Problem 40 - Intermediate Algebra

Question

For Exercises 37 through \(42,\) first divide the equation through by the coefficient of \(\left.x^{2} \text { (or } y^{2}\right)\) $$7(x-1)^{2}+7(y-3)^{2}=63$$

Step-by-Step Solution

Answer

The simplified equation is \((x-1)^{2} + (y-3)^{2} = 9\).

Step 1: Identify the equation to be simplified

We start with the equation \(7(x-1)^{2}+7(y-3)^{2}=63\). The task is to simplify it by dividing through by the coefficient of \(x^2\) and \(y^2\), which is \(7\).

Step 2: Divide the entire equation by 7

To simplify the equation, we divide both sides by \(7\). This will modify the equation:\[ \frac{7(x-1)^{2}}{7} + \frac{7(y-3)^{2}}{7} = \frac{63}{7} \]

Step 3: Simplify each term

The equation simplifies to:\( (x-1)^{2} + (y-3)^{2} = 9 \). This represents a circle's equation.

Key Concepts

Circle Equation

In mathematics, a circle equation is a beautiful representation of a circle in the coordinate plane. The standard form of a circle's equation is \[ (x-h)^2 + (y-k)^2 = r^2 \]In this form,

• \( (h,k) \) are the coordinates of the circle's center.
• \( r \) is the radius of the circle.

To identify the equation of a circle, we look for expressions with squares of terms in both \( x \) and \( y \). In the given exercise, after simplifying, the circle's equation becomes \((x-1)^2 + (y-3)^2 = 9\). Here, the center is at \((1, 3)\) and the radius is 3, since \(r^2 = 9\). Understanding the circle equation helps us locate the circle in space and comprehend its size.

Simplifying Equations

Simplifying equations makes math problems more manageable and often involves several techniques to reduce an equation to its simplest form. The process generally includes:

• Combining like terms
• Performing operations like division or multiplication to eliminate coefficients
• Reducing fractions when possible

In the given exercise, simplifying began by dividing each term by the coefficient 7. This step is crucial as it reduces the equation to a simpler form, making it easier to analyze or solve further. Simplifying also helps in deducing essential properties or characteristics of shapes described by the equation, such as a circle in this case.

Dividing Equations by Coefficients

Dividing equations by coefficients is a useful technique in algebra to simplify equations. The coefficient is the number located in front of variables, affecting the term's magnitude. By dividing all parts of an equation by a particular coefficient, you often simplify the solution path.For the exercise problem at hand, each term was divided by 7, the original coefficient of the squared terms. This division:

• Clarified the constants within the quadratic terms (\((x-1)^2\) and \((y-3)^2\))
• Simplified the operation by aligning the equation closely with the standard circle form

This step not only transforms the equation into a manageable form but also directly aligns it with standard forms like that of a circle or parabola, bringing clarity to its geometric interpretation.

Algebra

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating those symbols. It forms the foundation of solving equations and understanding abstract relationships between quantities. In exercises involving conic sections like circles, algebra comes into play by allowing us to manipulate equations into forms that we can analyze. Through steps like factoring, distributing, and simplifying, algebra enables us to solve for unknowns or establish geometrical interpretations. By using algebraic techniques, we can understand complex expressions, make substitutions, and apply multiple properties to reach solutions efficiently. In the current problem, algebra was employed to simplify the circle equation through division and simplification, demonstrating its powerful ability to reduce calculations and illuminate mathematical relationships.