To solve this, we need to propose two equations, one linear, and one quadratic.
We can propose:
[tex]x+y=12[/tex][tex]y=-2x^2+3x+20[/tex]The system is:
[tex]\begin{cases}x+y={12} \\ y={-2x^2+3x+20}\end{cases}[/tex]Now, we need to solve the system of equations. If we solve the first equation for y, then we can substitute in the second equation:
[tex]y=12-x[/tex]And substitute:
[tex]12-x=-2x^2+3x+20[/tex]Next, we want to leave a '0' on one side of the equation:
[tex]\begin{gathered} 0=-2x^2+3x+x+20-12 \\ . \\ 0=-2x^2+4x+8 \end{gathered}[/tex]We know how to find the solution to this equation, using the quadratic formula:
[tex]x_{1,2}=\frac{-4\pm\sqrt{4^2-4\cdot(-2)\cdot8}}{2\cdot(-2)}[/tex]And solve:
[tex]x_{1,2}=\frac{-4\pm\sqrt{16+64}}{-4}=\frac{4\pm\sqrt{80}}{4}=\frac{4\pm4\sqrt{5}}{4}=1\pm\sqrt{5}[/tex]Now, we can find the y-values of the solutions, using the first equation in the system:
[tex]\begin{gathered} (1+\sqrt{5})+y=12\Rightarrow y=12-(1+\sqrt{5})=11-\sqrt{5} \\ . \\ (1-\sqrt{5})+y=12\Rightarrow y=12-(1-\sqrt{5})=11+\sqrt{5} \end{gathered}[/tex]Thus, the two solutions to this system are:
[tex]\begin{gathered} Solution\text{ }1:(1+\sqrt{5},11-\sqrt{5})\approx(3.236,8.764) \\ Solution\text{ }2:(1-\sqrt{5},11+\sqrt{5})\approx(-1.236,13,236) \end{gathered}[/tex]And to verify, we can graph the system:
And see that our analytic result is correct.
