Chapter 11, Problem 70RE

To determine

Solve each system of equations.

Answer to Problem 70RE

  $\boxed{(-4,0)and(2,\pm 2\sqrt{3})}$

Explanation of Solution

Given information:

Solve each system of equations.

  $\{\begin{array}{l}{x}^2+y^2=16\\ 2x-y^2=-8\end{array}$

Calculation:

The given system of non-linear equations in two variables is given by the expressions given below.

  $\{\begin{array}{l}{x}^2+y^2=16\\ 2x-y^2=-8\end{array}$

The given equations can be solved method by substitution. We obtain a relation of $y$ in terms of

  $x$ from the second equation and substitute in the first equation. The relation can be obtained as follows:

  $\begin{array}{l}2x-y^2=-8\\ y^2=2x+8\end{array}$....(1)

Now we can substitute the above obtained relation in the first equation to obtain the possible values of $x$ .

  $x^2+2x+8=16$

  $x^2+2x+1=9$

  ${(x+1)}^2={(\pm 3)}^2$

  $x+1\pm 3$

  $x=-4,2$

The obtained value of $x$ can be substituted in the equation $(1)$ to obtain the value of $y$ .

  $y^2=2(-4)+8$

  $=0$

  $y=0$

  $y^2=2(2)+8$

  $=12$

  $y=\pm 2\sqrt{3}$

Hence, the solutions to give equations are $(-4,0)and(2,\pm 2\sqrt{3})$ .