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Chapter 13: Problem 32

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}+4 y^{2}=20 \\ x y=4 \end{array}\right.$$

Short Answer

Expert verified
The solutions to the system of equations are (2, 2), (-2, -2), (4, 1), and (-4, -1).

Step by step solution

01

Choose a method and an equation to solve for a variable

Decide to use substitution for this system because one of the equations is already solved for a variable. From the second equation, \(xy = 4\), it's straightforward to solve for \(x\): \(x = \frac{4}{y}\).
02

Substitute the expression of \(x\) into the first equation

Substitute \(x = \frac{4}{y}\) into the first equation \(x^{2}+4 y^{2}=20\). This gives \(\left(\frac{4}{y}\right)^{2} + 4y^{2} = 20\), which simplifies to \(16/y^{2} + 4y^{2} = 20\). Multiply every term by \(y^{2}\) to remove the fraction: \(16 + 4y^{4} = 20y^{2}\). This simplifies to \(4y^{4} - 20y^{2} + 16 = 0\).
03

Solve the resulted fourth degree equation

This is a quadratic in terms of \(y^{2}\), and it can be written as \(4(y^{2})^{2} - 20(y^{2}) + 16 = 0\). Let \(z = y^{2}\), substitute this into the equation to get \(4z^{2} - 20z + 16 = 0\). Then divide by 4 to simplify the equation to \(z^{2} - 5z + 4 = 0\). This factors to \((z - 4)(z - 1) = 0\). So \(z = 4\) or \(z = 1\).
04

Solve for \(y\)

Substitute \(z = y^{2}\) back into \(z = 4\) and \(z = 1\). This gives us \(y^{2} = 4\) which means \(y = 2\) or \(y = -2\), and \(y^{2} = 1\) which means \(y = 1\) or \(y = -1\).
05

Solve for \(x\)

Substitute these \(y\) values back into the equation \(x = \frac{4}{y}\) gives four solutions: \((x, y) = (2, 2), (-2, -2), (4, 1), (-4, -1)\).

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Most popular questions from this chapter

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$x=(y-4)^{2}-1$$

Will help you prepare for the material covered in the next section. Solve by the substitution method: $$\left\\{\begin{array}{c}4 x+3 y=4 \\\y=2 x-7\end{array}\right.$$

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Graph: \(f(x)=2^{1-x}\). (Section \(12.1,\) Example 4 )

(GRAPH CANT COPY) Find the coordinates of the vertex for the horizontal parabola defined by the given equation. $$x=y^{2}-6 y+6$$
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