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Chapter 10: Problem 40

Solve. Remember that graphs can be used to confirm all real solutions. $$\begin{aligned}&x^{2}+y^{2}=13\\\&x y=6\end{aligned}$$

Short Answer

Expert verified
The solutions are (3, 2), (-3, -2), (2, 3), and (-2, -3).

Step by step solution

01

Express One Variable in Terms of the Other

From the second equation, solve for one variable in terms of the other. We'll solve for y in terms of x from the equation: \[ xy = 6 \] This gives: \[ y = \frac{6}{x} \]
02

Substitute into the First Equation

Substitute \( y = \frac{6}{x} \) into the first equation \( x^2 + y^2 = 13 \). This gives: \[ x^2 + \left( \frac{6}{x} \right)^2 = 13 \] Simplify to: \[ x^2 + \frac{36}{x^2} = 13 \]
03

Multiply by \( x^2 \)

Multiply the entire equation by \( x^2 \) to clear the fraction: \[ x^4 + 36 = 13x^2 \] Rearrange to form a quadratic in terms of \( x^2 \): \[ x^4 - 13x^2 + 36 = 0 \]
04

Substitute \( u = x^2 \)

Let \( u = x^2 \). The equation becomes: \[ u^2 - 13u + 36 = 0 \]
05

Solve the Quadratic Equation

Solve the quadratic equation \( u^2 - 13u + 36 = 0 \) using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ u = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2} \] This gives two solutions: \[ u = 9 \] and \[ u = 4 \] Thus, \( x^2 = 9 \) or \( x^2 = 4 \).
06

Solve for x

Solve for \( x \) from \( x^2 = 9 \) and \( x^2 = 4 \): \[ x = \pm 3 \] and \[ x = \pm 2 \]
07

Solve for y

Using \( y = \frac{6}{x} \), find the corresponding \( y \) values: For \( x = 3 \), \( y = 2 \); For \( x = -3 \), \( y = -2 \); For \( x = 2 \), \( y = 3 \); For \( x = -2 \), \( y = -3 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Quadratic Equations
Quadratic equations are a type of polynomial equation where the highest power of the variable is squared (\(x^2\) or \(y^2\)). These equations usually take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. To solve these equations, we can use methods like:
  • Factoring
  • Completing the square
  • Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our exercise above, we transformed it into a quadratic by substituting \(u = x^2\). The resulting quadratic equation was then solved using the quadratic formula, yielding solutions for \(x^2\).
Substitution Method
The substitution method is a technique for solving systems of equations by expressing one variable in terms of another. Here's how it works:
  1. Solve one of the equations for one variable in terms of the other. In our example, we solved \(xy = 6\) for \(y\), yielding \(y = \frac{6}{x}\)
  2. Substitute this expression into the other equation. By doing this, the system of equations reduces to one equation with a single variable.
This method simplifies complex equations and makes it easier to find solutions. The next step often involves solving the transformed equation using another algebraic method, like the quadratic formula in this case.
Graphical Solution Verification
Graphing equations provides a visual way to confirm solutions. Here's how you can verify solutions graphically:
  • Plot each equation on the same graph. Each equation will produce a curve.
  • The points where the curves intersect represent the solutions to the system of equations. In our problem, plotting \(x^2 + y^2 = 13\) (a circle) and \(xy = 6\) (a hyperbola) reveals 4 intersection points: (3, 2), (-3, -2), (2, 3), and (-2, -3).
  • Check if these points satisfy both equations. True intersections confirm the algebraic solutions.
This approach not only verifies solutions but also helps understand the relationships between equations. For students, seeing the graphical representation can make the concepts more intuitive and comprehensible.

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