Solve each equation. Check your solution.

$2\left(4r+6\right)=\frac{2}{3}\left(12r+18\right)$

The given equation is “$2\left(4r+6\right)=\frac{2}{3}\left(12r+18\right)$”.

Solving,

$2\left(4r+6\right)=\frac{2}{3}\left(12r+18\right)$ (Given equation)

$8r+12=\frac{2}{3}\times 12r+\frac{2}{3}\times 18$ (Distributive property)

$8r+12=8r+12$ (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 r.

Put $r=1$in the given equation $2\left(4r+6\right)=\frac{2}{3}\left(12r+18\right)$to get,

$\begin{array}{l}2\left(4+6\right)=\frac{2}{3}\left(12+18\right)\\ 2\left(10\right)=\frac{2}{3}\left(30\right)\\ 20=2\left(10\right)\\ 20=20\end{array}$

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

Put $r=0$in the given equation $2\left(4r+6\right)=\frac{2}{3}\left(12r+18\right)$to get,

$\begin{array}{l}2\left(4\left(0\right)+6\right)=\frac{2}{3}\left(12\left(0\right)+18\right)\\ 2\left(0+6\right)=\frac{2}{3}\left(0+18\right)\\ 2\left(6\right)=\frac{2}{3}\left(18\right)\\ 12=2\left(6\right)\\ 12=12\end{array}$

Since the values on the left-hand side and the right-hand side are equal, it can be said that $r=0$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.