Q. 49 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

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

Step-by-Step Solution

Answer

The solution of system of equations are $\left(0,1\right),\left(0,-2\right),\left(2,-1\right)$.

Step 1. Given information.

Consider the given question,

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

Multiply equation (ii) with y,

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

Subtract equations (i) and (iii),

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

Step 2. Substitute the values in equation (iii).

Substitute $x=0$ in equation (iii),

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

Substitute $x=2$ in equation (iii),

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

$y=0$ cannot be considered as in equation (ii) y is in the denominator.

Therefore, the solution sets are  $\left(0,1\right),\left(0,-2\right),\left(2,-1\right)$.