Chapter 10.5, Problem 62E

To determine

To find: The points of intersection.

Answer to Problem 62E

The points of intersections of given equations are $\left(-3,0\right)$ and $\left(0,\frac{3}{2}\right)$ .

Explanation of Solution

Given information:

The given equations are as follows.

  $\begin{array}{c}xy+x-2y+3=0\\ x^2+4y^2-9=0\end{array}$

Calculation:

Using thegiven equations.

  $xy+x-2y+3=0$ ….(1)

  $x^2+4y^2-9=0$ ….(2)

Add the above both equations.

  $\begin{array}{c}xy+x-2y+3+\left(x^2+4y^2-9\right)=0\\ x^2+4y^2-xy+x-2y-6=0\end{array}$

Solve equation (1) for the values of $x$ .

  $\begin{array}{c}xy+x-2y+3=0\\ x=\frac{2y-3}{y+1}\end{array}$ …. (3)

Substitute the value of $x$ from equation (3) in equation (2).

  $\begin{array}{c}{\left(\frac{2y-3}{y+1}\right)}^2+4y^2-9=0\\ {\left(2y-3\right)}^2+4y^2{\left(y+1\right)}^2-9{\left(y+1\right)}^2=0\\ y\left(4y^3+8y^2-y-30\right)=0\end{array}$

The real solutions of he above equation are $y=0,\frac{3}{2}$

For the value of $y=0$ , the value of $x$ is.

  $\begin{array}{c}x=\frac{2\times 0-3}{0+1}\\ =-3\end{array}$

For the value of $y=\frac{3}{2}$ , the value of $x$ is.

  $\begin{array}{c}x=\frac{2\times \frac{3}{2}-3}{\frac{3}{2}+1}\\ =0\end{array}$

Then the points of intersections are $\left(-3,0\right)$ and $\left(0,\frac{3}{2}\right)$ .

The graph is shown in figure (1).

  

Figure (1)

The points of intersection are verified by moving cursor over the points of intersection in the graph.

Therefore, the points of intersections of given equations are $\left(-3,0\right)$ and $\left(0,\frac{3}{2}\right)$ .