Q. 43 - Precalculus Enhanced with Graphing Utilities

Question

In problems 39-43, Solve each system of equations 

$\left\{\begin{array}{l}{x}^2-3x+y^2+y=-2\\ \frac{x^2-x}{y}+y+1=0\end{array}\right.$

Step-by-Step Solution

Answer

Solutions of the system of equations$\left\{\begin{array}{l}{x}^2-3x+y^2+y=-2\\ \frac{x^2-x}{y}+y+1=0\end{array}\right.$ are $(1,0) \& (1,-1)$

Step 1. Given data

The given system of equation is  

$\left\{\begin{array}{l}{x}^2-3x+y^2+y=-2\\ \frac{x^2-x}{y}+y+1=0\end{array}\right.$

Step 2. Rearrange system of equation

multiply the equation $\frac{x^2-x}{y}+y+1=0$by y

$$\begin{gathered} \left(\frac{x^2-x}{y}+y+1\right)y=\left(0\right)y \\ {x}^2-x+y^2+y=0 \end{gathered}$$

so a new system of equation is

$\left\{\begin{array}{l}{x}^2-3x+y^2+y=-2\\ x^2-x+y^2+y=0\end{array}\right.$

Step 3. determine the values of variable

Determine the difference between both equations of a system of equation

$$\begin{gathered} \left(x^2-3x+y^2+y\right)-\left(x^2-x+y^2+y\right)=-2-0 \\ -2x=-2 \\ x=1 \end{gathered}$$

Step 4. Solution of the system of equations

Substitute $x=1$ in equation $x^2-3x+y^2+y=-2$

$$\begin{gathered} 1^2-3\left(1\right)+y^2+y=-2 \\ {y}^2+y=0 \\ y(y+1)=0 \\ y=0,-1 \end{gathered}$$

So solutions of the system of equations are $(1,0) \& (1,-1)$