Problem 11 - Precalculus with Limits

Question

Solving a System by Substitution In Exercises \(7-14,\) solve the system by the method of substitution. Check your solution(s) graphically. $$\left\\{\begin{aligned}-\frac{1}{2} x+y &=-\frac{5}{2} \\ x^{2}+y^{2} &=25 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 4, y = -3\) and \(x = -5, y = 0\).

Step 1: Solve the Linear Equation for y

From the first equation \(-\frac{1}{2}x + y = -\frac{5}{2}\), solve for y: \(y = \frac{1}{2}x - \frac{5}{2}\)

Step 2: Substitute y into the Circle Equation

Substitute \(y = \frac{1}{2}x - \frac{5}{2}\) into \(x^2 + y^2 = 25\): \(x^2 + \left(\frac{1}{2}x - \frac{5}{2}\right)^2 = 25\)

Step 3: Simplify the Equation

Simplify the above equation: \(x^2 + \frac{1}{4}x^2 - \frac{5}{2}x + \frac{25}{4} = 25\)

Step 4: Solve for x

Simplify the equation and solve it for x: \(1.25x^2 - \frac{5}{2}x - 25 = 0\). By using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the solutions for x are \(4\) and \(-5\)

Step 5: Find Corresponding y Values

Substitute the values of x into the equation \( y = \frac{1}{2}x - \frac{5}{2}\) to get the corresponding y values. For \(x = 4\), \(y = -3\), and for \(x = -5\), \(y = 0\).

Key Concepts

Substitution Method

The substitution method is a crucial technique for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation.
This way, you convert a system of equations into a single equation, making it easier to solve.

• Identify the easier equation: Typically start with the linear equation, because it's simpler to manipulate.
• Solve for one variable: In our exercise, we solved for \( y \) in the equation \(-\frac{1}{2}x + y = -\frac{5}{2}\), giving us \( y = \frac{1}{2}x - \frac{5}{2}\).
• Substitute into the other equation: Substitute the expression for \( y \) into the quadratic equation \( x^2 + y^2 = 25 \). This substitutes the variable \( y \) with the expression found in the linear equation.
• Solve the resulting equation: Once substituted, solve for the remaining variable to find its potential values.

After finding these values, substitute back to find the corresponding value for the other variable. This method effectively reduces complexity and provides a streamlined approach to solving systems of equations.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They take the form \(ax^2 + bx + c = 0\) and are solvable through various methods.
The quadratic in our system was formed by substituting \( y \) into the circle equation resulting in \( 1.25x^2 - \frac{5}{2}x - 25 = 0 \).

• Simplify first: Make sure the equation is in standard quadratic form.
• Use the quadratic formula: This formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is used effectively when factoring is not straightforward. Here, it helped calculate the values of \( x \).
• Discern real and complex solutions: The discriminant \( b^2 - 4ac \) will help indicate whether solutions are real or complex.

For our case, solving the quadratic gave us real solutions: \( x = 4 \) and \( x = -5 \). These are the potential \( x \)-values, which we then use to find \( y \). Quadratics naturally appear across many contexts, and mastering them allows effective progressions in problem-solving.

Graphical Solution

A graphical solution provides a visual representation of the system of equations. By graphing both equations on a coordinate plane, you can easily assess where they intersect.
The intersection points are simultaneous solutions to both equations.

• Graph each equation: In our problem, one equation represents a line (\(y = \frac{1}{2}x - \frac{5}{2}\)) and the other a circle (\(x^2 + y^2 = 25\)).
• Locate the intersections: Points where the line meets the circle are the solutions to the system. Upon graphing, these are \((4, -3)\) and \((-5, 0)\).
• Verification: This method verifies computation steps and offers visual confirmation of solutions found through algebraic means.

The graphical method fosters intuition about solutions and can validate solutions from substitution or elimination methods, ensuring accuracy and understanding of the problem's context.