Solve each system of equations by using substitution

$\begin{array}{c}2r+s=11\\ 6r-2s=-2\end{array}$

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 $2r+s=11$for sin terms of r, subtract 2r from both sides as shown below.

$\begin{array}{c}2r+s=11\\ s=11-2r\end{array}$

Now, substitute $s=11-2r$in the equation $6r-2s=- 2$ and solve for r.

$\begin{array}{c}6r-2s=-2\\ 6r-2\left(11-2r\right)=-2\\ 6r-22+4r=-2\\ 10r=20\end{array}$

Simplify it further by dividing 10 on both sides as

$\begin{array}{c}\frac{10r}{10}=\frac{20}{10}\\ r=2\end{array}$

Thus, the value of r is 2.

To find the value of s, substitute $r=2$in the equation $2r+s=11$ and then solve for s as shown.

$\begin{array}{c}2r+s=11\\ 2\left(2\right)+s=11\\ s=11-4\\ s=7\end{array}$

Thus, the value of s is 7.

Hence, the solution of the provided system of equations is $\left(2 , 7\right)$.