Solving Systems Of Equations Graphically X^2 + Y^2 = 4 And X - Y = 1

In the realm of mathematics, solving systems of equations graphically provides a powerful visual approach to understanding the solutions. This method is particularly insightful when dealing with systems involving both linear and non-linear equations. In this article, we will delve into the system:

$$\left\{ \begin{array}{r} x^2+y^2=4 \\ x-y=1 \end{array} \right.$$

and explore how to determine the graph that represents its solution. We'll break down each equation, understand their graphical representations, and then analyze their intersection points to find the solution set. This comprehensive exploration will not only help you solve this specific problem but also equip you with the skills to tackle similar problems in the future.

Understanding the Equations

To effectively solve this system graphically, we must first understand the individual equations and their graphical representations. Let's analyze each equation separately:

Equation 1: $x^2 + y^2 = 4$

This equation represents a circle in the Cartesian coordinate system. The general form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. In our case, the equation $x^2 + y^2 = 4$ can be rewritten as $(x - 0)^2 + (y - 0)^2 = 2^2$. Comparing this to the general form, we can identify the center as $(0, 0)$ and the radius as $2$. Therefore, the first equation represents a circle centered at the origin with a radius of 2. Graphing this circle involves plotting points that are 2 units away from the origin in all directions, creating a perfectly round shape.

Understanding this circular representation is crucial for visualizing the solution set of the system. The circle acts as one boundary within which the solutions must lie. By recognizing this equation as a circle, we can leverage our knowledge of circles to predict and interpret the graphical solution more effectively. The radius of 2 also gives us a scale for the graph, helping us to accurately plot the circle and identify potential intersection points with the other equation.

The ability to recognize the standard form of a circle's equation is a fundamental skill in analytic geometry. It allows us to quickly identify key features such as the center and radius, which are essential for graphing and solving related problems. This understanding will be invaluable as we move on to analyzing the second equation and finding the points where the two graphs intersect.

Equation 2: $x - y = 1$

This equation represents a straight line. To understand its graphical representation, we can rewrite the equation in the slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. Rearranging $x - y = 1$, we get $y = x - 1$. From this form, we can see that the slope ($m$) is $1$ and the y-intercept ($b$) is $-1$. This means the line rises one unit for every one unit it moves to the right, and it crosses the y-axis at the point $(0, -1)$.

Graphing this line involves plotting the y-intercept and then using the slope to find other points on the line. For example, since the slope is 1, we can move one unit to the right and one unit up from the y-intercept to find another point, such as $(1, 0)$. Connecting these points gives us the straight line represented by the equation. The line extends infinitely in both directions, indicating that there are an infinite number of points that satisfy the equation. However, only the points that also lie on the circle will be solutions to the system of equations.

Recognizing that this equation represents a line is essential for solving the system graphically. The line acts as another boundary, and the solutions to the system must lie on both the circle and the line. Understanding the slope and y-intercept allows us to accurately graph the line and visually identify the points where it intersects the circle. This graphical representation is key to finding the solution set of the system.

Graphical Representation of the System

Now that we understand the individual equations, let's visualize the system graphically. We have a circle centered at the origin with a radius of 2, and a straight line with a slope of 1 and a y-intercept of -1. To solve the system, we need to find the points where these two graphs intersect. These intersection points represent the solutions to the system because they are the only points that satisfy both equations simultaneously.

Visualizing the Intersection

Imagine plotting both the circle and the line on the same coordinate plane. The circle will be a smooth curve surrounding the origin, while the line will cut across the plane at an angle determined by its slope. The points of intersection are where the line crosses the circle. These points are crucial because their coordinates represent the $x$ and $y$ values that satisfy both the equation of the circle ($x^2 + y^2 = 4$) and the equation of the line ($x - y = 1$).

By visualizing the intersection, we can get a sense of how many solutions to expect. A line can intersect a circle at zero, one, or two points. If the line doesn't intersect the circle at all, the system has no real solutions. If the line is tangent to the circle, touching it at only one point, the system has one solution. If the line passes through the circle, crossing it at two distinct points, the system has two solutions. In our case, the line $y = x - 1$ appears to intersect the circle at two points, suggesting that the system has two solutions.

The graphical method provides a clear visual representation of the solutions, making it easier to understand the relationship between the equations. It also helps in verifying the solutions obtained algebraically. After finding the intersection points graphically, we can substitute their coordinates back into the original equations to confirm that they satisfy both equations.

Finding the Intersection Points

While the graphical method gives us a visual understanding of the solutions, we need to determine the exact coordinates of the intersection points. This can be done by carefully plotting the graphs and reading the coordinates directly, or by using algebraic methods to solve the system analytically.

Graphical Estimation

By plotting the circle $x^2 + y^2 = 4$ and the line $y = x - 1$ on a coordinate plane, we can visually estimate the coordinates of the intersection points. The accuracy of this method depends on the precision of the graph. However, it provides a quick way to get an approximate solution. Looking at the graph, the intersection points appear to be around $(1.6, 0.6)$ and $(-0.6, -1.6)$. These are rough estimates, and we will verify them using algebraic methods for exact solutions.

Estimating the solutions graphically is a valuable skill because it allows us to check the reasonableness of our algebraic solutions. If our algebraic solutions are significantly different from the graphical estimates, it suggests that we may have made an error in our calculations. The graphical method also helps in understanding the nature of the solutions. For example, if the line appears to be tangent to the circle, we would expect to find only one solution algebraically.

