Problem 11 - Algebra and Trigonometry

Question

\(9-12\) . Use the elimination method to find all solutions of the system of equations. $$ \left\\{\begin{array}{c}{x+2 y=5} \\ {2 x+3 y=8}\end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Write Down the Equations

Write the given system of equations clearly. The first equation is \( x + 2y = 5 \) and the second equation is \( 2x + 3y = 8 \).

Step 2: Multiply to Align Coefficients for Elimination

To eliminate one of the variables, multiply the first equation by 2 to align the coefficients of \( x \) in both equations. This gives us \( 2(x + 2y) = 2(5) \), which simplifies to \( 2x + 4y = 10 \). Now the system becomes \( 2x + 4y = 10 \) and \( 2x + 3y = 8 \).

Step 3: Subtract the Equations to Eliminate \( x \)

Subtract the second equation \( (2x + 3y = 8) \) from the first modified equation \( (2x + 4y = 10) \) to eliminate \( x \). This gives us: \[ (2x + 4y) - (2x + 3y) = 10 - 8 \] which simplifies to \( y = 2 \).

Step 4: Substitute Back to Find \( x \)

Using the value \( y = 2 \), substitute it back into the first original equation \( x + 2y = 5 \). Substitute \( y = 2 \) into the equation: \( x + 2(2) = 5 \), which simplifies to \( x + 4 = 5 \). Solving for \( x \), we get \( x = 1 \).

Step 5: Write the Solution

The solution to the system of equations is \( x = 1 \) and \( y = 2 \). Therefore, the solution point is \((1, 2)\).

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same set of variables. In this case, our system involves two linear equations:

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

These equations represent straight lines when plotted on a graph. The solution of a system is the point where the lines intersect, which gives us the values of \( x \) and \( y \) that satisfy both equations simultaneously. It's like finding a common destination from two different paths.
Solving these systems can involve different methods, such as substitution, elimination, or graphing. For this particular exercise, we are using the elimination method, which simplifies calculations and helps in easily determining the variables involved.

Solving Linear Equations

Solving linear equations is the process of finding the value of the variables that make the equation true. In linear equations, variables appear only to the first degree, meaning no exponents higher than one. The elimination method provides a systematic approach to isolate one variable.
In this process, we adjusted the coefficients of one variable to make them the same in both equations. For our problem:

• We multiplied the first equation by 2, changing it to \( 2x + 4y = 10 \), to align \( x \)'s coefficients with the second equation.

This step sets the stage for eliminating one variable by subtraction, simplifying the path to find the other variable's value.

Algebra Steps

The algebra steps in solving systems of equations using the elimination method involve several important actions.
Here's a concise breakdown for our example:

• **Align Coefficients**: Multiply one or both equations so the coefficients of one variable match. Here, we multiplied the first equation by 2.
• **Elimination**: Subtracting the equations to remove one variable. In this exercise, subtract the second equation from the newly formed first equation to eliminate \( x \).
• **Solve for Remaining Variable**: With \( x \) eliminated, solve for \( y \), giving us \( y = 2 \).
• **Back Substitution**: Insert \( y \) back into one of the original equations to find \( x \). Here, we found \( x = 1 \).

Following these steps helps in systematically approaching any system of linear equations, enabling accurate and efficient solutions.