Chapter 8, Problem 32T

Solution

To find the solution of the system $4x^2+y^2=16,x+y=2$.

The solutions of the system are $\left(1,-2\right)$ and $\left(1,2\right)$.

Given information:

The system, $4x^2+y^2=16,x+y=2$.

Calculation:

The equation $x^2+y^2-2x-3=0$ can be written in standard form as ${\left(x-1\right)}^2+{\left(y-0\right)}^2=2^2$ . This is an ellipse with center $\left(1,0\right)$.

The equation $y^2-x^2+2x-5=0$ can be written as $\frac{{\left(y-0\right)}^2}{2^2}-\frac{{\left(x-1\right)}^2}{2^2}=1$ . This is a hyperbola with center at $\left(0,1\right)$.

Use a graphing calculator to graph the equation and observe the points of intersection the graphs.

  

The points of intersection are $\left(1,-2\right)$ and $\left(1,2\right)$.

Conclusion:

The solutions of the system are $\left(1,-2\right)$ and $\left(1,2\right)$.