Problem 10 - 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) $$ \left(\begin{array}{l} y=x^{2}-4 x+5 \\ -x+y=1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solutions are \((4, 5)\) and \((1, 2)\).

Step 1: Identify the equations

The given system of equations is composed of a quadratic equation and a linear equation:\[ y = x^2 - 4x + 5 \] and \[-x + y = 1.\]

Step 2: Graph the equations

To better understand the system, graph both equations on the same coordinate plane. The quadratic \( y = x^2 - 4x + 5 \) is a parabola opening upwards, and \( -x + y = 1 \) or rearranged as \( y = x + 1 \) is a straight line with a slope of 1 that intersects the y-axis at 1.

Step 3: Use substitution method

From the linear equation, express \( y \) in terms of \( x \): \( y = x + 1 \). Substitute this into the quadratic equation: \[ x + 1 = x^2 - 4x + 5. \]

Step 4: Rearrange and solve for x

Rearrange the equation to set it to 0: \[ x^2 - 5x + 4 = 0. \] This quadratic equation can be factored into \((x - 4)(x - 1) = 0\). Solve to find \( x = 4 \) and \( x = 1 \).

Step 5: Solve for y using x-values

For \( x = 4 \), substitute back into \( y = x + 1 \) to get \( y = 5 \). For \( x = 1 \), \( y = 1 + 1 = 2 \).

Step 6: Conclusion: Solutions

The solutions to the system are the points where the graphs intersect, which are \((4, 5)\) and \((1, 2)\).

Key Concepts

Graphing Systems of Equations

Understanding how to graph systems of equations is an essential skill in solving these mathematical problems. When tackling a system like the given, one needs to visually interpret how the two different equations interact on a graph. Consider our system:

• A quadratic equation: \( y = x^2 - 4x + 5 \), represented as a parabola that opens upwards. The turning point or vertex of this parabola can be found using the formula \( x = -\frac{b}{2a} \), which gives us the vertex at the point \( (2, 1) \).
• A linear equation: \( y = x + 1 \). This is a straight line with a slope of 1, and it crosses the y-axis at 1.

To graph these, plot a few strategic points for each equation, and draw the parabola and the line on the same coordinate plane. Observing where they intersect reveals the solutions, which are the coordinates of the points of intersection. In this case, these points are \((4, 5)\) and \((1, 2)\). Graphing helps verify our solutions and understand the relationship visually.

Substitution Method

The substitution method is a powerful technique for solving systems of equations, where one equation is substituted into another to eliminate a variable. Let's break down how it's done:

• First, solve one of the equations for one of the variables. For our system, we rearrange the linear equation \(-x + y = 1\) to express \(y\) in terms of \(x\): \(y = x + 1\).
• Next, replace this expression in the quadratic equation, substituting \(y = x + 1\) into \(y = x^2 - 4x + 5\). This gives a new equation with just one variable: \(x + 1 = x^2 - 4x + 5\).
• Solve this single variable equation. Rearrange it into a standard quadratic form \(x^2 - 5x + 4 = 0\), and then factor it to find \(x = 4\) and \(x = 1\).
• Finally, substitute these \(x\) values back into the expression \(y = x + 1\) to find the corresponding \(y\) values: \(y = 5\) for \(x = 4\) and \(y = 2\) for \(x = 1\).

Using substitution helps to streamline finding exact solutions, especially when equations are compatible, as we see with our intersecting curves.

Quadratic and Linear Equations

When working with systems of equations that include both quadratic and linear equations, you are essentially comparing a curved graph with a straight line. Let's explore their characteristics:

• Quadratic Equations: These are equations in the form of \(y = ax^2 + bx + c\). The graph is a parabola, which if \(a > 0\), opens upwards, otherwise it opens downwards. Key features include the vertex, axis of symmetry, and whether it has real roots or vertices on the x-axis.
• Linear Equations: A simpler form, \(y = mx + b\), where \(m\) is the slope, showing the inclination, and \(b\) is the y-intercept, revealing where the line crosses the y-axis.

In our example, the quadratic equation \(y = x^2 - 4x + 5\) produces a parabola, while the linear \(y = x + 1\) results in a straight line. The solutions to their intersection - determined both graphically and algebraically - yield the points where these different shapes meet. Understanding both allows a comprehensive approach to solving and graphing systems that combine differing equation types.