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Chapter 13: Problem 41

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}+y^{2}+3 y=22 \\ 2 x+y=-1 \end{array}\right.$$

Short Answer

Expert verified
The solution of the system of equations will be the pair of ordered pairs (x, y) that satisfy both of the given equations.

Step by step solution

01

Isolate variable y in the second equation

Rearrange the second equation to isolate y. The equation becomes: \( y = -1 - 2x \).
02

Substitute the expression for y in the first equation

Substitute the expression of \( y = -1 - 2x \) in the first equation. The first equation \( x^{2}+y^{2}+3y=22 \) becomes: \( x^{2} + (-1 - 2x)^2 + 3(-1 - 2x) = 22 \). Simplify this expression.
03

Solve the quadratic equation

Simplify the equation to form a quadratic equation in terms of x and solve for x. This will give two potential values of x.
04

Substitute x values into the equation for y

Substitute the obtained x values separately in the rearranged equation from Step 1 to get the corresponding y values.

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Most popular questions from this chapter

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

How can you distinguish parabolas from other conic sections by looking at their equations?

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. $$x=-3 y^{2}-6 y$$

Will help you prepare for the material covered in the next section. Solve: \(x^{2}=2(3 x-9)+10\)

Find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. $$\left\\{\begin{array}{l}x=(y+2)^{2}-1 \\\ (x-2)^{2}+(y+2)^{2}=1\end{array}\right.$$
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