Solve each system of equations by using either substitution or elimination.

$\begin{array}{c}3g-2h=-1\\ 8h=5+12g\end{array}$

The algebraic method of elimination involves adding or subtracting the equations to eliminate one of the variables and forming new equation that is true. Sometimes, direct addition or subtraction of equations does not eliminate the variable then one equation requires formation of equivalent equation through multiplication so that one of the two variables has the same or opposite coefficient in both the equations. Multiplying the equation by a nonzero number, resulting new equation has same set of solutions.

To solve the equations, multiply $3g-2h=- 1$ by 4 and then add the resulting equation into other equation as shown below.

$\begin{array}{c}4\left(3g-2h\right)=4\left(-1\right)\\ 12g-8h=-4\end{array}$

Now, add $12g-8h=- 4$and $-12g+8h=5$.

$\begin{array}{ccccccc}& 12g& -& 8h& =& -& 4\\ -& 12g& +& 8h& =& & 5\\ & & & & & & \\ & 0& +& 0& =& & 1\end{array}$

Since, the equation $0=1$ is not true, so the given system of equations is an inconsistent system and no solution exist.

An independent system of equations is one in which the result on solving the equations, is an equation that can never be true, therefore no solution exists for such system of equations.

Hence, there is no solution for the provided system of equations.