Problem 6 - Precalculus

Question

Solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=16 \\ x-y &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution points are \((1+\sqrt{7}, -1+\sqrt{7})\) and \((1-\sqrt{7}, -1-\sqrt{7})\).

Step 1: Identify the Equations

We have two equations to solve. The first equation is a circle: \( x^2 + y^2 = 16 \). The second equation is a line: \( x - y = 2 \). Our goal is to find the points where these two equations intersect, which will give us the solution of the system.

Step 2: Solve the Linear Equation for One Variable

From the linear equation \( x - y = 2 \), solve for \( x \):\[ x = y + 2 \] This expression for \( x \) will be substituted into the equation of the circle.

Step 3: Substitute Expression into Circle Equation

Substitute \( x = y + 2 \) into the circle equation \( x^2 + y^2 = 16 \):\[ (y + 2)^2 + y^2 = 16 \] Simplify and solve for \( y \).

Step 4: Simplify and Solve the Quadratic Equation

Expand and simplify the expression:\[ y^2 + 4y + 4 + y^2 = 16 \]Combine like terms:\[ 2y^2 + 4y + 4 = 16 \]Rearrange into standard quadratic form:\[ 2y^2 + 4y - 12 = 0 \] Divide through by 2 to simplify:\[ y^2 + 2y - 6 = 0 \]

Step 5: Factor the Quadratic Equation

The quadratic equation does not factor nicely, so use the quadratic formula. For \( ay^2 + by + c = 0 \), use:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 1, b = 2, c = -6 \).\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 24}}{2} \= \frac{-2 \pm \sqrt{28}}{2} \= \frac{-2 \pm 2\sqrt{7}}{2} \= -1 \pm \sqrt{7} \]

Step 6: Solve for x Using Values of y

With \( y = -1 + \sqrt{7} \) and \( y = -1 - \sqrt{7} \), substitute into \( x = y + 2 \):- For \( y = -1 + \sqrt{7} \), \( x = -1 + \sqrt{7} + 2 = 1 + \sqrt{7} \).- For \( y = -1 - \sqrt{7} \), \( x = -1 - \sqrt{7} + 2 = 1 - \sqrt{7} \).

Step 7: Verify the Solution Points

The solutions are \((1 + \sqrt{7}, -1 + \sqrt{7})\) and \((1 - \sqrt{7}, -1 - \sqrt{7})\). Substitute back into the original equations to verify. Both points satisfy both equations.

Step 8: Sketch the Graphs

Sketch the circle \( x^2 + y^2 = 16 \) centered at the origin with a radius of 4. Also, sketch the line \( x - y = 2 \), which has a slope of 1 and y-intercept of -2. The intersections of these two will verify the calculated solutions.

Key Concepts

Circle Equation

A circle equation in the Cartesian coordinate system is generally written as \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle and the equation specifies that the center is at the origin \((0,0)\). In the given exercise, the equation \(x^2 + y^2 = 16\) corresponds to a circle centered at the origin.
When you see this equation:

• Recognize the constant on the right side, \(16\), as the square of the radius.
• The radius is therefore \(\sqrt{16} = 4\).

This tells us that the circle is drawn with a radius of 4 units from the center. By sketching this circle, you have a perfect path for where points \((x, y)\) could lie.
Remember, a circle is the set of all points in a plane that are a fixed distance from a center. It's a basic shape in geometry with some profound properties.

Linear Equation

A linear equation is represented generally as \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. For this system, you have an equation \(x - y = 2\). This can be rewritten to express \(y\) explicitly:

• First, isolate \(y\): \(x = y + 2\) becomes \(y = x - 2\).
• The equation now shows that the line has a slope of 1 with a y-intercept of -2.

Understanding the properties of linear equations helps gauge how the line will appear on a graph. With a slope of 1, the line rises consistently at the same rate it runs. This gives us a straightforward diagonal line, easily visualized on both graphs and calculations.

Quadratic Formula

The quadratic formula is a crucial tool for solving equations of the form \(ax^2 + bx + c = 0\). It can also be adapted to find roots in terms of any variable, such as \(y\). For our specific quadratic case after substitution, \(y^2 + 2y - 6 = 0\), we solve using:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plugging in the coefficients:

• \(a = 1\), \(b = 2\), and \(c = -6\).
• Sum in the values: \( y = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-6)}}{2 \times 1} \).
• Simplify to \(y = -1 \pm \sqrt{7}\).

This calculation provides the \(y\) coordinates where the circle and line intersect. The quadratic formula is critical because it offers a straightforward way to find roots of quadratic equations even when factoring isn't apparent or possible.

Graphing Equations

Graphing equations is a visual method to analyze and verify the results of mathematical systems, especially helpful for nonlinear systems. In the given exercise, both a circle and a line need to be drawn on the same plane.

• Start with the circle equation \(x^2 + y^2 = 16\), plotting points that form a circle centered at \((0,0)\) with a radius of 4.
• Next, add the linear equation \(x - y = 2\) which, rewriting, you get \(y = x - 2\). This is a line with slope 1 and y-intercept of -2.

By graphing both the circle and the line, their intersections visually provide the solution to the simultaneous equation. Double-checking through graphing can enhance understanding and confidence, confirming algebraic solutions in a clear and precise way.