Problem 9 - Precalculus: Mathematical for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{array}{r}x+2 y=5 \\\2 x+3 y=8\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = 1\) and \(y = 2\).

Step 1: Label the Equations

The system of equations we need to solve is: \(x + 2y = 5\) and \(2x + 3y = 8\). Let's call the first equation Equation (1) and the second equation Equation (2).

Step 2: Multiply to Eliminate x

Our goal is to eliminate one of the variables, so we will manipulate the equations accordingly. Multiply Equation (1) by 2 to make the coefficient of \(x\) in Equation (1) the same as in Equation (2). This gives us the new Equation (3): \(2x + 4y = 10\).

Step 3: Subtract to Eliminate x

Now subtract Equation (2) from Equation (3) to eliminate \(x\):\[(2x + 4y) - (2x + 3y) = 10 - 8\]This simplifies to \(y = 2\).

Step 4: Substitute the Value of y

With \(y = 2\) found, substitute this value back into Equation (1): \[x + 2(2) = 5\]Simplify this to find \(x\):\[x + 4 = 5\]Thus, \(x = 1\).

Step 5: Verify the Solution

Substitute \(x = 1\) and \(y = 2\) into both original equations to ensure they are satisfied. Check Equation (1):\[1 + 2(2) = 5\Rightarrow 1 + 4 = 5\] which is true.Check Equation (2):\[2(1) + 3(2) = 8\Rightarrow 2 + 6 = 8\] which is also true.

Key Concepts

Elimination Method

The elimination method is a straightforward technique for solving a system of linear equations. The main idea is to eliminate one variable so that you're left with a single-variable equation, which is easier to solve. Here's how it works: you manipulate the equations so that adding or subtracting them results in the cancellation of one variable. This involves equating the coefficients of the desired variable to effectively "eliminate" it.

Let's look at the original system:

• Equation (1): \(x + 2y = 5\)
• Equation (2): \(2x + 3y = 8\)

In the given solution, Equation (1) was multiplied by 2 to obtain Equation (3): \(2x + 4y = 10\). By doing this, we effectively matched the coefficient of \(x\) between Equation (2) and Equation (3). Subtracting Equation (2) from Equation (3) eliminated \(x\), leaving us with one clean equation: \(y = 2\).

This method is powerful because, once you've simplified the system to one equation, the path to finding the solution becomes much clearer.

Linear Equations

Linear equations form the backbone of algebraic problem-solving and are defined by their expressions, which are linear polynomials. In simpler terms, they have no variables raised to a power other than one. A standard linear equation in two variables looks like this: \(ax + by = c\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables.Each equation in a system of linear equations represents a line on a graph. The solution to the system is the point where the lines intersect, representing the values where both equations are satisfied simultaneously.

In our case, the system:

• \(x + 2y = 5\)
• \(2x + 3y = 8\)

represents two lines that meet at exactly one point. This point, which is the solution of the system, was found using the elimination method, giving us \(x = 1\) and \(y = 2\). Linear equations highlight the relationship between variables and show how they change together in a linear, predictable way.

Solution Verification

After solving a system of equations, it's critical to verify that the obtained solution satisfies all original equations. This step ensures accuracy and confirms that no mistakes were made during the calculation process.

Verification involves substituting the found values of the variables back into the original equations. For our problem, substituting \(x = 1\) and \(y = 2\) into both equations yields:

• In Equation (1): \(1 + 2(2) = 5\), which simplifies to \(5 = 5\), confirming that \(x = 1\) and \(y = 2\) satisfies Equation (1).
• In Equation (2): \(2(1) + 3(2) = 8\), simplifying to \(8 = 8\), which confirms it also satisfies Equation (2).

This verification step is essential because it confirms that the solution is indeed correct. It not only serves as a confidence boost for students but also as a critical check to ensure precise results in practical applications.