Problem 24 - Intermediate Algebra

Question

For Problems \(1-28\), (a) graph each system so that approximate real number solutions (if there are any) can be predicted, and (b) solve each system using the substitution method or the elimination-by-addition method. (Objectives 1 and 2) $$ \left(\begin{array}{l} y=x^{2}+2 \\ y=2 x^{2}+1 \end{array}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Given Problem

We have a system of equations: 1) \( y = x^2 + 2 \) and 2) \( y = 2x^2 + 1 \). Our task is to graph these equations to predict approximate solutions and then solve them using substitution or elimination.

Step 2: Graph the Equations

Plot the two equations on the same set of axes. The first equation \( y = x^2 + 2 \) is a parabola that opens upwards, shifted 2 units up. The second equation \( y = 2x^2 + 1 \) is another upward-opening parabola but narrower and 1 unit up. The intersection points of these curves represent the solutions to the system.

Step 3: Solve Using Substitution Method

Set the equations equal to each other to find the x-coordinates of intersection: \( x^2 + 2 = 2x^2 + 1 \). Simplify to get \( 2 = x^2 + 1 \). Then subtract 1 from both sides: \( 1 = x^2 \). This gives \( x = \pm 1 \).

Step 4: Find Corresponding y-Values

Substitute \( x = 1 \) into the first equation to find \( y \): \( y = 1^2 + 2 = 3 \). Substitute \( x = -1 \) into the first equation to find \( y \): \( y = (-1)^2 + 2 = 3 \). Both solutions satisfy both equations.

Step 5: Check Solutions

Substitute both (1, 3) and (-1, 3) back into the second equation. For both, \( y = 2x^2 + 1 \) holds true, confirming that they are indeed solutions of the system.

Key Concepts

Graphing Systems of Equations

When solving systems of equations, graphing is one of the fundamental techniques we use to visually interpret potential solutions. In this method, we plot each equation on a shared set of axes. The visual intersection points of these graphs indicate the solutions to the system.
In the original exercise, we are dealing with the equations \( y = x^2 + 2 \) and \( y = 2x^2 + 1 \). Both are parabolas, which are curves that open upwards because they contain \( x^2 \) terms.

• The first equation, \( y = x^2 + 2 \), is a parabola shifted 2 units up, meaning every point on the parabola has 2 added to its y-value compared to \( y = x^2 \).
• The second equation, \( y = 2x^2 + 1 \), is a narrower parabola, affected by the coefficient 2. This makes it steeper, and it is shifted 1 unit up.

By graphing these equations, you will find out where they intersect on the grid. These intersection points provide the approximate solutions to the system.
Graphing gives an intuitive picture, but it's often used best as a preliminary step before verifying solutions through algebraic methods.

Substitution Method

The substitution method is an algebraic technique used to solve systems of equations by expressing one variable in terms of another. This is particularly useful when one equation is simpler or easily rearranged to isolate a variable.
In our case, we have:

• Equation 1: \( y = x^2 + 2 \)
• Equation 2: \( y = 2x^2 + 1 \)

To use substitution, we set the equations equal to each other, since they both equal \( y \). This results in the equation \( x^2 + 2 = 2x^2 + 1 \). We can then simplify:

• Subtract \( x^2 \) from both sides: \( 2 = x^2 + 1 \)
• Subtract 1 from both sides: \( 1 = x^2 \)
• Taking the square root gives us \( x = \pm 1 \).

Once the \(x\)-values are found, substitute them back into one of the original equations to find the corresponding \(y\)-values. In this example, for both \( x = 1 \) and \( x = -1 \), \( y = 3 \). So, our solutions are the points \((1, 3)\) and \((-1, 3)\).
Substitution is valuable because it provides exact solutions, ensuring precision in cases where graphing might only suggest approximate locations.

Elimination Method

The elimination method, sometimes called the addition method, involves combining equations to eliminate one of the variables. This method is particularly useful when both equations are in linear form, but can be adapted for non-linear like our current equations too.
In our example, eliminating variables directly using addition might not seem straightforward because both equations are in quadratic form. However, focusing on eliminating common terms or coefficients by combining equations can simplify the process. Unfortunately, elimination is not the most effective standalone method for these specific non-linear equations, as substitution and graphing are evidently more direct.
However, in general situations involving linear terms or more complex systems where substitution is cumbersome, adjusting the coefficients and combining the equations cleverly can simplify finding solutions. Recognizing when to use each method helps solve a variety of systems efficiently. While the elimination method wasn't primarily used in solving the given exercise, understanding its functionality is essential for tackling a wide range of problems, especially when dealing with multiple linear equations.