Solving Systems Of Equations Graphically Plotting Real Solutions

In mathematics, solving a system of equations involves finding the values that satisfy all equations simultaneously. This article delves into solving a system of equations graphically, focusing on the specific example:

\left\{
    \begin{array}{c}
        y^2 + x^2 = 20 \\
        y = -x^2
    \end{array}
\right.

We will explore the steps involved in plotting the real solutions to this system on a coordinate grid.

Understanding the Equations

Before plotting, it's crucial to understand the nature of each equation. The first equation, y² + x² = 20, represents a circle. To recognize this, recall the standard equation of a circle centered at the origin (0, 0) with radius r: x² + y² = r². In our case, r² = 20, so the radius r is √20, which simplifies to 2√5. This means we have a circle centered at the origin with a radius of approximately 4.47 units.

The second equation, y = -x², represents a parabola. This is a quadratic function, and the negative sign in front of the x² term indicates that the parabola opens downwards. The vertex of this parabola is at the origin (0, 0).

Graphical Solution: Plotting and Finding Intersections

To solve the system graphically, we need to plot both the circle and the parabola on the same coordinate plane. The real solutions to the system are the points where the two graphs intersect. These intersection points represent the (x, y) pairs that satisfy both equations simultaneously.

Step 1: Plotting the Circle

To plot the circle x² + y² = 20, we can identify key points along the circle. Since the radius is 2√5 (approximately 4.47), we know the circle intersects the x-axis at (2√5, 0) and (-2√5, 0), and the y-axis at (0, 2√5) and (0, -2√5). These four points provide a good starting point for sketching the circle. We can also find additional points by choosing some x-values and solving for y (or vice versa), but it's often easier to sketch the circle using the radius and the center.

Step 2: Plotting the Parabola

To plot the parabola y = -x², we can create a table of values. Choose a few x-values, both positive and negative, and calculate the corresponding y-values.

• If x = 0, then y = -(0)² = 0.
• If x = 1, then y = -(1)² = -1.
• If x = -1, then y = -(-1)² = -1.
• If x = 2, then y = -(2)² = -4.
• If x = -2, then y = -(-2)² = -4.

Plot these points (0, 0), (1, -1), (-1, -1), (2, -4), and (-2, -4) on the coordinate plane. Connect the points with a smooth curve to form the parabola. Notice that the parabola is symmetric about the y-axis.

Step 3: Identifying Intersection Points

Once both the circle and the parabola are plotted, visually identify the points where the two graphs intersect. These points represent the real solutions to the system of equations. From the graph, it appears there are two intersection points. Estimating the coordinates of these points can give us approximate solutions. To find the exact solutions, we will need to use algebraic methods.

Algebraic Solution: Substitution Method

The graphical method provides a visual representation of the solutions and helps us estimate their values. However, to find the exact solutions, we can use algebraic methods such as substitution or elimination. In this case, substitution is a straightforward approach.

Step 1: Substitute

Since we have y = -x² from the second equation, we can substitute this expression for y into the first equation:

• (-x²)² + x² = 20

Step 2: Simplify and Rearrange

Simplify the equation:

• x⁴ + x² = 20

Rearrange the equation to form a quadratic-like equation:

• x⁴ + x² - 20 = 0

Step 3: Solve the Quadratic-like Equation

This equation is quadratic in form. Let u = x². Then the equation becomes:

• u² + u - 20 = 0

Factor the quadratic equation:

• (u + 5)(u - 4) = 0

Solve for u:

• u = -5 or u = 4

Step 4: Solve for x

Since u = x², we have:

• x² = -5 or x² = 4

For x² = -5, there are no real solutions for x, as the square of a real number cannot be negative. For x² = 4, we have two real solutions:

• x = 2 or x = -2

Step 5: Solve for y

Now, substitute the x-values back into the equation y = -x² to find the corresponding y-values:

• When x = 2, y = -(2)² = -4
• When x = -2, y = -(-2)² = -4

Therefore, the real solutions to the system of equations are (2, -4) and (-2, -4).

Plotting the Real Solutions

Now that we have the real solutions (2, -4) and (-2, -4), we can plot these points on the coordinate grid. These points are the exact intersection points of the circle and the parabola.

Coordinate Grid Plot

1. Locate the point (2, -4) on the coordinate grid. This point is 2 units to the right of the origin and 4 units down.
2. Locate the point (-2, -4) on the coordinate grid. This point is 2 units to the left of the origin and 4 units down.

Mark these points clearly on the grid. These are the real solutions to the system of equations.

Conclusion

Solving a system of equations graphically involves plotting the equations on a coordinate plane and identifying the intersection points. In this case, the system of equations y² + x² = 20 and y = -x² represents a circle and a parabola, respectively. By plotting these graphs, we visually identified two intersection points. To find the exact solutions, we used the substitution method, which led us to the real solutions (2, -4) and (-2, -4). These solutions were then plotted on the coordinate grid, confirming the graphical analysis. Understanding both graphical and algebraic methods provides a comprehensive approach to solving systems of equations.

This detailed exploration illustrates how combining graphical representation with algebraic techniques provides a robust method for solving systems of equations. The graphical method offers a visual understanding, while algebraic methods provide precise solutions. For the given system, we found that the circle and parabola intersect at two points, which we accurately determined using both methods. This approach is fundamental in various mathematical and scientific applications, making it an essential skill for students and professionals alike.