Q. 50 - Precalculus Enhanced with Graphing Utilities

Question

Solve the system. Use any method you wish.

$\left\{\begin{array}{l}{x}^3-2x^2+y^2+3y-4=0\\ x-2+\frac{y^2-y}{x^2}=0\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given information.

Consider the given question,

$\left\{\begin{array}{l}{x}^3-2x^2+y^2+3y-4=0 ...... \left(i\right)\\ x-2+\frac{y^2-y}{x^2}=0 ...... \left(ii\right)\end{array}\right.$

Multiply equation (ii) by $x^2$,

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

Step 2. Substitute the value of y in equation (iii).

Subtract equations (i) and (iii),

$$\begin{gathered} \left(x^3-2x^2+y^2+3y-4\right)-\left(x^3-2x^2+y^2-y\right)=0 \\ 4y-4=0 \\ y-1=0 \\ y=1 \end{gathered}$$

Substitute the value of y in equation (iii),

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

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

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