Problem 1 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} y=x^{2}-4 \\ y=2 x-1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are (3, 5) and (-1, -3).

Step 1: Identify the Equations

The system of equations is given as: \( y = x^2 - 4 \) and \( y = 2x - 1 \). Both equations are equal to \( y \), which allows us to set them equal to each other for substitution.

Step 2: Substitute and Set Equations Equal

Since both equations equal \( y \), we set them equal to each other: \( x^2 - 4 = 2x - 1 \). We now have a single equation in terms of \( x \).

Step 3: Rearrange the Equation

Rearrange the equation to set it to zero: \( x^2 - 2x - 3 = 0 \). This is a quadratic equation.

Step 4: Factor the Quadratic

Factor the quadratic equation: \( x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \). This gives the solutions for \( x \).

Step 5: Solve for x

Set each factor equal to zero and solve for \( x \):- From \( x - 3 = 0 \), we find \( x = 3 \).- From \( x + 1 = 0 \), we find \( x = -1 \).

Step 6: Solve for Corresponding y Values

Substitute \( x = 3 \) and \( x = -1 \) back into \( y = 2x - 1 \) to find corresponding \( y \) values:- For \( x = 3 \): \( y = 2(3) - 1 = 5 \)- For \( x = -1 \): \( y = 2(-1) - 1 = -3 \)

Step 7: State the Solutions

The solutions to the system are the points where the two curves intersect: \( (3, 5) \) and \( (-1, -3) \).

Key Concepts

System of Equations

A system of equations is a set of equations with multiple variables that are solved simultaneously. In this exercise, we are dealing with two equations, both involving the variables \( x \) and \( y \). The equations given are: \( y = x^2 - 4 \) and \( y = 2x - 1 \). Each equation forms a curve when graphed, and solving the system means finding the points where these curves intersect. An effective method for solving this is the substitution method, where one equation is substituted into another to eliminate a variable, simplifying the solution process.

Quadratic Equation

A quadratic equation is a second-degree polynomial, usually written in the form \( ax^2 + bx + c = 0 \). In our exercise, substituting \( y = 2x - 1 \) into \( y = x^2 - 4 \) resulted in the quadratic equation \( x^2 - 2x - 3 = 0 \). Solving this equation involves finding values of \( x \) that make the equation true. To solve, we factor the equation: \( x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \). We then set each factor equal to zero and solve for \( x \); this can give us two solutions, which represent the potential \( x \)-values where the two curves might intersect.

Intersecting Points

Intersecting points between two curves are the solutions that satisfy both equations in a system. You can think of them as the common points between the graphs of the two equations. In this exercise, after factoring the quadratic equation, we found that \( x = 3 \) and \( x = -1 \) are possible solutions. We substitute these \( x \)-values back into one of the original equations to find the corresponding \( y \)-values: \( y = 5 \) when \( x = 3 \) and \( y = -3 \) when \( x = -1 \). Therefore, the points \( (3, 5) \) and \( (-1, -3) \) are where the parabola and the line intersect. These points are the answer to the system of equations, showing exactly where these two mathematical constructs meet on a graph.