Solve each system of equations by using substitution

$\begin{array}{c}6c+3d=12\\ 2c=8-d\end{array}$

The algebraic method of substitution involves solving the one of the two equations for one variable in terms of other variable and then substituting the expression so formed for the variable in the second equation.

To solve the equation $2c=8-d$ for cin terms of d, divide 2both sides as shown below.

$\begin{array}{c}2c=8-d\\ \frac{2c}{2}=\frac{8-d}{2}\\ c=\frac{8-d}{2}\end{array}$

Now, substitute $c=\frac{8-d}{2}$in the equation $6c+3d=12$ and solve for d.

$\begin{array}{c}6c+3d=12\\ 6\left(\frac{8-d}{2}\right)+3d=12\\ 3\left(8-d\right)+3d=12\\ 24-3d+3d=12\end{array}$

Simplify it further as

$\begin{array}{c}24-3d+3d=12\\ 24=12\end{array}$

Since, the equation $24=12$ 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.