Answer:
• Equations
[tex]y=x+2[/tex][tex]y=x^2[/tex]• Solutions (2, 4) and (–1, 1)
Explanation
System of equations
• Linear equation
[tex]y=x+2[/tex]• Quadratic equation
[tex]y=x^2[/tex]We have to set both equations to 0 and equalize them:
• 1. Setting them to 0:
[tex]0=x+2[/tex][tex]0=x^2[/tex]• 2. Equalizing them:
[tex]x+2=x^2[/tex]To solve the system using the General Quadratic Formula, we have to set the equation in the form ax² + bx+ c = 0:
[tex]x^2-x-2=0[/tex]Thus, in this case a = 1, b = -1 and c = -2. Using the formula:
[tex]x_{1,2}=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-2)}}{2(1)}[/tex][tex]x_{1,2}=\frac{1\pm\sqrt{1+8}}{2}[/tex][tex]x_{1,2}=\frac{1\pm\sqrt{9}}{2}[/tex]Finding both solutions:
[tex]x_1=\frac{1+3}{2}=\frac{4}{2}=2[/tex][tex]x_2=\frac{1-3}{2}=\frac{-2}{2}=-1[/tex]Finally, replacing these values in the linear equation to find y:
[tex]y_1=2+2=4[/tex][tex]y_2=-1+2=1[/tex]Therefore, the solutions are (2, 4) and (–1, 1).
