Chapter 11, Problem 41RE

To determine

To find: The solution of system of equations.

Answer to Problem 41RE

The solution of the system of equations can be $x=0$

  $x=\pm 3$ and $y=0\text{or}3$

Explanation of Solution

Given information:

The equations are $x^2+y^2=6y$ and $x^2=3y$

Calculation:

Consider the equations.

  $x^2+y^2=6y$ …… (1)

  $x^2=3y$ …… (2)

Substitute $3y$ for $x^2$ in equation (1).

  $\begin{array}{l}3y+y^2=6y\\ y^2-3y=0\\ y\left(y-3\right)=0\\ y=0\text{}\text{or}y=3\end{array}$

Substitute $y=0\text{}$ in equation (1).

  $\begin{array}{l}{x}^2+0^2=6\left(0\right)\\ x=0\end{array}$

Substitute $y=3$ in equation (1).

  $\begin{array}{l}{x}^2+3^2=6\left(3\right)\\ x^2=18-9\\ x=\sqrt{9}\\ x=\pm 3\end{array}$

Therefore, the solution of the system of equations can be $x=0$

  $x=\pm 3$ and $y=0\text{or}3$