Answer
Both points $(\dfrac{38}{25}, \dfrac{41}{25})$ and $(-2,-1)$ are solution of the system.
Work Step by Step
We are given two system of equations: $x^2+y^2=5 \ and \ -3x+4y=2$
As depicted in the graph, there are two points $(\dfrac{38}{25}, \dfrac{41}{25})$ and $(-2,-1)$.
Our next step is to check the points satisfy both equations.
Plug $x=\dfrac{38}{25}$ and $y=\dfrac{41}{25}$ in the given equations of system to obtain:
$(\dfrac{38}{25})^2+(\dfrac{41}{25})^2=5\implies 5=5 (True)\quad and \quad , -3(-2)+(4)(-1)=2 \implies 2 = 2 (True)$
This implies that the point $(\dfrac{38}{25}, \dfrac{41}{25})$ is the solution or point of intersection of graphs.
Now, plug $x=-2$ and $y=-1$ in the given equations to obtain:
$ (2)^2+(-1)^2=5\implies 5=5 (True)\quad and \quad 6-4=2 \implies 2 = 2 (True)$
This implies that the point $(4, 3)$ is the solution or point of intersection of graphs.
Therefore, both points $(\dfrac{38}{25}, \dfrac{41}{25})$ and $(-2,-1)$ are solution of the system.
