Q. 41 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

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 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 $y=3$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 $(0,0), (-3,3), \& (3,3)$