Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{r} x+3 y=5 \\ x^{2}+y^{2}=25 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((5, 0)\) and \((-4, 3)\).

Step 1: Solve for x in terms of y

From the first equation, \( x + 3y = 5 \), solve for \( x \). Subtract \( 3y \) from both sides to get \( x = 5 - 3y \).

Step 2: Substitute x in the second equation

Take the expression for \( x \) from Step 1, \( x = 5 - 3y \), and substitute it into the second equation \( x^2 + y^2 = 25 \). This gives \((5 - 3y)^2 + y^2 = 25\).

Step 3: Expand and Simplify

Expand the squared term: \((5 - 3y)^2 = (5)^2 - 2 \cdot 5 \cdot 3y + (3y)^2 = 25 - 30y + 9y^2 \). Substitute this back into the equation: \( 25 - 30y + 9y^2 + y^2 = 25 \), which simplifies to \( 10y^2 - 30y = 0 \).

Step 4: Factor and Solve for y

Factor out a \( y \) in the equation \( 10y^2 - 30y = 0 \) to get \( y(10y - 30) = 0 \). This gives the solutions \( y = 0 \) and \( 10y - 30 = 0 \). Solving \( 10y - 30 = 0 \) gives \( y = 3 \).

Step 5: Find corresponding x values

Substitute \( y = 0 \) and \( y = 3 \) back into \( x = 5 - 3y \) to find the corresponding \( x \) values.For \( y = 0 \), \( x = 5 - 3(0) = 5 \).For \( y = 3 \), \( x = 5 - 3(3) = -4 \).

Step 6: Verify the solutions

Check each pair \((x, y)\) in the original equations.For \((5, 0)\): Equation 1: \( 5 + 3(0) = 5 \) ✓Equation 2: \( 5^2 + 0^2 = 25 \) ✓For \((-4, 3)\):Equation 1: \(-4 + 3(3) = 5 \) ✓Equation 2: \((-4)^2 + 3^2 = 16 + 9 = 25 \) ✓

Step 7: Final Solution

The system has two solutions: \((5, 0)\) and \((-4, 3)\).

Key Concepts

System of Equations

A "system of equations" consists of a set of two or more equations with the same set of variables. The primary goal when working with a system of equations is to find the values for the variables that satisfy all equations in the system simultaneously.
In this exercise, we have two equations:

• Linear equation: \( x + 3y = 5 \)
• Quadratic equation: \( x^2 + y^2 = 25 \)

Solving a system of equations typically involves finding a point or points \((x, y)\) that solve both equations at the same time. In this exercise, the method of substitution is used to find these solutions by reducing the number of variables involved in the equations.

Quadratic Equations

A "quadratic equation" is a polynomial equation in which the highest degree of the variable is 2. It generally takes the form \( ax^2 + bx + c = 0 \). In this exercise, the second equation, \( x^2 + y^2 = 25 \), represents a circle's equation in the coordinate plane.
This quadratic equation is a little different from the standard form because it has two variables, making it more complex as it involves finding a solution that accommodates both variables.

• It showcases how two variables can interact within a quadratic framework.
• Solving involves using algebraic techniques to simplify and manage multiple solutions.

Algebraic Solution

An "algebraic solution" involves manipulating the given equations using algebraic rules to find the values of variables. This process often includes steps such as substitution, simplification, and solving equations.
This exercise utilizes the substitution method for the algebraic solution:

• First, solve the linear equation for one variable in terms of the other, here \( x = 5 - 3y \).
• Next, substitute this expression into the other equation, reducing it to a single variable. This step transforms the quadratic equation into a simpler form that is solvable using algebraic techniques.
• Finally, find each variable’s value that satisfies the conditions set by the equations, checking your work as you solve.

Polynomial Factoring

"Polynomial factoring" is an algebraic technique used to express a polynomial as the product of its factors. It is a critical skill in solving polynomial equations, particularly when you reach a simplified form needing further breakdown.
During the solution process of this exercise, you arrive at \( 10y^2 - 30y = 0 \) after substitution and simplification. Factoring is an essential step to finding the roots of the equation:

• Factor out the common element \( y \), yielding \( y(10y - 30) = 0 \).
• Set each factor equal to zero to solve for \( y \), which leads to \( y = 0 \) and \( y = 3 \).

Factoring not only simplifies equations but also reveals crucial information about the solutions or roots of the given polynomial equation.