a linear equation, we have to take all the variable terms to one side of the equation, and constant terms to another side of the equation. We can do this in the given equation as follows,
  \[7v-\left( 6-2v \right)=12\]
Expanding the bracket in the equation, we get
\[\begin{align}
  & \Rightarrow 7v-6+2v=12 \\
 & \Rightarrow 9v-6=12 \\
\end{align}\]
Adding 6 to both sides of the above equation, we get
\[\Rightarrow 9v-6+6=12+6\]
\[\Rightarrow 9v=18\]
Dividing both sides of the above equation by 9, we get
\[\begin{align}
  & \Rightarrow \dfrac{9v}{9}=\dfrac{18}{9} \\
 & \therefore v=2 \\
\end{align}\]
Hence, the solution of the given equation is \[v=2\].

Note: before solving an equation, first check the degree of the equation. The method to solve an equation depends on the degree of the equation. For example, linear equations can be solved by taking all variable terms to one side, and constant to another side of the equation. Whereas with quadratic equations we can use different methods to find the roots of quadratic equations, the method includes formula method, factorization method, and completing the square method.