Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Solve each nonlinear system of equations analytically. $$\begin{aligned}x^{2}+y^{2} &=5 \\\\-3 x+4 y &=2\end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \((x_1, y_1)\) and \((x_2, y_2)\) obtained from solving the quadratic equation.

Step 1: Identify the System of Equations

The given system consists of two equations: 1. \(x^2 + y^2 = 5\) which represents a circle centered at the origin with radius \(\sqrt{5}\) 2. \(-3x + 4y = 2\) which is a linear equation representing a line.

Step 2: Solve the Linear Equation for One Variable

From the linear equation \(-3x + 4y = 2\), isolate \(y\):\[ 4y = 3x + 2 \]\[ y = \frac{3}{4}x + \frac{1}{2} \]

Step 3: Substitute into the Nonlinear Equation

Substitute \(y = \frac{3}{4}x + \frac{1}{2}\) into the circle equation:\[ x^2 + \left(\frac{3}{4}x + \frac{1}{2}\right)^2 = 5 \]

Step 4: Simplify the Substitution

Expand \(\left(\frac{3}{4}x + \frac{1}{2}\right)^2\):\[ \left(\frac{3}{4}x + \frac{1}{2}\right)^2 = \left(\frac{3}{4}x\right)^2 + 2\cdot\frac{3}{4}x\cdot\frac{1}{2} + \left(\frac{1}{2}\right)^2 \]\[ = \frac{9}{16}x^2 + \frac{3}{4}x + \frac{1}{4} \]

Step 5: Formulate a Quadratic Equation

Substitute the expanded form into the circle equation:\[ x^2 + \frac{9}{16}x^2 + \frac{3}{4}x + \frac{1}{4} = 5 \]Combine like terms to form a quadratic equation:\[ \frac{25}{16}x^2 + \frac{3}{4}x - \frac{19}{4} = 0 \]

Step 6: Solve the Quadratic Equation

Multiply the entire equation by 16 to clear the fractions:\[ 25x^2 + 12x - 76 = 0 \]Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 25\), \(b = 12\), and \(c = -76\):\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 25 \cdot (-76)}}{2 \cdot 25} \]Simplify to find \(x\).

Step 7: Calculate Corresponding y-values

Once you have the values for \(x\), substitute them back into the equation \(y = \frac{3}{4}x + \frac{1}{2}\) to find the corresponding \(y\)-values.

Step 8: Verify the Solutions

Verify the solutions by substituting both \(x\) and \(y\) back into the original system to ensure both equations are satisfied.

Key Concepts

Circle Equations

A circle equation is a special type of nonlinear equation defining all the points equidistant from a central point. The standard form of a circle's equation is \[x^2 + y^2 = r^2\]where

• \(x\) and \(y\) are the coordinates of any point on the circle
• \(r\) represents the radius of the circle

In this system, the equation \(x^2 + y^2 = 5\) describes a circle centered at the origin \((0,0)\) with a radius of \(\sqrt{5}\).
This means any point on this circle is exactly \(\sqrt{5}\) units away from the center. By understanding this geometric property, one can visualize the circle's expansive boundary which forms a perfect round shape.
Solving such equations involves considering both algebraic manipulation and geometric interpretation to find possible points lying on the circle's circumference.

Linear Equations

Linear equations are equations of the first degree, meaning they form a straight line when graphed. Their standard form is expressed as \[ax + by = c\] where

• \(a\), \(b\), and \(c\) are constants
• \(x\) and \(y\) are variables

For the given system, the linear equation is \(-3x + 4y = 2\).
To solve systems involving non-linear equations, it's often effective to isolate one of the variables in the linear equation, making the other variable easier to handle.
By rearranging, we determined \(y\) as \(y = \frac{3}{4}x + \frac{1}{2}\). This conversion simplifies the problem, allowing substitution into other equations, such as quadratic or circle equations, to solve the overall system.
This linear equation essentially cuts across the circle, and the points of intersection are the solutions to the system. Understanding how to manipulate and graph these types of equations can provide vital clues in finding intersections of linear and nonlinear curves.

Quadratic Equations

Quadratic equations form another class of non-linear equations that appear as parabolas when graphed. The standard form of a quadratic equation is \[ax^2 + bx + c = 0\]where

• \(a\), \(b\), and \(c\) are constants
• \(x\) is the variable

In our exercise, a quadratic equation arises when substituting the expression of \(y\) obtained from the linear equation into the circle equation to eliminate \(y\), leading to \[\frac{25}{16}x^2 + \frac{3}{4}x - \frac{19}{4} = 0\]Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can solve for \(x\).
This formula provides the roots of the quadratic equation, which are the x-values where the parabola intersects the x-axis.
The solutions give potential x-values for our original system of equations. Each x-value found can be used to determine the corresponding y-value, revealing the intersection points on the circle and the line. Grasping quadratic equations and their properties opens up the understanding of solving complex systems involving both linear and non-linear equations.