Problem 17 - 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) $$ \left(\begin{array}{l} y=x^{2}+2 x-1 \\ y=x^{2}+4 x+5 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \((-3, 2)\).

Step 1: Graph the System

To find approximate real-number solutions graphically, plot each equation on the same coordinate plane.- For the first equation \( y = x^2 + 2x - 1 \), plot key points and sketch the parabola.- For the second equation \( y = x^2 + 4x + 5 \), similarly plot key points and sketch this parabola.Observe where these parabolas intersect–the intersection points give us approximate coordinates for the solution(s).

Step 2: Set Equations Equal

To solve the system algebraically using substitution, set the equations equal to each other since both equations are equal to \( y \).Set \( x^2 + 2x - 1 = x^2 + 4x + 5 \).

Step 3: Simplify and Solve for x

Subtract \( x^2 + 2x - 1 \) from both sides to simplify the equation.This gives \( 0 = 2x + 6 \).

Step 4: Solve for x

Solve \( 0 = 2x + 6 \) for \( x \).- Subtract 6 from both sides: \( -6 = 2x \)- Divide by 2 to isolate \( x \): \( x = -3 \).

Step 5: Substitute x Back into One Equation

Use either equation to find \( y \). Substitute \( x = -3 \) into the first original equation.Calculate: \( y = (-3)^2 + 2(-3) - 1 = 9 - 6 - 1 = 2 \).

Step 6: Verify the Solution

Double-check the solution \((x, y) = (-3, 2)\) by substituting back into the second original equation to ensure it satisfies both equations.\( y = (-3)^2 + 4(-3) + 5 \) gives:\( y = 9 - 12 + 5 = 2 \). This matches, confirming the solution.

Key Concepts

Graphing Methods

Graphing methods are a visual way to solve systems of equations. In graphing, each equation is plotted on the same set of axes. This method helps to find approximate solutions by identifying intersections of the graphs.

For example, in this exercise, we have two quadratic equations: \(y = x^2 + 2x - 1\) and \(y = x^2 + 4x + 5\). Both graphs are parabolas. Here are the steps to graph these equations:

• Identify key points for each equation, such as the vertex and x-intercepts, if any.
• Use these points to sketch the parabola on a graph.
• Look for the points where the two graphs intersect, as these intersections indicate the solutions to the system.

By graphing, you obtain a visual representation that can suggest where solutions lie before confirming them through other methods like substitution or elimination.

Substitution Method

The substitution method is a powerful algebraic technique to solve systems of equations. It's especially useful when one equation is solved for a single variable.

In this exercise, both equations are already set equal to \(y\), so we can set the equations equal to each other: \[x^2 + 2x - 1 = x^2 + 4x + 5\]
This method involves solving for one variable and then plugging it back into the original equation to find the other variable. Here's how it works:

• Eliminate like terms from both sides of the equation, simplifying it to \(0 = 2x + 6\).
• Solve for \(x\) by isolating it. Re-arrange the simplified equation to find \(x = -3\).
• Use the value of \(x\) to find \(y\) by substituting back into one of the original equations.

Substitution simplifies the process by focusing on one variable at a time, making problem-solving more straightforward.

Elimination by Addition

The elimination by addition method involves adding or subtracting equations to cancel out one of the variables. This method is most efficient when one of the coefficients of the variables matches in two equations.

Although this method wasn't used in the solution provided, it's essential to understand its application. For example, if we modify the initial system to make coefficients align, we could apply it as follows:

• Adjust one or both equations so that the coefficients of one variable are opposites.
• Add or subtract the equations to eliminate the targeted variable entirely.
• Solve the resulting single-variable equation.
• Substitute back to find the value of the eliminated variable.

Using elimination helps when direct substitution isn't straightforward, and it provides a clear path to simplify multi-variable problems.

Quadratic Equations

Quadratic equations come in the form of \(ax^2 + bx + c = 0\) and have a distinctive U-shaped graph called a parabola. Solving systems involving quadratic equations often requires a mix of graphing and algebraic approaches like substitution or elimination.

When dealing with quadratics, keep in mind the following:

• The solutions to the equation are where the parabola intersects the x-axis (these are the roots).
• In systems of equations, the focus is often on where the two graphs intersect.
• Quadratics can have zero, one, or two real solutions based on the discriminant \(b^2-4ac\).

In our exercise, both equations are quadratic. By graphing both, you found that they intersect only once, at \(x = -3\), \(y = 2\). This confirmed that there is a unique solution to the system, showing the nature of quadratic intersections.