Chapter 11, Problem 78RE

To determine

Solve each system of equation.

Answer to Problem 78RE

  $\boxed{\begin{array}{l}x=0,-2,-1,1\\ y=0\end{array}}$

Explanation of Solution

Given information:

Solve each system of equation.

  $\left\{\begin{array}{l}{\chi }^2+\chi +y^2=y+2\\ \chi +1=\frac{2-y}{\chi }\end{array}\right\}$

Calculation:

The given system of equations in two variables is given by the expressions given below.

  $\left\{\begin{array}{l}{\chi }^2+\chi +y^2=y+2\\ \chi +1=\frac{2-y}{\chi }\end{array}\right\}$

The given equation can be solved method by substitution. We obtain a relation of $y$ in terms of $x$ from the second equation and substitute in the first equation. The relation can be obtained as follows.

  $\begin{array}{l}x+1=\frac{2-y}{x}\\ (x+1)x=2-y\\ y=2-x-x^2\\ =-(x^2+x-2)\\ =-(x-1)(x+2)\mathrm{.........}(1)\end{array}$

Now we can substitute the above obtained relation in the first equation to obtain the possible values of $y$ .

  $\begin{array}{}$ x^2+x+{(-(x-1)(x+2))}^2=-(x-1)(x+2)+2$\end{array}$

  $x^2+x-2+{(x^2+x-2)}^2=-(x^2+x-2)$

  ${(x^2+x-2)}^2+2(x^2+x)=0$(x2+x−2)(x2+x−2+2)=0(x−1)(x+2)(x2+x)=0x(x−1)(x+2)(x+1)=0x=−2,−1,0,1

The second given non-linear equation has $x$ in its denominator in factorised form. Therefore $x=0$ cannot be a solution to the given non-linear equation. The corresponding values of ordinates can be determined by substituting the abscissas of the solutions in the equation $(1).$

  $\begin{array}{l}y=-(-2-1)(-2+2)\\ =0\\ y=-(-1-1)(-1+2)\\ =2\\ y=-(1-1)(1+2)\\ =0\end{array}$

Hence, the solutions to the given set of non-linear equations are $\boxed{\begin{array}{l}x=0,-2,-1,1\\ y=0\end{array}}$