Problem 67 - 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. $$\begin{aligned} &x-y=10\\\ &x+2 y=4 \end{aligned}$$

Step-by-Step Solution

Answer

The point of intersection of the two lines given by the equations \(x-y=10\) and \(x+2y=4\) is \((8.5, -3)\).

Step 1: Solve Algebraically

Use elimination method to solve the system of equations. Add the two equations together: \((x-y) + (x+2y) = 10 + 4\), which simplifies to \(2x+y=14\). Solve for \(x\) by subtracting \(y\) from both sides, yielding \(x=(14-y)/2\). Now, substitute \(x\) into the first equation to find \(y\) . This gives \([(14-y)/2] - y = 10\) which simplifies to \(14 - 2y = 20\), thus \(y=-3\). Substituting \(y=-3\) back into \(x = (14-y)/2\) provides the \(x\) value, \(x = 8.5\).

Step 2: Verify Algebraic Solution

To verify the solution, plug in the values of \(x\) and \(y\) you found into the original equations. In the first equation \(x-y=10\), substitute \(x = 8.5\) and \(y = -3\), the equation simplifies to \(8.5 - (-3) = 10\), which is true as \(11.5 = 11.5\). Similarly, in the second equation \(x+2y=4\), substituting \(x = 8.5\) and \(y = -3\), we get \(8.5 + 2*(-3) = 4\), which simplifies to \(8.5 - 6 = 4\), which is also true as \(2.5 = 2.5\). So, the solution is correct.

Step 3: Create the Table of Values

Creating a table of values for each linear function allows us to see the relationship between \(x\) and \(y\) and provides us with numerical evidence of the point of intersection. For the first equation \(x-y = 10\), the table of values could be: for \(y = -3\), \(x = 8.5\); for \(y = 1\), \(x = 11\); for \(y = 0\), \(x = 10\). We see when \(y=-3\), \(x=8.5\) is exactly our point of intersection. For the second equation \(x+2y = 4\), the table of values could look like this: for \(y = -3\), \(x = 8.5\); for \(y = 1\), \(x = 2\); for \(y = 0\), \(x = 4\). Again, when \(y=-3\), \(x=8.5\) is our point of intersection. Thus, the point of intersection of the two lines is \((8.5, -3)\).

Key Concepts

Elimination Method

When solving systems of equations, the elimination method is often a go-to strategy. This method eliminates one variable by adding or subtracting the equations, making it easier to solve for the remaining variable.
In our example, we're given the equations:

• \( x - y = 10 \)
• \( x + 2y = 4 \)

To eliminate \( x \), add the two equations together. This cancellation gives you a new equation with only \( y \):
\[ (x - y) + (x + 2y) = 10 + 4 \]
Simplifying, we find \( 2x + y = 14 \). Now, solve for one variable, say \( x \), in terms of \( y \). This step helps break down complex problems into simpler tasks. Once \( x \) is expressed as \( x=(14-y)/2 \), substitute this expression back into either original equation to further simplify and find \( y \). Through such systematic steps, the elimination method becomes a powerful tool to easily tackle linear equations.

Substitution Method

The substitution method is another effective way to solve systems of linear equations. This technique involves solving one of the equations for one of the variables, and then substituting this expression into the other equation. Let's take a closer look at how this works.
Begin with the equations:

• \( x - y = 10 \)
• \( x + 2y = 4 \)

From the first equation, solve for \( x \):\[ x = y + 10\]
Next, substitute this expression for \( x \) into the second equation:
\[(y + 10) + 2y = 4 \]
This equation now only contains \( y \), making it easier to solve. Solving gives us \( y = -3 \).
Substitute \( y = -3 \) back into \( x = y + 10 \) to find \( x = 7 \).

Using substitution is especially helpful when one equation is already solved for a single variable, reducing the complexity of the system.

Tables of Values

Creating tables of values for linear equations offers a visual perspective on how variables relate to each other. This approach not only confirms solutions found algebraically, but also provides an educational insight into the behavior of the equations.
For our example, let's examine the tables for both equations:

• For the first equation \( x - y = 10 \):
  • If \( y = -3 \), \( x = 7 \)
  • If \( y = 0 \), \( x = 10 \)
• For the second equation \( x + 2y = 4 \):
  • If \( y = -3 \), \( x = 7 \)
  • If \( y = 0 \), \( x = 4 \)

These tables indicate that when \( y = -3 \), \( x = 7 \) appears in both tables, confirming our point of intersection. Such numerical evidence can be particularly powerful for visual learners.

Linear Equations

Linear equations are a fundamental part of algebra and appear as straight lines when graphed. Each equation can be written in the standard form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. The goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.

The linear nature of these equations means each solution corresponds to a unique point on a plane where the lines intersect. In our initial problem:

• \( x - y = 10 \)
• \( x + 2y = 4 \)

The two lines intersect at exactly one point. Discovering this intersection involves finding the single \( x \) and \( y \) pair that satisfies both equations. This demonstrates how linear equations provide a clear and predictable representation of relationships between variables in a system.