Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Solve the system graphically. $$\left\\{\begin{array}{r}2 x-y+3=0 \\ x^{2}+y^{2}-4 x=0\end{array}\right.$$

Step-by-Step Solution

Answer

To find the exact points of intersection, one may need to use algebraic methods in combination with graphical methods. However, graphically, there should be two points of intersection representing the solutions to the system of equations.

Step 1: Rewriting both equatioms in a more recognizable form

Let's first re-arrange the linear equation by solving for y to get it in the form y = mx + b . Thus, the linear equation becomes y = 2x + 3. For the circular equation, complete the square for the x terms to turn it into the form \((x-a)^2+(y-b)^2=r^2\). This gives us \((x-2)^2 + y^2 =4\) which is a circle centered at (2, 0) with a radius of 2.

Step 2: Plotting the graphical representations

Next, plot both the line y = 2x + 3 and the circle \((x-2)^2 + y^2 =4\) on the same graph. The line will intersect the circle at the solution points to the system.

Step 3: Finding the points of intersection

Determine the points of intersection, which are the solutions for the system of equations. These will be the points where the plotted line cuts through the plotted circle.

Key Concepts

Linear Equations

Linear equations form the foundation of algebra and appear in a basic form as \( y = mx + b \). Here, \( m \) represents the slope, and \( b \) is the y-intercept, where the line crosses the y-axis. This equation describes a straight line on a coordinate plane. In our exercise, the linear equation derived from the system is \( y = 2x + 3 \).

This means the slope \( m \) is 2, indicating the line moves up two units on the y-axis for every one unit it moves to the right on the x-axis. The y-intercept, \( b = 3 \), shows the line crosses the y-axis at (0,3). By graphing such lines, you can visualize solutions to equations and understand relationships between x and y.

To graph a linear equation like \( y = 2x + 3 \):

• Start at the y-intercept \( b = 3 \) on the y-axis.
• Use the slope to find another point. From (0,3), move 2 units up and 1 unit right to (1,5).
• Draw a straight line through these points. This line represents all possible solutions to the equation.

Quadratic Equations

Quadratic equations involve terms of \( x^2 \) and, when plotted, form a distinct curve called a parabola or, in specific situations like our problem, they can define a circle. In standard form, a quadratic equation is \( ax^2 + bx + c = 0 \), but variations exist based on the context.

In the exercise, the rearranged equation \((x-2)^2 + y^2 = 4\) takes the form of a circle rather than a standard parabola. This equation represents a circle centered at (2, 0), with a radius of 2, deviating from the usual parabolic shape of quadratic equations due to its distinct structure.

When converting to this form, recognize:

• The expression \( (x-a)^2+(y-b)^2=r^2 \) represents a circle centered at \((a,b)\) with radius \( r \).
• In our exercise, \((x-2)^2 + y^2 = 4\), showing a center at (2,0) with a radius of \( \sqrt{4} = 2\).

This unique quadratic form helps visualize more complex relationships graphically, such as where a linear equation intersects with a circle.

Completing the Square

Completing the square is a fundamental technique to simplify quadratic equations, especially for circles or determining minimum or maximum values. It transforms expressions from \( ax^2 + bx + c \) to a structured form \((x-h)^2 + k\), making them easier to graph or analyze.

In our problem, we complete the square for the quadratic equation part \( x^2 - 4x \) to transform it into \((x-2)^2 + y^2 =4\). Here’s how it works:

• Take the \(x\) term \( 4x \), halve the coefficient: \( 4/2 = 2\), and square it: \( 2^2 = 4\).
• Add \(4\) inside the equation: \((x^2 - 4x + 4)\), then balance it to maintain equality.
• Reexpress the squared term: \((x-2)^2\).

Thus, this conversion shows the circle’s center shifted to (2,0) with radius 2. Completing the square is crucial in algebra to simplify, solve, and graph complex equations accurately.