Answer: (-2, -1) and (3, 14).
Step-by-step explanation:
[tex]\left \Bigg \{ { \bigg{y=x^{2} +2x-1} \atop \bigg {y-3x=5}} \right. ; \left \Bigg \{ { \bigg{y=x^{2} +2x-1} \atop \bigg {y=5+3x}} \right.[/tex]
x² + 2x - 1 = 5 + 3x
x² + 2x - 1 - 5 - 3x = 0
x² - x - 6 = 0
[tex]General formula\\\\x=\dfrac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\\\a=1; \: \: b=-1; \: \: c=-6\\\\x=\dfrac{1 \pm \sqrt{(-1)^{2}-4 \cdot1 \cdot (-6)} }{2 \cdot 1}=\dfrac{1 \pm\sqrt{25} } {2} =\dfrac{1 \pm5}{2} \\\\x_{1} = -2\\x_{2} =3[/tex]
y = 5 + 3x
y₁ = 5 + 3 * (-2) = -1
y₂ = 5 + 3 * 3 = 14
The pair of points representing the solution set of this system of equations is (-2, -1) and (3, 14).
