Chapter 3.5, Problem 60E

To determine

To calculate: The solution of the system of equations.

Answer to Problem 60E

The required solutions of the system of equations are $x=2,y=0$ and $x=0,y=2$ .

Explanation of Solution

Given information:

The system of equations:

  $x^2+y^2=4$

  $2y=x^2-2x+4$

  $y=-x+2$

Calculation:

The given equations are

  $x^2+y^2=4$........................... (1)

  $2y=x^2-2x+4$........................... (2)

  $y=-x+2$........................... (3)

Substitute $-x+2$ for y in the equation (1).

  $\begin{array}{c}{x}^2+{\left(-x+2\right)}^2=4\\ x^2+x^2-4x+4-4=0\\ 2x^2-4x=0\\ 2x\left(x-2\right)=0\end{array}$........................... (4)

Substitute $-x+2$ for y in the equation (2).

  $\begin{array}{c}2\left(-x+2\right)=x^2-2x+4\\ -\cancel{2x}+\cancel{4}=x^2-\cancel{2x}+\cancel{4}\\ x^2=0\\ x=0\end{array}$

From equation (4) solve for x .

  $\begin{array}{c}2x=0\\ x=0\end{array}$

And, $\begin{array}{c}x-2=0\\ x=2\end{array}$

The values of x are 0, 2.

Substitute 0 for x in the equation (3).

  $\begin{array}{c}y=-\left(0\right)+2\\ =2\end{array}$

Substitute 2 for x in the equation (3).

  $\begin{array}{c}y=-\left(2\right)+2\\ =-2+2\\ =0\end{array}$

Hence, the required solutions of the system of equations are $x=2,y=0$ and $x=0,y=2$ .