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Chapter 13: Problem 33

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}+4 y^{2}=20 \\ x+2 y=6 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 2\), \(y = 2\)

Step by step solution

01

Express one variable in terms of the other

In the second equation, express \(x\) as \(6 - 2y\)
02

Substitute in the first equation

Replace \(x\) in the first equation with \(6 - 2y\), yielding \((6 - 2y)^{2} + 4y^{2} = 20\)
03

Simplify the equation

Expand and simplify the equation to get a quadratic equation in terms of \(y\), \(-4y^{2} + 24y - 16 = 0\)
04

Solve the quadratic equation

Solving the quadratic equation we get \(y = 2\) or \(y = 2\). As result is same, only one value of \(y\) is valid.
05

Substitute y in the second equation

Then, substitute \(y = 2\) into the second equation to get \(x = 2\)
06

Check the solution

Substitute the values \(x = 2\) and \(y = 2\) in the original system of equations to verify that the solution is correct.

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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. $$x^{2}+4 y^{2}=16$$

Use the vertex and intercepts to sketch the graph of each equation. If needed, find additional points on the parabola by choosing values of y on each side of the axis of symmetry. $$x=(y-2)^{2}-4$$

If \(f(x)=\frac{1}{3} x-5,\) find \(f^{-1}(x) .\) (Section 8.4, Example 4)

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y=-x^{2}-4 x+4$$

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$4 x^{2}+y^{2}=16$$
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