Problem 4 - Algebra and Trigonometry

Question

\(3-8=\) Use the substitution method to find all solutions of the system of equations. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=25 \\ y &=2 x \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

Step 1: Understand the Problem

You need to solve the system of equations, which consists of a circle equation \(x^2 + y^2 = 25\) and a linear equation \(y = 2x\). Use substitution to solve the equations.

Step 2: Substitute the Linear Equation into the Circle Equation

Given \(y = 2x\), substitute \(2x\) for \(y\) in the first equation \(x^2 + y^2 = 25\). This gives \(x^2 + (2x)^2 = 25\), which simplifies to: \(x^2 + 4x^2 = 25\).

Step 3: Solve for x

Simplify the equation from Step 2: \(5x^2 = 25\). Divide both sides by 5 to get \(x^2 = 5\). Taking square roots of both sides gives \(x = \sqrt{5}\) or \(x = -\sqrt{5}\).

Step 4: Solve for y using x values

Use \(y = 2x\) to find \(y\) corresponding to each \(x\). If \(x = \sqrt{5}\), then \(y = 2\sqrt{5}\). If \(x = -\sqrt{5}\), then \(y = -2\sqrt{5}\).

Step 5: Write Down the Solutions

The solutions to the system are two points: \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

Key Concepts

System of Equations

A system of equations involves finding values for variables that satisfy all given equations simultaneously. Such systems can contain any number of equations, but they share a common goal: every equation must hold true using the same values for those variables.

In the featured problem, we deal with a system comprising a circle equation and a linear equation. Here are a few steps that can guide you when solving systems of equations:

• Identify the equations involved and understand their individual meanings.
• Decide on a method to solve them, like substitution or elimination.
• Solve step by step, ensuring all solutions satisfy the original system.

When using the substitution method, you choose one equation to express one variable in terms of another, then substitute this expression into the other equation(s). This helps eliminate one variable, making it easier to solve the remaining equation.

Circle Equation

Circle equations often appear in the form: \[ x^2 + y^2 = r^2 \] where \( (x, y) \) are the coordinates of any point on the circle and \( r \) is the radius of the circle.

In our problem, the circle equation is \( x^2 + y^2 = 25 \), which tells us about all the points \((x, y)\) that lie 5 units (since \(r^2 = 25\), so \(r = \sqrt{25} = 5\)) from the origin (0,0).

This equation graphically represents a perfect circle centered at the origin with a radius of 5. Each solution pair \((x, y)\) satisfies this circle equation, meaning—geometrically—they lie on the circle's boundary. When combined with another equation, like a line, we find the points where the line intersects this circle.

Linear Equation

A linear equation represents a straight line in the coordinate plane. The general form is \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.

In this case, the linear equation is \(y = 2x\). Here, \(m = 2\) and \(c = 0\), which shows two important characteristics:

• The slope \(m = 2\) indicates the line rises two units for every unit it moves to the right.
• Since there's no y-intercept term, the line passes through the origin \((0,0)\).

When solving the system, substituting this linear equation into the circle equation involves replacing \(y\) with \(2x\). This substitution method helps to eliminate \(y\), making it a single-variable equation easy to solve. Solving gives us values of \(x\), which we then use to find corresponding \(y\) values for complete solutions.