Problem 12 - 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) $$ \left(\begin{array}{l} y=-x^{2}+3 \\ y=x^{2}+1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solutions are \((1, 2)\) and \((-1, 2)\).

Step 1: Graph the Equations

First, graph both equations on the same coordinate plane to predict approximate solutions. For the equation \( y = -x^2 + 3 \), sketch a downward-facing parabola that opens downwards and whose vertex is at \((0,3)\). For the equation \( y = x^2 + 1 \), graph an upward-facing parabola with its vertex at \((0,1)\). The approximate points where these parabolas intersect give us estimated solutions for the system.

Step 2: Set the Equations Equal to Each Other

To solve the system algebraically using the substitution method, set the expressions for \( y \) in both equations equal to each other: \( -x^2 + 3 = x^2 + 1 \). This creates an equation with only one variable, \( x \), which allows you to find the x-values where the graphs intersect.

Step 3: Solve the Equation for x

Combine and simplify the equation obtained: \( -x^2 + 3 = x^2 + 1 \) becomes:\( 3 - 1 = x^2 + x^2 \)\( 2 = 2x^2 \).Divide both sides by 2: \( x^2 = 1 \).Solve for \( x \): take the square root of both sides to get \( x = \pm 1 \).

Step 4: Substitute x-values back into One of the Original Equations

Use the \( x \) values found in Step 3 and substitute them back into one of the original equations to find the corresponding \( y \) values. Let's use \( y = x^2 + 1 \): - For \( x = 1 \), \( y = 1^2 + 1 = 2 \).- For \( x = -1 \), \( y = (-1)^2 + 1 = 2 \).The points of intersection are \((1, 2)\) and \((-1, 2)\).

Step 5: Verify the Solutions

Check the solutions \((1, 2)\) and \((-1, 2)\) in both original equations to ensure correctness. Substitute \((1, 2)\) into \( y = -x^2 + 3 \) and \( y = x^2 + 1 \) and verify both yield \( y = 2 \). Likewise, substitute \((-1, 2)\) into both equations, and verify \( y = 2 \). Both points satisfy each original equation, confirming they are correct solutions to the system.

Key Concepts

Graphing Systems of Equations

Graphing systems of equations gives a visual way to understand the potential solutions of two or more equations. For each equation, plot its graph on the same coordinate plane. The points where these graphs intersect represent the solutions to the system. Consider the following example:

• For the equation \( y = -x^2 + 3 \), you'll graph a downward-facing parabola. Its vertex is at the point \((0,3)\).
• The second equation, \( y = x^2 + 1 \), creates an upward-facing parabola, with its vertex at \((0,1)\).

By plotting both, you can visually estimate the intersection points. This initial step is crucial for understanding how the equations interact and to predict solutions before using algebraic methods.

Substitution Method

The substitution method is a powerful algebraic technique for solving systems of equations. It involves substituting one equation into another to eliminate one of the variables. Here's how you use this method effectively:
- Start by expressing one variable in terms of the other using one of the equations. For example, if you have \( y = -x^2 + 3 \) and \( y = x^2 + 1 \), you can substitute one into the other: set \(-x^2 + 3 = x^2 + 1\).- This substitution transforms the problem into an equation with only one variable: \( -x^2 + 3 = x^2 + 1 \).- Solve this simplified equation: combine like terms to get \( 2 = 2x^2 \), then solve for \( x \) to find \( x = \pm 1 \).This method simplifies the process by focusing on one equation at a time, allowing you to solve complex systems more efficiently.

Elimination Method

The elimination method is another algebraic technique to solve systems of equations. Unlike substitution, it involves adding or subtracting equations to eliminate a variable. While we used substitution in the problem above, knowing about elimination helps tackle different systems. Here’s a basic rundown:
- Consider two linear equations. You adjust terms so that adding or subtracting the equations will cancel out one of the variables. - Solve for the remaining variable. - Once you have the value of one variable, substitute it back into any of the original equations to find the other. Although not used with quadratic equations directly, the elimination method is crucial for systems involving linear equations, offering a straightforward way to isolate variables.

Quadratic Equations

Quadratic equations are expressed in the standard form \( ax^2 + bx + c = 0 \). These types of equations create parabolas when graphed on a coordinate plane. Understanding their properties is essential for dealing with systems including quadratic forms. Key features include:

• The shape is either a `U` or an upside-down `U` depending on the sign of \( a \).
• The vertex is the turning point, which can be found using \( x = -b/(2a) \).
• Intercepts are found by solving the equation \( y=0 \) for roots, using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\].

In our system, each equation is quadratic, providing two potential intersection points, found both graphically and algebraically. Understanding parabolas and their intersections allows you to predict and verify solutions effectively in multi-step solution problems like these.