ANSWER
The solutions are (-2, 1) and (1, 4)
EXPLANATION
(3) To solve the system algebraically, we have to graph both equations. The solutions of the system are the points where the graphs of the equations intersect. In this case, the first equation is a parabola pointing downward and the second equation is a line,
The solutions to the system are the points (-2, 1) and (1, 4)
(4) To solve the same system algebraically, we can use the elimination method. First, solve the second equation for y,
[tex]y=x+3[/tex]
Subtract the first equation from the second,
Now we have to solve this equation for x, using the quadratic formula,
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]
In this case, a = 1, b = 1 and c = -2,
[tex]x=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1}=\frac{-1\pm\sqrt[]{1+8}}{2}=\frac{-1\pm\sqrt[]{9}}{2}=\frac{-1\pm3}{2}[/tex]
The x-coordinates of the solutions are,
[tex]\begin{gathered} x_1=\frac{-1+3}{2}=\frac{2}{2}=1 \\ x_2=\frac{-1-3}{2}=\frac{-4}{2}=-2 \end{gathered}[/tex]
Now, to find the y-coordinates, we have to replace each of these values in one of the equations from the system. Replacing in the linear equation,
[tex]\begin{gathered} y=x_1+3=1+3=4 \\ y=x_2+3=-2+3=1 \end{gathered}[/tex]
Hence, the solutions are, as found before, (-2, 1) and (1, 4).
