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Chapter 13: Problem 38

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}-y^{2}-4 x+6 y-4=0 \\ x^{2}+y^{2}-4 x-6 y+12=0 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are (2, 3) and (-2, -3).

Step by step solution

01

Rearrange the Equations

Rewrite these equations in terms of complete squares to match them to the format \((x - h)^2 + (y - k)^2 = r^2\). The given equations when rearranged yield: \((x - 2)^2 - (y - 3)^2 = 7\) and \((x - 2)^2 + (y + 3)^2 = 7\).
02

Substitute. Set the two equations equal to each other

Since we have two expressions for \((x - 2)^2\), we can set them equal to each other: \(-(y - 3)^2 = (y + 3)^2\)
03

Simplify and Solve

After expanding and simplifying, the equation can be given as: \(-y^2 + 6y - 9 = y^2 + 6y + 9\). Subtract \(y^2 + 6y + 9\) from both sides to get \(2y^2 = 18\), from where, \(y = \pm 3\)
04

Solve for x

Substitute \(y = 3\) and \(y = -3\) into the first given equation, we have \(x = 2, -2\) respectively.
05

Check the Solutions

It's always a good practice to check the solutions in both equations to make sure they are correct. Substitute values of x and y into the original equations to check the solutions. Doing so for \(x = 2, y = 3\), and \(x = -2, y = -3\) confirms that they are indeed solutions for the system.

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Most popular questions from this chapter

Solve the systems. $$\left\\{\begin{array}{l} \log x^{2}=y+3 \\ \log x=y-1 \end{array}\right.$$

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$4 x^{2}=36+y^{2}$$

Will help you prepare for the material covered in the next section. Solve by the addition method: $$\left\\{\begin{array}{l}2 x+4 y=-4 \\\3 x+5 y=-3\end{array}\right.$$

How can you distinguish ellipses from circles by looking at their equations?

The equation of a parabola is given. Determine: a. if the parabola is horizontal or vertical. b. the way the parabola opens. c. the vertex. $$x=-(y+3)^{2}+4$$
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