Solve each system of equations by using elimination

$\begin{array}{c}2c+6d=14\\ \frac{1}{2}c-3d=8\end{array}$

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 $\frac{1}{2}c-3d=8$by 2 then add the resulting equation in the first equation as shown below.

$\begin{array}{c}2\left(\frac{1}{2}c-3d\right)=2\left(8\right)\\ c-6d=16\end{array}$

Now, add $c-6d=16$and $2c+6d=14$.

$\begin{array}{ccccccc}& c& -& 6d& =& & 16\\ & 2c& +& 6d& =& & 14\\ & & & & & & \\ & 3c& +& 0& =& & 30\end{array}$

Simplify it further as

$\begin{array}{c}3c=30\\ c=10\end{array}$

Thus, the value of c is $10$.

To find the value of d, substitute $c=10$in the equation $2c+6d=14$and then solve as shown.

$\begin{array}{c}2c+6d=14\\ 2\left(10\right)+6d=14\\ 20+6d=14\\ 6d=-6\end{array}$

Simplify it further as

$\begin{array}{c}6d=-6\\ d=-1\end{array}$

Thus, the value of d is $- 1$.

Hence, the solution of the provided system of equations is $\left(10 , - 1\right)$.