Problem 65 - 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. $$\begin{aligned} &2 x+y=6\\\ &-x+y=0 \end{aligned}$$

Step-by-Step Solution

Answer

The point of intersection of the two functions \(2x + y = 6\) and \(-x + y = 0\) is \((2, 2)\). Verified by creating a table of values for each function.

Step 1: Identify Two Equations

First, identify the two equations from the problem. They are \(2x + y = 6\) and \(-x + y = 0\). The goal is to solve for \(x\) and \(y\).

Step 2: Solve using Elimination or Substitution

Solving using the elimination method involves making the coefficients of either \(x\) or \(y\) the same in both equations, then subtracting one equation from the other to eliminate that variable. Alternatively, you can solve one equation for one variable, and then substitute this into the other equation. Here, the second equation \(-x + y = 0\) can be rewritten as \(y = x\). Substituting this into the first equation gives: \(2x + x = 6\), leading to \(x = 2\). Substituting \(x = 2\) into \(y = x\) gives \(y = 2\). Hence, the point of intersection is \((2, 2)\).

Step 3: Verification through Table of Values

Create two tables of values for each function, and verify that they intersect at the point \((2, 2)\). For \(2x + y = 6\), choose values of \(x\) and calculate corresponding \(y\) values. Repeat this process for \(-x + y = 0\). The table for \(2x + y = 6\) could have: \((1, 4), (2, 2), (3, 0)\) and for \(-x + y = 0\): \((1, 1), (2, 2), (3, 3)\). Agreeing pair of \((x, y)\) values verifies the calculated intersection.

Key Concepts

Elimination Method

The elimination method is a powerful tool for solving a system of equations. It involves manipulating the two equations so that one of the variables is eliminated when the equations are combined. This allows you to solve for the remaining variable easily.

The basic steps for using the elimination method are:

• Choose a variable to eliminate. Look at the coefficients of the variables in both equations.
• Multiply one or both equations, if needed, so that the coefficients of the chosen variable are equal or opposite.
• Add or subtract the equations to eliminate the chosen variable.
• Solve the resulting equation for the remaining variable.
• Substitute the solution back into one of the original equations to find the other variable.

In the exercise given, elimination is not explicitly used, but you can implement it easily. If you were to eliminate \( y \), you could modify the coefficients and subtract the equations. However, sometimes substitution is more straightforward, like in this problem.

Substitution Method

The substitution method involves substituting a known expression for one variable into another equation. This approach is often simpler when one of the equations is already solved for one variable or can easily be rearranged.

• Start by solving one of the equations for one variable. In this exercise, the equation \(-x + y = 0\) is rearranged to find that \(y = x\).
• Substitute this expression into the other equation. For the problem at hand, you substitute \(y = x\) into \(2x + y = 6\), resulting in \(2x + x = 6\).
• Solve the new equation for the remaining variable, \(3x = 6\), which quickly gives \(x = 2\).
• Now, substitute \(x = 2\) back into \(y = x\) to find \(y = 2\).

This method provides a quick way to find the point of intersection \((2, 2)\), confirming the solution processed algebraically.

Point of Intersection

The point of intersection is a central concept in systems of equations, as it represents the solution where both equations are satisfied simultaneously. Graphically, it is where the lines represented by the equations cross each other.

To find this point algebraically:

• Solve the system of equations using either the elimination or substitution method.
• Determine the values of \(x\) and \(y\) that make both equations true.
• In the exercise provided, the point \((2, 2)\) was found, indicating that both lines intersect at this coordinate on a graph.
• This point is also the only solution to the system when dealing with two distinct equations representing straight lines.

Verification can be done numerically by creating tables of values for both equations. When you find the same pair in both tables as in \((2, 2)\), it reaffirms that this is indeed the correct point of intersection.