Problem 63 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine any point(s) of intersection algebraically. Then verify your result numerically by creating a table of values for each function. $$\begin{aligned} &y=6-x\\\ &y=3 x-2 \end{aligned}$$

Step-by-Step Solution

Answer

The point of intersection of the equations \(y = 6 - x\) and \(y = 3x - 2\) is (2, 4)

Step 1: Set the Expressions Equal

In order to find the point(s) of intersection of the two functions \(y = 6-x\) and \(y = 3x-2\), set the two expressions for y equal to each other: \(6 - x = 3x - 2\)

Step 2: Solve for x

Now, take \(x\) terms to one side and constants to the other side to solve for \(x\). Add \(x\) to both sides and add 2 to both sides to get \(4x = 8\). Divide both sides by 4 to isolate \(x\), so \(x = 2\).

Step 3: Substitute x into one of the Original Equations

Substitute \(x = 2\) into the equation \(y = 6 - x\) to solve for \(y\): \(y = 6 - 2 = 4\)

Step 4: Verifying the solution Numerically

Create a table of values around \(x = 2\) for each function, substitute \(x\) into each of the functions and observe that when \(x = 2\), \(y = 4\) for both functions, which confirms the point of intersection as (2, 4).

Key Concepts

Point of Intersection

A point of intersection is where two graphs, in this case linear equations, meet on the coordinate plane. It's essentially a solution that satisfies both equations simultaneously. When finding this point for the equations \( y = 6 - x \) and \( y = 3x - 2 \), you're looking for the values of \( x \) and \( y \) where both expressions yield the same result. Visually, think about two lines crossing at a point that they both share. Mathematically, this means substituting values into each expression results in equality. Identifying this point algebraically involves setting the equations equal to find \( x \), then using this value in one of the initial equations to determine \( y \). For these specific equations, the point of intersection is \((2, 4)\). This indicates that when you plug \( x = 2 \) into either equation, you get \( y = 4 \).

Algebraic Solution

Solving systems of equations algebraically involves using mathematical operations to find where both equations are satisfied at the same time. This typically means you are solving for both \( x \) and \( y \). Let's break down the steps:- **Set the equations equal:** Since both functions equal \( y \), you can set them equal to each other: \( 6 - x = 3x - 2 \).- **Solve for \( x \):** Rearrange terms to group all \( x \) terms on one side. Adding \( x \) to both sides, you have \( 6 = 4x - 2 \).- **Isolate \( x \):** Add 2 to both sides to give \( 8 = 4x \), then divide by 4 to find \( x = 2 \).- **Find \( y \):** Substitute \( x = 2 \) back into one of the original equations, like \( y = 6 - x \), to get \( y = 4 \).This systematic approach helps ensure the solution satisfies both equations.

Verification by Table of Values

Creating a table of values is a useful method to numerically verify that the solution is correct. Start by calculating several \( y \) values for various \( x \) values in both equations, particularly around the identified point of intersection:- **Select a range of \( x \):** Choose values close to \( x = 2 \). For example, try \( x = 1, 2, 3 \).- **Substitute into each equation:** Calculate \( y \) for each \( x \) in both \( y = 6 - x \) and \( y = 3x - 2 \).For the given point of intersection \( (2, 4) \):

• In \( y = 6 - x \), when \( x = 2 \), \( y = 4 \).
• In \( y = 3x - 2 \), when \( x = 2 \), \( y = 4 \).

When both tables yield the same \( y \) value at \( x = 2 \), this confirms that our algebraic solution, the point \( (2, 4) \), is indeed correct.