Chapter 11, Problem 43RE

To determine

To find: The solution of system of equations.

Answer to Problem 43RE

The solution of the system of equations can be $x=1$ and $y=0\text{or}-1$

Explanation of Solution

Given information:

The equations are $x^2-3x=y^2+y=-2$ and $\frac{x^2-x}{y}+y+1=0$

Calculation:

Consider the equations.

  $x^2-3x+y^2+y=-2$ …… (1)

  $\frac{x^2-x}{y}+y+1=0$ …… (2)

Multiply both sides of Equation (2) by $y$ .

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

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

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

Divide by $-2$ both the sided of above equation.

  $x=1$ .

Substitute $x=1$ in Equation (1).

  $\begin{array}{l}{1}^2-3\left(1\right)+y^2+y=-2\\ y^2+y=0\\ y\left(y+1\right)=0\\ y=0\text{or}y=-1\end{array}$

Therefore, the solution of the system of equations can be $x=1$ and $y=0\text{or}-1$