Question

In Problems 39–43, solve each system of equations.

$\left\{\begin{array}{l}{x}^2+y^2=6y\\ x^2=3y\end{array}\right.$

Short answer

Solutions of the system of equations$\left\{\begin{array}{l}{x}^2+y^2=6y\\ x^2=3y\end{array}\right.$ are$(0,0),(-3,3),\&(3,3)$

Step-by-step solution

Step 1. Given data 

The given system of equation is

$\left\{\begin{array}{l}{x}^2+y^2=6y\\ x^2=3y\end{array}\right.$

Step 2. Formation of the single-variable equation 

Substitute $x^2=3y$in equation $x^2+y^2=6y$

$$\begin{gathered} x^2+y^2=6y \\ 3y+y^2=6y \\ {y}^2-3y=0 \\ y(y-3)=0 \end{gathered}$$

Step 3. Solution of the single variable equation 

Use zero product property for $y(y-3)=0$

$$\begin{gathered} y-3=0 \\ y=3 \end{gathered}$$

and$y=0$

Step 4. Solution of the system of equations 

Substitute $y=0$in equation$x^2=3y$

$$\begin{gathered} x^2=3y \\ {x}^2=3\left(0\right) \\ {x}^2=0 \\ x=0 \end{gathered}$$

Substitute role="math" localid="1646948862177" $y=3$in equation$x^2=3y$

$$\begin{gathered} x^2=3y \\ {x}^2=3\left(3\right) \\ {x}^2=9 \\ x=\pm 3 \end{gathered}$$

So solutions of the system of equations are$(0,0),(-3,3),\&(3,3)$