Problem 35 - Precalculus: Mathematics for Calculus

Question

Use the graphical method to find all solutions of the system of equations, rounded to two decimal places. $$\left\\{\begin{array}{l} x^{2}+y^{2}=25 \\ x+3 y=2 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are approximately \((3.58, -0.53)\) and \((-3.58, 1.86)\).

Step 1: Understand the equations

The system consists of two equations: 1. \( x^2 + y^2 = 25 \) is the equation of a circle centered at the origin (0,0) with radius 5.2. \( x + 3y = 2 \) is a linear equation with a slope of -1/3 that crosses the y-axis at (0,2/3).

Step 2: Graph the Circle

Plot the circle defined by \( x^2 + y^2 = 25 \). This circle is centered at the origin (0,0) and has a radius of 5, reaching points like (5,0), (-5,0), (0,5), and (0,-5) on the coordinate axes.

Step 3: Graph the Line

Plot the line from the equation \( x + 3y = 2 \). This can be rewritten as \( y = -\frac{1}{3}x + \frac{2}{3} \). Begin plotting at y-intercept (0, 2/3) and use the slope of -1/3 to find another point such as (3, 0). Draw the line through these points.

Step 4: Determine Intersection Points

Find where the circle and line intersect. The intersections of these graphs are the solution to the system. By inspecting the graph, approximate the intersection points.

Step 5: Solve Algebraically for Precision

For precision, solve the equations simultaneously:- Substitute \( y = -\frac{1}{3}x + \frac{2}{3} \) into \( x^2 + y^2 = 25 \): \[ x^2 + \left(-\frac{1}{3}x + \frac{2}{3}\right)^2 = 25 \]- Simplify and solve the quadratic equation for \( x \), then substitute back to find \( y \). Approximating for decimal values is necessary.

Step 6: Solve the Quadratic

Solving the above quadratic equation results in two potential solutions for \( x \). Approximating using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), yields specific values for \( x \). Corresponding \( y \) values are obtained by substitution back into the line equation.

Key Concepts

Understanding System of Equations

A system of equations is a set of two or more equations with the same variables. The main goal in solving such a system is to find the values of the variables that satisfy all the given equations simultaneously. In this exercise, we have a system consisting of a circle equation and a linear equation. By using both graphical and algebraic methods, we aim to find the exact points where both equations hold true. This intersection reflects the solution of the system. Solving systems graphically allows you to visually see where these conditions are met. However, algebraic solutions provide more precision, especially when rounding numbers to decimal places.

Exploring Circle Equations

The equation of the circle given is \(x^2 + y^2 = 25\). This is a standard form of a circle equation, where \(x\) and \(y\) are variables representing points on the coordinate plane. The equation tells us that the circle is centered at the origin (i.e., the point (0, 0)) and has a radius of 5, since the square root of 25 is 5.
To graph this, start by marking the center of the circle at the origin. From there, measure a distance of 5 units in all directions to identify points like (5,0), (-5,0), (0,5), and (0,-5). Draw a smooth curve through these points to complete the circle. This visual representation helps identify possible regions of intersection with other graphs, such as a line.

Insight into Linear Equations

The linear equation part of the system is \(x + 3y = 2\). Linear equations like these describe straight lines on the graph. To make graphing easier, it may help to convert the equation into the slope-intercept form. By rearranging, you get \(y = -\frac{1}{3}x + \frac{2}{3}\).
Here, \(-\frac{1}{3}\) is the slope, which indicates that for every unit the line goes right on the x-axis, it falls \(\frac{1}{3}\) units down on the y-axis. The y-intercept is \(\frac{2}{3}\), meaning this is the point where the line crosses the y-axis. Start plotting from (0, \(\frac{2}{3}\)) and use the slope to find another point like (3,0), then draw the line through these plotted points.

Finding Intersection Points

Intersection points are crucial as they represent solutions to our system of equations. Graphically, these are the points where the line and the circle meet on the graph. Initially, you can make an educated guess as to where these points lie by inspecting the graph. Analytical methods, like substitution and solving quadratic equations, then help us find these points more precisely.
By substituting the linear equation \(y = -\frac{1}{3}x + \frac{2}{3}\) into the circle equation \(x^2 + y^2 = 25\), we can solve for \(x\) mathematically. This substitution leads to a quadratic expression that can be solved using the quadratic formula, providing exact x-coordinates for our solution points. Plug these x-values back into the linear equation to get the corresponding y-values, completing the intersection points as solutions.