Answer
$$\left\{ {\left( {\frac{{38}}{{25}},\frac{{41}}{{25}}} \right),\left( { - 2, - 1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\,\,\,\,\,{x^2} + {y^2} = 5\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \, - 3x + 4y = 2\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{2}} \right){\text{ for }}y \cr
& y = \frac{{3x + 2}}{4} \cr
& {\text{Substitute }}\frac{{3x + 2}}{4}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& {x^2} + {\left( {\frac{{3x + 2}}{4}} \right)^2} = 5 \cr
& {x^2} + \frac{{9{x^2} + 12x + 4}}{{16}} = 5 \cr
& 16{x^2} + 9{x^2} + 12x + 4 = 80 \cr
& 25{x^2} + 12x - 76 = 0 \cr
& \left( {25x - 38} \right)\left( {x + 2} \right) = 0 \cr
& {x_1} = \frac{{38}}{{25}},\,\,\,{x_2} = - 2 \cr
& {\text{Substitute }}{x_1} = \frac{{38}}{{25}}{\text{ into the equation }}y = \frac{{3x + 2}}{4}{\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = \frac{{3\left( {38/25} \right) + 2}}{4} \cr
& y = \frac{{41}}{{25}} \cr
& {\text{The first solution is }}\left( {\frac{{38}}{{25}},\frac{{41}}{{25}}} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = - 2{\text{ into the equation }}y = \frac{{3x + 2}}{4}{\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = \frac{{3\left( { - 2} \right) + 2}}{4} \cr
& y = - 1 \cr
& {\text{The second solution is }}\left( { - 2, - 1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {\frac{{38}}{{25}},\frac{{41}}{{25}}} \right),\left( { - 2, - 1} \right)} \right\} \cr} $$
