Problem 1 - College Algebra Essentials

Question

In Exercises \(1-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} x+y=2 \\ y=x^{2}-4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are (2,0) and (-3,5).

Step 1: Isolate the Variable in the First Equation

From the first equation \(x + y = 2,\) isolating 'y', we get \(y = 2 - x\).

Step 2: Substitute for 'y' in the Second Equation

We substitute \(y\) in the second equation \(y = x^2 - 4\), giving us \(2 - x = x^2 - 4\). This simplifies to a quadratic equation \(x^2 + x - 6 = 0\).

Step 3: Solve the Quadratic Equation

By factorizing the quadratic equation, we have \((x - 2)(x + 3) = 0\). So, the solutions for 'x' are \(x=2\) and \(x=-3\).

Step 4: Substitute 'x' Back into the Isolated First Equation

Solving for 'y' using \(y = 2 - x\), if \(x = 2\), we get \(y = 2 - 2 = 0\). If \(x = -3\), we get \(y = 2 - (-3) = 5\).

Key Concepts

System of Equations

A system of equations consists of two or more equations that share the same set of variables. In this exercise, the system is:

• \(x + y = 2\)
• \(y = x^2 - 4\)

The goal is to find values for 'x' and 'y' that satisfy both equations simultaneously. There are several methods to solve a system of equations, such as graphing, elimination, and substitution. Here, we'll focus on the substitution method.

In the substitution method, one equation is solved for one variable, and then that expression is substituted into the other equation. This method works particularly well when one of the equations is linear, like the first equation here. By isolating 'y' in the first equation \(x + y = 2\), we get \(y = 2 - x\). We then substitute this expression for 'y' in the second equation. This allows us to solve for 'x'. Only after finding 'x', we re-substitute to find 'y'. This stepwise approach is logical and eliminates one variable at a time.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation, and it is typically of the form \(ax^2 + bx + c = 0\). In this system, substituting for 'y' results in the quadratic equation \(x^2 + x - 6 = 0\).

Quadratic equations can be solved using various methods such as:

• Factoring
• Completing the square
• Using the quadratic formula

In our exercise, the equation is solved by factoring, since the quadratic is factorable over real numbers. This method is often straightforward when the equation can be easily decomposed into a product of linear binomials.

Understanding the nature of quadratic equations is crucial because it aids in identifying the patterns and structures that allow for different solving techniques. It also helps in recognizing when a quadratic equation might need more advanced methods, such as the quadratic formula.

Factoring Quadratics

Factoring is a technique that involves rewriting a polynomial as the product of simpler polynomials. For the equation \(x^2 + x - 6 = 0\), we look for two numbers that multiply to -6 and add to +1, which are +3 and -2. Therefore, the quadratic equation factors as \((x - 2)(x + 3) = 0\).

After factoring, we use the Zero-Product Property, which tells us that if the product of two numbers is zero, then at least one of the numbers must be zero. Therefore, setting each factor equal to zero gives us:

• \(x - 2 = 0\)
  Leading to \(x = 2\)
• \(x + 3 = 0\)
  Leading to \(x = -3\)

Factoring quadratics is a handy and efficient method when the quadratic is easily factorable. However, not all quadratic equations are factorable by simple methods, and in such cases, other methods like completing the square or using the quadratic formula might be necessary. Understanding when and how to use factoring can save time and simplify solving systems of equations.