Problem 10 - Intermediate Algebra

Question

Find a substitution equation that can be used to solve the system: $$\left\\{\begin{array}{l}x^{2}+y^{2}=9 \\ 2 x-y=3\end{array}\right.$$

Step-by-Step Solution

Answer

Substitute \(y = 2x - 3\) into \(x^2 + y^2 = 9\).

Step 1: Identify Equations in the System

The system consists of two equations: 1. \(x^2 + y^2 = 9\) 2. \(2x - y = 3\)

Step 2: Solve One Equation for a Variable

Choose the simpler linear equation \(2x - y = 3\) and solve for \(y\). Rearrange this equation to isolate \(y\): \(y = 2x - 3\).

Step 3: Substitute into the Other Equation

Now that we have \(y = 2x - 3\) from Step 2, substitute this expression for \(y\) into the first equation \(x^2 + y^2 = 9\). This substitution gives us: \(x^2 + (2x - 3)^2 = 9\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when graphed on a coordinate plane. They are typically written in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. The equations are called 'linear' because the solutions to these equations form a straight line.

In our exercise, one of the equations given is a linear equation: \(2x - y = 3\). This equation describes a line where every point on the line is a solution to the equation.

• The coefficient of \(x\) is 2, which indicates the slope of the line.
• The coefficient of \(y\) is -1, showing how \(y\) changes in response to changes in \(x\).

Understanding linear equations is crucial, as they form the basis for solving more complex equations and systems, such as this one with a quadratic component. By rearranging this linear equation to solve for \(y\), we find that \(y = 2x - 3\), simplifying the process of substitution into the other equation.

Quadratic Equations

Quadratic equations, as opposed to linear, involve terms squared, which means they graph as parabolas rather than straight lines. A basic quadratic equation has the format \(ax^2 + bx + c = 0\). In our problem, \(x^2 + y^2 = 9\) is not a typical quadratic equation because it involves both \(x^2\) and \(y^2\) on the same level, creating a circular graph rather than a parabola.

This equation defines a circle with radius 3, centered at the origin, as the sum of squares of \(x\) and \(y\) equals a constant (the square of the radius). Parsing quadratic equations necessitates understanding the relationship between their components:

• The term \(x^2\) indicates that changes in \(x\) affect the equation quadratically.
• Similarly, the \(y^2\) term affects the equation based on the \(y\) variable's square.

By substituting the expression derived from the linear equation \(y = 2x - 3\) into \(x^2 + y^2 = 9\), you convert it into a single variable quadratic, \(x^2 + (2x - 3)^2 = 9\), which becomes simpler to solve for \(x\).

Solving Systems of Equations

Solving systems of equations involves finding the set of values that satisfy all equations simultaneously. The substitution method is a popular technique used for this, especially when dealing with one linear and one non-linear equation, like in our example.

The key steps to follow in this method are:

• Identify each equation.
• Solve one of the equations, usually the simpler linear one, for one variable.
• Substitute this expression into the other equation, allowing you to work with a single variable.

This allows for simplification, turning a system of two equations into a more manageable single equation. In our given system, by substituting \(y = 2x - 3\) into \(x^2 + y^2 = 9\), we effectively reduce two variables in two equations down to one variable, \(x\), in a transformed quadratic equation. This approach simplifies the solution process as you can then solve this quadratic equation using methods appropriate for single-variable quadratics, such as factoring or using the quadratic formula.