Question

In Exercises 47–52, solve the nonlinear system. Justify your answer with a graph. $$ \begin{aligned} &x^2+y^2=25 \\ &y=-\frac{3}{4} x+\frac{25}{4} \end{aligned} $$

Short answer

The solution to the system of equations are two points: (-3, 4) and (4, -3).

Step-by-step solution

Substitute the second equation into the first one

Replace \( y \) in the first equation \( x^2 + y^2 = 25 \) with the second equation \( y = -\frac{3}{4}x + \frac{25}{4} \). This gives \( x^2 + (-\frac{3}{4}x + \frac{25}{4})^2 = 25. \)

Solve for x

First expand the terms and simplify the equation to a quadratic equation. Then solve for \( x \). This should give two solutions for \( x \). To solve for these values use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Solve for y

Substitute each solution for \( x \) into the second equation to get corresponding \( y \) values. This should give two corresponding \( y \) values.

Justify with a graph

Graph both original equations in the same coordinate plane. The points of intersection should match the solutions found.

Key concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). In the given exercise, the equation \( x^2 + y^2 = 25 \) represents a circle, which is a form of a quadratic equation in two variables. In any system involving quadratic equations, solving these involves finding the values of \( x \) and \( y \) that satisfy the equation.
This equation can sometimes result in two solutions, no real solutions, or infinite solutions depending on the discriminant \( b^2 - 4ac \). A positive discriminant indicates two distinct solutions, zero means one solution, and a negative discriminant indicates no real solution.
When solving, you might use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] to find these values.

System of Equations

A system of equations involves solving two or more equations simultaneously to find common solutions. In this exercise, you have a nonlinear system because it includes a quadratic equation \( x^2 + y^2 = 25 \) and a linear equation \( y = -\frac{3}{4}x + \frac{25}{4} \).
To solve a system of equations, you aim to find the values of variables that satisfy all given equations at the same time. Nonlinear systems can have a finite number of solutions, like two points of intersection, or other complex numbers of solutions.

• The process often involves using algebraic methods like substitution or elimination.
• In this exercise, substitution is employed to solve for the common points of intersection between the curve (circle) and the line.

Graphing Solutions

Graphing is a visual way to solve or confirm solutions to a system of equations. For the given equations \( x^2 + y^2 = 25 \) and \( y = -\frac{3}{4}x + \frac{25}{4} \), graphing helps to visually demonstrate the points of intersection. It is these intersection points that are the solution to the system.
Typically, you will:

• Plot each equation on the same graph.
• Observe where the graphs intersect since those points are solutions to the system.
• Confirm algebraic solutions graphically by checking if they correspond to intersection points.

Graphing is advantageous because it provides an intuitive understanding of how different equations relate to each other visually in the coordinate plane.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves solving one equation for a single variable and substituting this solution into the other equations. In this exercise, \( y = -\frac{3}{4}x + \frac{25}{4} \) is substituted into \( x^2 + y^2 = 25 \).
This substitution leads to a single equation in one variable \( x \), which can then be solved using techniques like factoring or using the quadratic formula.
After finding the value(s) of \( x \), plug these back into the linear equation to find the corresponding \( y \) values. The substitution method simplifies complex systems by reducing the number of variables in the initial equations, making them more manageable.