Problem 60 - College Algebra with Modeling and Visualization

Question

If possible, solve the nonlinear system of equations. $$ \begin{aligned} &x^{2}+y^{2}=9\\\ &x+y=3 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((0, 3)\) and \((3, 0)\).

Step 1: Understand the System of Equations

We have two equations: \( x^2 + y^2 = 9 \) and \( x + y = 3 \). Our goal is to find values of \( x \) and \( y \) that satisfy both equations simultaneously. The first equation represents a circle, and the second a line. We need to find their points of intersection.

Step 2: Express One Variable in Terms of the Other

Using the second equation, we can express \( y \) in terms of \( x \). From \( x + y = 3 \), we get \( y = 3 - x \). This expression for \( y \) will be substituted into the first equation.

Step 3: Substitute into the First Equation

Substitute \( y = 3 - x \) into the first equation \( x^2 + y^2 = 9 \): \[x^2 + (3 - x)^2 = 9\]This becomes:\[x^2 + (9 - 6x + x^2) = 9\]which simplifies to \[2x^2 - 6x + 9 = 9\]by combining like terms.

Step 4: Solve the Quadratic Equation

Simplify the equation to solve for \( x \):\[2x^2 - 6x = 0\]Factor out the common term:\[2x(x - 3) = 0\]This gives the solutions \( x = 0 \) and \( x = 3 \) by setting each factor equal to zero.

Step 5: Find Corresponding Values of y

Substitute \( x = 0 \) and \( x = 3 \) back into \( y = 3 - x \) to find corresponding \( y \) values. - If \( x = 0 \), then \( y = 3 - 0 = 3 \).- If \( x = 3 \), then \( y = 3 - 3 = 0 \).The solutions are \( (0, 3) \) and \( (3, 0) \).

Step 6: Verify Solutions

Let's verify the solutions by checking them in both original equations.- For \( (0, 3) \), checking: \( 0^2 + 3^2 = 9 \) and \( 0 + 3 = 3 \).- For \( (3, 0) \), checking: \( 3^2 + 0^2 = 9 \) and \( 3 + 0 = 3 \).Both solutions satisfy the original equations, so they are correct.

Key Concepts

Circle Equation

A circle equation usually appears in the form \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle. In this specific problem, the circle is represented by the equation \(x^2 + y^2 = 9\). Here, \(r^2 = 9\), which means the radius \(r\) is 3. This tells us that any point \((x, y)\) that satisfies this equation will be exactly 3 units away from the origin \((0,0)\).

Understanding the equation of a circle is crucial for determining where a straight line might intersect it. In graphing terms, you'd draw a perfect circle centered at the origin with a radius of 3, which helps you visualize the possible points at which another function, such as a line, could intersect the circle.

Linear Equation

A linear equation like \(x + y = 3\) represents a straight line when plotted on a graph. In this equation, we have both \(x\) and \(y\) as variables whose sum always equals 3. This means that any point \((x, y)\) that lies on this line will satisfy this condition.

The line equation can be rearranged to express one variable in terms of another, which is particularly useful when solving systems of equations. For example, solving the line equation for \(y\) gives \(y = 3 - x\). This makes it easier to substitute into another equation, like a quadratic or circle equation, to find where they intersect.

Quadratic Equation

Quadratic equations come into play when substituting one expression into another equation, resulting in a new polynomial equation. In this problem, after substituting \(y = 3 - x\) into \(x^2 + y^2 = 9\), we end up with the quadratic equation \(2x^2 - 6x = 0\).

Quadratics typically have the form \(ax^2 + bx + c = 0\). Here, it's \(2x^2 - 6x + 0 = 0\), making it somewhat simpler since \(c = 0\). Solving the quadratic can be accomplished by factoring: \(2x(x - 3) = 0\), yielding solutions \(x = 0\) and \(x = 3\).

Quadratic equations often have two solutions, which correspond to the points of intersection with another graph like a circle or line.

Intersection Points

Intersection points are where two different equations meet or cross each other on a graph. In solving systems of equations, the points \((0, 3)\) and \((3, 0)\) are the solutions that make both the circle equation and linear equation true simultaneously.

To find them, we solve the system of equations and then verify each solution. For example, if \(x = 0\), substituting into \(y = 3 - x\) gives \(y = 3\). If \(x = 3\), \(y\) becomes 0. Thus \((0, 3)\) and \((3, 0)\) are not just algebraic solutions, but confirmed intersection points.

These points on the graph represent where the circle and the line cross each other. Verifying these intersections means checking both equations to confirm that each point satisfies them, ensuring accuracy in the solution.