Chapter 11, Problem 77RE

To determine

Solve each system of equations.

Answer to Problem 77RE

  $\boxed{(1,-1)}$

Explanation of Solution

Given information:

Solve each system of equations.

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

Calculation:

Multiply both the sides of the second equation in the system by $y$ to eliminate the fraction.

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

An equivalent system of equation is formed.

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

Subtract $x^2-x+y^2+y=0$ from $x^2-3x+y^2+y=-2$ to eliminate $x$ .

  $\begin{array}{l}\{\begin{array}{l}{x}^2-3x+y^2+y=-2\\ \begin{array}{c}{x}^2-x+y^2+y=0\end{array}\end{array}\\ \begin{array}{ccccc}& & & -2x=-2& \end{array}\end{array}$

Divide by $-2$ on both the sides.

  $\frac{-2}{-2}x=\frac{-2}{-2}$

  $x=1$

Back-substitute the value of $x$ in equation $1$ to find $y$ .

  ${\left(1\right)}^2-3\left(1\right)+y^2+y=-2$

  $-2+y^2+y=-2$

Add 2 to both the sides of the equation.

  $-2+y^2+y+2=-2+2$

  $y^2+y=0$

Factor the expression.

  $y(y+1)=0$

By the zero product property, we have either $Y=0$ or $Y+1=0$ . So, $Y=0$ or $-1$ .

Hence, the solutions are $x=1$ and $y=-1$ or the ordered pair $(1,-1)$ .