Problem 8 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Solve the First Equation for y

We start by solving the first equation for \(y\):\[x^2 - y = 1\]Rearrange to express \(y\) in terms of \(x\):\[y = x^2 - 1\]

Step 2: Substitute y in the Second Equation

Substitute \(y = x^2 - 1\) into the second equation:\[2x^2 + 3(x^2 - 1) = 17\]Simplify the expression:\[2x^2 + 3x^2 - 3 = 17\]

Step 3: Solve the Simplified Equation for x

Combine like terms:\[5x^2 - 3 = 17\]Add 3 to both sides:\[5x^2 = 20\]Divide both sides by 5:\[x^2 = 4\]Take the square root of both sides:\[x = 2 \quad \text{or} \quad x = -2\]

Step 4: Find y for Each x-Value

For \(x = 2\), substitute back to find \(y\):\[y = (2)^2 - 1 = 4 - 1 = 3\]For \(x = -2\), substitute back to find \(y\):\[y = (-2)^2 - 1 = 4 - 1 = 3\]Thus, the solutions are \((2, 3)\) and \((-2, 3)\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find the variable values that satisfy all equations simultaneously. In this exercise, the system consists of two equations:

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

These equations are considered nonlinear because they involve the variable \(x\) raised to the power of two. To solve systems like this, a common approach is to use substitution or elimination methods.
The substitution method involves solving one equation for one variable and using that expression to substitute into the other equation. This reduces the system to a single equation with one variable, making it easier to solve. In our example, we solve for \(y\) in terms of \(x\) in the first equation, and substitute that into the second equation.

Solving Quadratic Equations

Solving quadratic equations is an essential skill in algebra. Quadratic equations are equations that can be written in the form \(ax^2 + bx + c = 0\). They involve a variable raised to the second power, known as a quadratic term.
In the original problem, once we substituted for \(y\), the second equation simplified to a quadratic equation:

• \(5x^2 - 3 = 17\)

To solve it:

• Add 3 to both sides to isolate the quadratic term: \(5x^2 = 20\)
• Divide by 5 to further isolate \(x^2\): \(x^2 = 4\)
• Take the square root of both sides to solve for \(x\), giving \(x = 2\) or \(x = -2\).

Quadratics can have two solutions, as seen here, or sometimes just one or none at all depending on the discriminant value, which is part of the quadratic formula.

Algebraic Manipulation

Algebraic manipulation is a fundamental process in solving equations. It involves changing the form of equations using mathematical operations to isolate variables and simplify expressions.
In our exercise, we used algebraic manipulation in several ways:

• Solving \(x^2 - y = 1\) for \(y\). This involved rearranging the terms to express \(y\) as \(y = x^2 - 1\).
• Substituting the expression for \(y\) into the second equation and simplifying: \(2x^2 + 3(x^2 - 1) = 17\) turns into \(5x^2 - 3 = 17\).
• Isolating \(x^2\) by combining like terms and balancing the equation, resulting in the simplified quadratic equation \(x^2 = 4\).

These steps allowed us to break down and solve the system effectively, demonstrating the power of algebraic techniques in problem-solving. These skills are not only useful in theoretical math but also in real-world scenarios where modeling scenarios mathematically becomes necessary.