Algebraic Verification

To verify these estimates and find the exact solutions, we can substitute the points we identified graphically back into the original equations. This is a crucial step in ensuring that our solutions are correct. If a point satisfies both equations, it is indeed a solution to the system.

Let's take our first estimated point, $(1.6, 0.6)$. Substituting these values into the equation of the circle, we get:

$$(1.6)^2 + (0.6)^2 = 2.56 + 0.36 = 2.92$$

This is close to 4, but not exactly equal, suggesting that our estimate is not perfect. Now, let's substitute the same point into the equation of the line:

$$1.6 - 0.6 = 1$$

This equation is satisfied, which means our point lies on the line, but not perfectly on the circle. This discrepancy highlights the limitations of graphical estimation and the need for algebraic solutions for precision.

Algebraic Solution: Substitution Method

To find the exact solutions, we will use the substitution method. This involves solving one equation for one variable and substituting that expression into the other equation. This transforms the system into a single equation with one variable, which can then be solved using algebraic techniques.

Steps for Substitution

1. Solve one equation for one variable: From the equation $x - y = 1$, we can easily solve for $x$ in terms of $y$:

   $$x = y + 1$$

   This gives us an expression for $x$ that we can substitute into the other equation.

2. Substitute into the other equation: Now, substitute the expression $x = y + 1$ into the equation of the circle $x^2 + y^2 = 4$:

   $$(y + 1)^2 + y^2 = 4$$

   This equation now involves only the variable $y$, making it solvable.

3. Simplify and solve the resulting equation: Expand and simplify the equation:

   $$y^2 + 2y + 1 + y^2 = 4$$

   Combine like terms:

   $$2y^2 + 2y + 1 = 4$$

   Subtract 4 from both sides to set the equation to zero:

   $$2y^2 + 2y - 3 = 0$$

   This is a quadratic equation in the form $ay^2 + by + c = 0$. We can solve it using the quadratic formula.

4. Apply the quadratic formula: The quadratic formula is given by:

   $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

   In our equation, $a = 2$, $b = 2$, and $c = -3$. Substituting these values into the formula, we get:

   $$y = \frac{-2 \pm \sqrt{2^2 - 4(2)(-3)}}{2(2)}$$

   $$y = \frac{-2 \pm \sqrt{4 + 24}}{4}$$

   $$y = \frac{-2 \pm \sqrt{28}}{4}$$

   $$y = \frac{-2 \pm 2\sqrt{7}}{4}$$

   $$y = \frac{-1 \pm \sqrt{7}}{2}$$

   This gives us two solutions for $y$:

   $$y_1 = \frac{-1 + \sqrt{7}}{2}$$

   $$y_2 = \frac{-1 - \sqrt{7}}{2}$$

5. Find the corresponding $x$ values: Substitute each $y$ value back into the equation $x = y + 1$ to find the corresponding $x$ values:

   For $y_1 = \frac{-1 + \sqrt{7}}{2}$:

   $$x_1 = \frac{-1 + \sqrt{7}}{2} + 1 = \frac{1 + \sqrt{7}}{2}$$

   For $y_2 = \frac{-1 - \sqrt{7}}{2}$:

   $$x_2 = \frac{-1 - \sqrt{7}}{2} + 1 = \frac{1 - \sqrt{7}}{2}$$

   Thus, we have two solution points:

   $$\left(\frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2}\right)$$

   $$\left(\frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2}\right)$$

The Solution Set and Graph Representation

Now that we have found the exact solutions algebraically, we can represent them graphically. The solution set consists of the two points where the line $x - y = 1$ intersects the circle $x^2 + y^2 = 4$. These points are:

$$\left(\frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2}\right) \approx (1.82, 0.82)$$

$$\left(\frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2}\right) \approx (-0.82, -1.82)$$

Graphing the Solutions

The graph that represents the solution of the system will show a circle centered at the origin with a radius of 2, and a line that intersects the circle at the two points we calculated. This graphical representation is a visual confirmation of our algebraic solutions.

If we were presented with multiple graphs, the correct one would be the one that accurately depicts these features: a circle centered at (0, 0) with a radius of 2, a line with a slope of 1 and a y-intercept of -1, and the intersection points at approximately (1.82, 0.82) and (-0.82, -1.82).

Understanding how to translate algebraic solutions into graphical representations is a fundamental skill in mathematics. It allows us to visualize the solutions and gain a deeper understanding of the relationship between equations and their graphical counterparts. The combination of algebraic and graphical methods provides a powerful approach to solving systems of equations and understanding their solutions.

Conclusion

In conclusion, solving the system of equations

$$\left\{ \begin{array}{r} x^2+y^2=4 \\ x-y=1 \end{array} \right.$$

requires a combination of graphical understanding and algebraic techniques. By recognizing the equations as a circle and a line, we can visualize the solution set as the intersection points of these graphs. Using the substitution method, we found the exact solutions:

$$\left(\frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2}\right)$$

$$\left(\frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2}\right)$$

These solutions are represented graphically by the points where the line $x - y = 1$ intersects the circle $x^2 + y^2 = 4$. The ability to solve systems of equations graphically and algebraically is a crucial skill in mathematics, providing a comprehensive understanding of the solutions and their representations. This approach not only helps in solving specific problems but also builds a strong foundation for tackling more complex mathematical challenges in the future.