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  

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. Formation of the single-variable equation

Multiply y to both sides of the equation $\frac{x^2-x}{y}+y+1=0$

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

so the new system of equation is $\left\{\begin{array}{l}{x}^2-3x+y^2+y=-2 \cdots i\\ x^2-x+y^2+y=0 \cdots ii\end{array}\right.$

Step 3. Formation of the single-variable equation

Subtract equation ii  from equation i 

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

Step 4. Solution of the system of equations

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

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