Solve each equation. Check your solution.

$14v+6=2\left(5+7v\right)-4$

The given equation is “$14v+6=2\left(5+7v\right)-4$”.

Solving,

$14v+6=2\left(5+7v\right)-4$ (Given equation)

$14v+6=10+14v-4$ (Distributive property)

$14v+6=14v+6$ (Simplify)

The expressions on both sides are identical.

This implies that the given equation is an identity. So, it is true for all values of v.

Put $v=1$in the given equation $14v+6=2\left(5+7v\right)-4$to get,

$\begin{array}{l}14+6=2\left(5+7\right)-4\\ 20=2\left(12\right)-4\\ 20=24-4\\ 20=20\end{array}$

Since the values on the left-hand side and the right-hand side are equal, it can be said that $v=1$is a solution of the given equation.

Put $v=0$in the given equation $14v+6=2\left(5+7v\right)-4$to get,

$\begin{array}{l}14\left(0\right)+6=2\left(5+7\left(0\right)\right)-4\\ 0+6=2\left(5+0\right)-4\\ 6=2\left(5\right)-4\\ 6=10-4\\ 6=6\end{array}$

Since the values on the left-hand side and the right-hand side are equal, it can be said that $v=0$is a solution of the given equation.

Thus, it can be seen that a linear equation has at least two different solutions. This verifies that the given equation is an identity.