Problem 9 - Algebra and Trigonometry

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 \((x, y) = (1, 2)\).

Step 1: Prepare the System for Elimination

Identify which variable to eliminate. Here, eliminating \(x\) is straightforward by aligning the coefficients. Multiply the first equation by 2 to get compatible coefficients for \(x\). The transformed system is: \[ \begin{align*} 2(x + 2y) &= 2(5) \ 2x + 3y &= 8 \ \end{align*} \] which simplifies to: \[ \begin{align*} 2x + 4y &= 10 \ 2x + 3y &= 8 \ \end{align*} \]

Step 2: Eliminate \(x\) by Subtraction

Subtract the second equation from the first equation to eliminate \(x\): \[ \begin{align*} (2x + 4y) - (2x + 3y) &= 10 - 8 \ 4y - 3y &= 2 \ y &= 2 \end{align*} \] Thus, \(y = 2\).

Step 3: Substitute to Find \(x\)

Substitute \(y = 2\) back into the first original equation to find \(x\): \[ x + 2(2) = 5 \ x + 4 = 5 \ x = 5 - 4 \ x = 1 \]

Step 4: Verify the Solution

Check the solution \((x, y) = (1, 2)\) in the second original equation: \[ 2(1) + 3(2) = 2 + 6 = 8 \] The equation holds true, so \((x, y) = (1, 2)\) is indeed the solution.

Key Concepts

System of Equations

When tackling a system of equations, you're working with multiple equations that share two or more variables. The objective is to find values for these variables that satisfy all given equations simultaneously. In the example explored, our system is:

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

These equations form a system because they involve the same variables: \(x\) and \(y\). In practical applications, systems of equations can be seen in areas like simultaneous financial calculations or physics, where different equations describe related phenomena.
To solve such a system, you can use various methods, with substitution and elimination being the most common. The goal is to find the intersection point of the lines represented by each equation, which gives you the solution.

Solving Equations

Solving equations is about finding values that make the equation true. In our system, the elimination method was chosen. Here's why this method works:First, the elimination method helps you remove one variable, making it possible to solve for the other. In our system, we aimed to eliminate the \(x\) variable. We modified Equation 1 by multiplying every term by 2, allowing the coefficients of \(x\) to match in both equations:

• Original: \(x + 2y = 5\)
• Modified: \(2x + 4y = 10\)

The second equation was left as it was. Now both equations have the same \(x\) term, \(2x\). Next, subtracting these equations cancels out \(x\), enabling us to solve for \(y\):\[4y - 3y = 2 \ y = 2\]By focusing on one equation at a time, we make the problem easier, solving for one variable before going back to find the other.

Algebra Solutions

Algebra solutions provide a systematic approach to finding variable values that satisfy equation conditions. Once you have solved for one variable, as we did with \(y = 2\), you substitute this value back into one of the original equations to find the other variable.Let's substitute \(y = 2\) back into the first equation:

• \(x + 2(2) = 5\)
• \(x + 4 = 5\)
• \(x = 5 - 4\)
• \(x = 1\)

The solution \((x, y) = (1, 2)\) satisfies both equations in the system. A quick substitution back into the second equation verifies this:\[2(1) + 3(2) = 8\]Therefore, the elimination method not only helps find a solution but also ensures it satisfies the conditions given by all equations in the system. This step is crucial in confirming our solution is correct.