Problem 40 - College Algebra Essentials

Question

In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} (x-1)^{2}+(y+1)^{2}=5 \\ 2 x-y=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = 1, y = -1 \).

Step 1: Rearrange the linear equation

Firstly, the linear equation will be rearranged to solve for \( y \). The equation is \( 2x - y = 3 \), so rearranging it to isolate \( y \) gives us \( y = 2x - 3 \).

Step 2: Substitute in the circle equation

Substitute \( y \) in the circle equation \( (x-1)^2 + (y+1)^2 = 5 \) with \( 2x -3 \). The equation will look as follows: \( (x-1)^2 + [(2x - 3) + 1]^2 = 5 \). Thus we get \( (x-1)^2 + (2x - 2)^2 = 5 \).

Step 3: Simplify and Solve

Expand and collect like terms: \(x^{2} - 2x + 1 + 4x^{2} - 8x + 4 = 5\). This simplifies to \( 5x^{2} - 10x \) which equals \( 5x^{2} - 10x -5 = 0 \). This equation is a quadratic equation and can be factored into \( 5(x - 1) (x - 1) = 0 \). Hence, \( x = 1 \).

Step 4: Find the corresponding y values

Substitute \( x = 1 \) into the rearranged linear equation to find the corresponding \( y \) values. So \( y = 2(1) - 3 = -1 \). Therefore, the solution of the system of equations is \( x = 1 \) and \( y = -1 \).

Key Concepts

Linear Equations

Linear equations are equations that represent lines in a two-dimensional space. They are usually expressed in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In the context of a system of equations, linear equations often provide constraints that must be satisfied simultaneously.
Linear equations are called such because their graph is always a straight line. This makes them easy to analyze and solve, especially using algebraic techniques like solving for one variable in terms of the other.
For instance, in the original exercise, the equation \( 2x - y = 3 \) is a linear equation. By rearranging it to solve for \( y \) (i.e., \( y = 2x - 3 \)), we make it suitable for substitution into the circle equation, simplifying the process of finding where the two equations intersect.

Quadratic Equations

Quadratic equations are polynomials where the highest power of the variable is 2. The general form is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations can produce parabolas when graphed.
In the exercise, as shown in the solution, a quadratic equation appears after substituting and simplifying the terms from the circle equation. The resulting equation \( 5x^2 - 10x - 5 = 0 \) describes a parabola.
The unique feature of quadratic equations is that they can be solved using various methods, including factoring, completing the square, or the quadratic formula. In this case, we factored it to \( 5(x - 1)(x - 1) = 0 \), where \( x = 1 \) provides the solution. Quadratic equations often yield two solutions due to their parabolic nature, but the exercise shows having a repeated root here.

Substitution Method

The substitution method is a powerful algebraic technique employed to solve systems of equations. It involves solving one equation for one variable and replacing that variable in the other equation.
Using the substitution method simplifies the system to an equation in one variable, making it more straightforward to solve. For instance, in the provided solution, we solve the linear equation \( 2x - y = 3 \) to express \( y \) as \( y = 2x - 3 \).
This expression is then substituted into the circle equation \((x-1)^2 + (y+1)^2 = 5\), transforming it into a single-variable quadratic equation. The substitution method reduces the complexity of solving multiple equations at once by focusing on one variable at a time. This strategy is incredibly efficient when dealing with linear and nonlinear equations together, like a line and a circle as seen here.

Circle Equation

A circle equation generally demonstrates all the points on a plane that are equidistant from a center. The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). Here, \( (h, k) \) denotes the circle's center and \( r \) its radius.
In the exercise, the circle equation \((x-1)^2 + (y+1)^2 = 5\) is given, where the center is \((1, -1)\) and the radius squared equals 5. This equation showcases the set of points forming a circle in the coordinate plane.
When solving systems involving a circle equation and a line, we're essentially looking for the points of intersection. After substituting the linear expression into the circle equation, we discovered that the line only touches the circle at one point \((1, -1)\), indicating a tangent scenario. This concept not only helps in solving systems but also in understanding geometric properties and intersections.