Problem 8 - Algebra and Trigonometry

Question

\(3-8=\) Use the substitution method to find all solutions of the system of equations. $$ \left\\{\begin{array}{c}{x^{2}-y=1} \\ {2 x^{2}+3 y=17}\end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((2, 3)\) and \((-2, 3)\).

Step 1: Solve First Equation for y

Start with the equation \(x^2 - y = 1\). Add \(y\) to both sides and subtract 1 from both sides to solve for \(y\). This gives \(y = x^2 - 1\).

Step 2: Substitute Expression for y into Second Equation

Take the expression for \(y\) from Step 1: \(y = x^2 - 1\) and substitute it into the second equation \(2x^2 + 3y = 17\). This gives \(2x^2 + 3(x^2 - 1) = 17\).

Step 3: Simplify and Solve Equation for x

Distribute the 3 into the expression \(3(x^2 - 1)\) to get \(3x^2 - 3\). Substitute this back into the equation: \(2x^2 + 3x^2 - 3 = 17\). Combine like terms to get \(5x^2 - 3 = 17\). Add 3 to both sides to obtain \(5x^2 = 20\). Divide by 5 to solve for \(x^2\), giving \(x^2 = 4\).

Step 4: Solve for x

Take the square root of both sides of \(x^2 = 4\). Remember there are two possible solutions for \(x\): \(x = 2\) or \(x = -2\).

Step 5: Solve for y Using Values of x

Now, substitute the values of \(x\) back into the expression for \(y = x^2 - 1\). For \(x = 2\), \(y = 2^2 - 1 = 4 - 1 = 3\). For \(x = -2\), \(y = (-2)^2 - 1 = 4 - 1 = 3\).

Step 6: Solution to the System of Equations

The solutions to the system of equations are \((2, 3)\) and \((-2, 3)\). These are the points where both equations are satisfied.

Key Concepts

System of Equations

A "system of equations" refers to a collection of two or more equations that share common variables. The goal is to find a set of variable values that satisfy all the equations at the same time.

In the original exercise, we have a system consisting of two equations:

• \( x^2 - y = 1 \)
• \( 2x^2 + 3y = 17 \)

These equations share the variables \( x \) and \( y \). By solving this system, we find specific values of \( x \) and \( y \) that work in both equations simultaneously.

The method chosen to solve these types of problems can greatly affect the ease and effectiveness of finding the solution. The substitution method, outlined in the exercise, is one of the most straightforward approaches for handling such systems.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, generally taking the form \( ax^2 + bx + c = 0 \). In our problem, we encounter quadratic terms, such as \( x^2 \), which play a pivotal role.

One equation in our system was transformed into a quadratic form during the solution process:

• \( x^2 = 4 \) resulted from the substitution and simplification steps.

Quadratic equations typically have two potential solutions because of the symmetric properties of parabolas. Remember, solving \( x^2 = 4 \) gives us two solutions: \( x = 2 \) and \( x = -2 \). These outputs underscore the necessity of considering all possible roots of a quadratic equation, reflecting the parabola's intersection points with the x-axis.

Understanding how quadratic equations behave is essential in algebra. Being able to recognize and solve them, particularly when they're part of systems, is a key skill in solving more complex algebraic problems.

Algebraic Solutions

"Algebraic solutions" refer to finding values of variables that satisfy an equation using algebraic manipulations. This process often involves strategies like factoring, using the quadratic formula, or employing methods such as substitution or elimination in systems of equations.

The original step-by-step solution made use of the substitution method to find the algebraic solutions. Here's how that method worked:

• First, solve one of the equations for a variable (in our case, \( y = x^2 - 1 \)).
• Substitute this expression into the other equation.
• Simplify and solve for the remaining variable.
• Back-substitute to find the value of the initial variable.

This method is especially useful when one equation can be easily solved for a single variable. The precision of algebraic methods allows for finding exact solutions, which are particularly beneficial in theoretical contexts where approximation isn't sufficient.